Given triangle $ABC$ the point $J$ is the centre of the excircle opposite the vertex $A.$ This excircle is tangent to the side $BC$ at $M$, and to the lines $AB$ and $AC$ at $K$ and $L$, respectively. The lines $LM$ and $BJ$ meet at $F$, and the lines $KM$ and $CJ$ meet at $G.$ Let $S$ be the point of intersection of the lines $AF$ and $BC$, and let $T$ be the point of intersection of the lines $AG$ and $BC.$ Prove that $M$ is the midpoint of $ST$.

Solution:
First, $BK = BM$ because $BK$ and $BM$ are both tangents from $B$ to the excircle $J$.  Then $BJ \bot KM$.  Call the $X$ the intersection between $BJ$ and $KM$.  Similarly, let the intersection between the perpendicular line segments $CJ$ and $LM$ be $Y$.  We have $\angle XBM = \angle XBK = \angle FBA$ and $\angle XMB = \angle XKB$.  We then have, $\angle XBM + \angle XBK + \angle XMB + \angle XKB = \angle MBK + \angle XMB + \angle XKB$ $= \angle MBK + \angle KMB + \angle MKB = 180^{\circ}$.  So $\angle XBM = 90^{\circ} - \angle XMB$.  We also have $180^{\circ} = \angle FBA + \angle ABC + \angle XBM = 2\angle XBM + \angle ABC = 180^{\circ} - 2\angle XMB$ $+ \angle ABC$.  Then $\angle ABC = 2\angle XMB$.  Notice that $\angle XFM = 90^{\circ} - \angle XMB - \angle BMF = 90^{\circ} - \angle XMB - \angle YMC$.  Then, $\angle ACB = 2\angle YMC$.  $\angle BAC = 180^{\circ} - \angle ABC - \angle ACB = 180^{\circ} - 2(\angle XMB + \angle YMC)$ $= 2(90^{\circ} - (\angle XMB + \angle YMC) = 2\angle XFM$.  Similarly, $\angle BAC = 2\angle YGM$.  Draw the line segments $FK$ and $GL$.  $\triangle FXK$ and $\triangle FXM$ are congruent and $\triangle GYL$ and $\triangle GYM$ are congruent.  Quadrilateral $AFJL$ is cyclic because $\angle JAL = \frac{\angle BAC}{2} = \angle XFM = \angle JFL$.  Quadrilateral $AFKJ$ is also cyclic because $\angle JAK = \frac{\angle BAC}{2} = \angle XFM = \angle XFK = \angle JFK$.  The circumcircle of $\triangle AFJ$ also contains the points $K$ and $J$ because there is a circle around the quadrilaterals $AFJL$ and $AFKJ$.  Therefore, pentagon $AFKJL$ is also cyclic.  Finally, quadrilateral $AGLJ$ is cyclic because $\angle JAL = \frac{\angle BAC}{2} = \angle YGM = \angle YGL = \angle JGL$.  Again, $\triangle AJL$ is common in both the cyclic pentagon $AFKJL$ and cyclic quadrilateral $AGLJ$, so the circumcircle of $\triangle AJL$ also contains the points $F$, $K$, and $G$.  Therefore, hexagon $AFKJLG$ is cyclic.  Since $\angle AKJ$ and $\angle ALJ$ are both right angles, $AJ$ is the diameter of the circle around cyclic hexagon $AFKJLG$.  Therefore, $\angle AFJ$ and $\angle AGJ$ are both right angles.  $\triangle BFS$ and $\triangle BFA$ are congruent by ASA congruency, and so are $\triangle CGT$ and $\triangle CGA$. We have $SB = AB$, $TC = AC$, $BM = BK$, and $CM = CL$.  Since $AK$ and $AL$ are tangents from $A$ to the circle $J$, $AK = AL$.  Then, we have $AK = AL$, which becomes $AB + BK = AC + CL$, which is $SB + BM = TC + CM$, or $SM = TM$.  This means that $M$ is the midpoint of $ST$.
QED
--Aopsqwerty 21:19, 19 July 2012 (EDT)

For simplicity, let $A, B, C$ written alone denote the angles of triangle $ABC$, and $a$, $b$, $c$ denote its sides.
Let $R$ be the radius of the A-excircle. Because $CM = CL$, we have $CML$ isosceles and so $\angle{CML} = \dfrac{\angle{C}}{2}$ by the Exterior Angle Theorem. Then because $\angle{FBS} = 90^\circ - \dfrac{B}{2}$, we have $\angle{BFM} = \dfrac{\angle{A}}{2}$, again by the Exterior Angle Theorem.
Notice that $\angle{BJM} = \dfrac{\angle{B}}{2}$ and $\angle{CJM} = \dfrac{\angle{C}}{2}$, and so

after converting tangents to sine and cosine. Thus,

It follows that $BM = a \sin \dfrac{B}{2} \cos \dfrac{C}{2} \sec \frac{A}{2}$. By the Law of Sines on triangle $BFM$ and $ABC$ and the double-angle formula for sine, we have

Therefore, triangle $BFA$ is congruent to a right triangle with hypotenuse length $c$ and one angle of measure $90^\circ - \dfrac{B}{2}$ by SAS Congruence, and so $\angle{BFA} = 90^\circ$. It then follows that triangles $BFS$ and $BFA$ are congruent by $ASA$, and so $AF = FS$. Thus, $J$ lies on the perpendicular bisector of $AS$. Similarly, $J$ lies on the perpendicular bisector of $AT$, and so $J$ is the circumcenter of $ATS$. In particular, $J$ lies on the perpendicular bisector of $ST$, and so, because $JM$ is perpendicular to $ST$, $M$ must be the midpoint of $ST$, as desired.
--Suli 17:53, 8 February 2015 (EST)

Same as Solution 2, except noticing that (letting $s = \dfrac{a + b + c}{2}$ be the semi-perimeter):
--Suli 18:21, 8 February 2015 (EST)

As before in Solution 2, we find that $\angle{JFL} = \dfrac{\angle{A}}{2}.$ But it is clear that $AJ$ bisects $\angle{KAL}$, so $\angle{JAL} = \dfrac{\angle{A}}{2} = \angle{JFL}$ and hence $AFJL$ is cyclic. In particular, $\angle{AFJ} = \angle{ALJ} = 90^\circ$, and continue as in Solution 2.


Prove that for any pair of positive integers $k$ and $n$, there exist $k$ positive integers $m_1,m_2,...,m_k$ (not necessarily different) such that
$1+\frac{2^k-1}{n}=(1+\frac{1}{m_1})(1+\frac{1}{m_2})...(1+\frac{1}{m_k})$.

Solution:
We prove the claim by induction on $k$.
Base case: If $k = 1$ then $1 +\frac{2^1-1}{n} = 1 + \frac{1}{n}$, so the claim is true for all positive integers $n$.
Inductive hypothesis: Suppose that for some $m \in \mathbb{Z}^{+}$ the claim is true for $k = m$, for all $n \in \mathbb{Z}^{+}$.
Inductive step: Let $n$ be arbitrary and fixed. Case on the parity of $n$:
[Case 1: $n$ is even] $1 + \frac{2^{m+1} - 1}{n} = \left( 1 + \frac{2^{m} - 1}{\frac{n}{2}} \right) \cdot \left( 1 + \frac{1}{n + 2^{m+1} - 2} \right)$
[Case 2: $n$ is odd] $1 + \frac{2^{m+1} - 1}{n} = \left( 1 + \frac{2^{m}-1}{\frac{n+1}{2}} \right) \cdot \left( 1 + \frac{1}{n} \right)$
In either case, $1 + \frac{2^{m+1} - 1}{n} = \left( 1 + \frac{2^m - 1}{c} \right) \cdot \left( 1 + \frac{1}{a_{m+1}} \right)$ for some $c, a_{m+1} \in \mathbb{Z}^+$.
By the induction hypothesis we can choose $a_1, ..., a_m$ such that $\left( 1 + \frac{2^m - 1}{c} \right) = \prod_{i=1}^{m} (1 + \frac{1}{a_i})$.
Therefore, since $n$ was arbitrary, the claim is true for $k = m+1$, for all $n$. Our induction is complete and the claim is true for all positive integers $n$, $k$.


Let $a_0<a_1<a_2<\cdots \quad$ be an infinite sequence of positive integers, Prove that there exists a unique integer $n\ge1$ such that

Solution:
Define $f(n) = a_0 + a_1 + \dots + a_n - n a_{n+1}$. (In particular, $f(0) = a_0.$) Notice that because $a_{n+2} \ge a_{n+1}$, we have

Thus, $f(n) > f(n+1)$; i.e., $f$ is monotonic decreasing. Therefore, because $f(0) > 0$, there exists a unique $N$ such that $f(N-1) > 0 \ge f(N)$. In other words,

This rearranges to give

Define $g(n) = a_0 + a_1 + \dots + a_n - n a_n$. Then because $a_{n+1} > a_n$, we have

Therefore, $g$ is also monotonic decreasing. Note that $g(N+1) = a_0 + a_1 + \dots + a_{N+1} - (N+1) a_{N+1} \le 0$ from our inequality, and so $g(k) \le 0$ for all $k > N$. Thus, the given inequality, which requires that $g(n) > 0$, cannot be satisfied for $n > N$, and so $N$ is the unique solution to this inequality.
--Suli 22:38, 7 February 2015 (EST)


We say that a finite set $\mathcal{S}$ in the plane is  balanced
if, for any two different points $A$, $B$ in $\mathcal{S}$, there is
a point $C$ in $\mathcal{S}$ such that $AC=BC$. We say that
$\mathcal{S}$ is centre-free if for any three points $A$, $B$, $C$ in
$\mathcal{S}$, there is no point $P$ in $\mathcal{S}$ such that
$PA=PB=PC$.

Solution:
Part (a): We explicitly construct the sets $\mathcal{S}$. For
odd $n$, $\mathcal{S}$ can be taken to be the vertices of
regular polygons $P_n$ with $n$ sides: given any two vertices $A$ and
$B$, one of the two open half-spaces into
which $AB$ divides $P_n$ contains an odd number of $k$ of vertices of
$P_n$. The $((k+1)/2)^{th}$ vertex encountered while moving from $A$
to $B$ along the circumcircle of $P_n$ is therefore equidistant from
$A$ and $B$.
If $n \geq 4$ is even, choose $m\geq 0$ to be the largest
integer such that   Hence
$x < 4\pi/3 < 2\pi$. Consider a circle
$K$ with centre $O$, and let
$A_1, \ldots, A_{n-1}$ be distinct points placed counterclockwise
(say) on $K$ such that $\angle A_iOA_{i+1}=\pi/3/2^m$ (for
$i=1,\ldots,n-2$). Hence for any line
$OA_i$, there is a line $OA_j$ such that $\angle A_iOA_j=\pi/3$
(using the facts that
$2\pi > x=\angle A_1OA_{n-1} \geq 2\pi/3$, and $n-1$ odd). Thus
$O$, $A_i$ and $A_j$ form an equilateral triangle. In other words, for
arbitrary $A_i$, there exists $A_j$ equidistant to
$O$ and $A_i$. Also given any $i,j$ such that
$1 \leq i, j \leq n-1$, $O$ is equidistant to $A_i$ and $A_j$. Hence
the $n$ points $O, A_1, \ldots, A_{n-1}$ form a balanced set.
Part (b): Note that if $n$ is odd, the set $\mathcal{S}$ of
vertices of a regular polygon $P_n$ of $n$ sides forms a balanced set
(as above) and a centre-free set (trivially, since the centre of the
circumscribing circle of $P_n$ does not belong to $\mathcal{S}$).
For $n$ even, we prove that a balanced, centre free set consisting of $n$
points does not exist.  Assume that
$\mathcal{S}=\{A_i: 1\leq i \leq n\}$ is centre-free. Pick an
arbitrary $A_i \in \mathcal{S}$, and let $n_i$ be the number of
distinct non-ordered pairs of points $(A_j,A_k)$ ($j\neq k$) to
which $A_i$ is equidistant. Any
two such pairs are disjoint (for, if there were two such pairs $(A_r,A_s)$
and $(A_r, A_t)$ with $r, s, t$ distinct, then $A_i$ would be
equidistant to $A_r$, $A_s$, and $A_t$, violating the centre-free
property). Hence $n_i \leq (n-2)/2$
(we use the fact that $n$ is even here), which means
$\sum_i n_i \leq n(n-2)/2 = n(n-1)/2 -n/2$. Hence there are at least
$n/2$ non-ordered pairs $(A_j, A_k)$ such that no point in
$\mathcal{S}$ is equidistant to $A_j$ and $A_k$.


Triangle $BCF$ has a right angle at $B$. Let $A$ be the point on line $CF$ such that $FA=FB$ and $F$ lies between $A$ and $C$. Point $D$ is chosen so that $DA=DC$ and $AC$ is the bisector of $\angle{DAB}$. Point $E$ is chosen so that $EA=ED$ and $AD$ is the bisector of $\angle{EAC}$. Let $M$ be the midpoint of $CF$. Let $X$ be the point such that $AMXE$ is a parallelogram. Prove that $BD,FX$ and $ME$ are concurrent.

Solution: This problem needs a solution. If you have a solution for it, please help us out by adding it.


For each integer $a_0 > 1$, define the sequence $a_0, a_1, a_2, \ldots$ for $n \geq 0$ as
Determine all values of $a_0$ such that there exists a number $A$ such that $a_n = A$ for infinitely many values of $n$.

Solution:
First we observe the following:
When we start with $a_0=3$, we get $a_1=6$, $a_2=9$, $a_3=3$ and the pattern $3,6,9$ repeats.
When we start with $a_0=6$, we get $a_1=9$, $a_2=3$, $a_3=6$ and the pattern $3,6,9$ repeats.
When we start with $a_0=9$, we get $a_1=3$, $a_2=6$, $a_3=9$ and the pattern $3,6,9$ repeats.
When we start with $a_0=12$, we get $a_1=15$, $a_2=15$,..., $a_8=36$, $a_9=6$, $a_{10}=9$, $a_{11}=3$ and the pattern $3,6,9$ repeats.
When this pattern $3,6,9$ repeats, this means that there exists a number $A$ such that $a_n = A$ for infinitely many values of $n$ and that number $A$ is either $3,6,$ or $9$.
When we start with any number $a_0\not\equiv 0\; mod\; 3$, we don't see a repeating pattern.
Therefore the claim is that $a_0=3k$ where $k$ is a positive integer and we need to prove this claim.
When we start with $a_0=3k$, the next term if it is not a square is $3k+3$, then $3k+6$ and so on until we get $3k+3p$ where $p$ is an integer and $(k+p)=3q^2$ where $q$ is an integer.  Then the next term will be $\sqrt{9q^2}=3q$ and the pattern repeats again when $q=k$ or when $q=3$ or $6$.
In order for these patterns to repeat, any square in the sequence need to be a multiple of 3.
To try the other two cases where $a_0\not\equiv 0\; mod\; 3$, we can try $a_0=3k\pm 1$ then the next terms will be in the form  $3k+3p\pm 1 = 3(k+p) \pm 1$.
When $3(k+p) \pm 1$ is a square, it will not be a multiple of $3$ because $3(k+p) \pm 1$ is not a multiple of $3$ and $3(k+p) \pm 1 \ne 9q^2$ because $3(k+p) \pm 1 \equiv \pm 1\; mod\; 3$ and $q^2$ would have to be $\frac{(k+p)}{3} \pm \frac{1}{9}$ which is not an integer even if $k+p$ is a multiple of $3$.
Therefore the pattern doesn't repeat for any of the other cases where $a_0=3k\pm 1$ and only repeats when $a_0\equiv 0\; mod\; 3$
So, the answer to this problem is $a_0=3k\;\forall k \in \mathbb{Z}^{+}$ and $A=3,6,$ or $9$.
~Tomas Diaz. orders@tomasdiaz.com
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.


Let $\Gamma$ be the circumcircle of acute triangle $ABC$. Points $D$ and $E$ are on segments $AB$ and $AC$ respectively such that $AD = AE$. The perpendicular bisectors of $BD$ and $CE$ intersect minor arcs $AB$ and $AC$ of $\Gamma$ at points $F$ and $G$ respectively. Prove that lines $DE$ and $FG$ are either parallel or they are the same line.

Solution:
http://wiki-images.artofproblemsolving.com/5/5d/FB_IMG_1531446409131.jpg
The diagram is certainly not to scale, but the argument is sound (I believe) and involves re-ordering the construction as specified in the original problem so that an identical state of affairs results, yet in so doing differently it is made clear that the line segments in question are parallel.
Construct a right-angled triangle ABC'. Select an arbitrary point H along the segment BC', and from point H select an arbitrary point F such that the segment HF is perpendicular to the segment AB. Mark the distance from the intersection of HF and AB to B at B" (i.e., HF is a perpendicular bisector). It follows that the triangle B"FB is isosceles. Construct an isosceles triangle FHG. Mark the distance of AB" along AC' at C". (From here, a circle can be constructed according to the sets of points A, B, F, and A, B, G. Points F and G may be repositioned to allow for these circles to coincide; also, point H may be repositioned so that point C falls on the coinciding circle, understood that HG is the other perpendicular bisector.)
Assign the angle BAC the value α. Hence, the angle FHG has the value 180°-α, and the angle HFG (also, HGF) has the value α/2. Assign the angle BFH the value β. Hence, the angle B'FB" has the value  β-α/2. Consequently, the angle FB'B" has the value 180°-(β-α/2)-(90°-β) =  90°+α/2, and so too its vertical angle BB'G. As the triangle B"AC" is isosceles, and its subtended angle has the value α, the angles BB"C" and CC"B" both have the value 90°+α/2. It follows therefore that segments B"C" and FG are parallel.
(N.B. Points D and E, as given in the wording of the original problem, have been renamed B" and C" here.)

The essence of the proof is using a rhombus formed by the perpendicular bisectors of the segments $BD, CE, AB$ and $AC$  and the parallelism of its diagonal and the base of the triangle formed by the perpendiculars from one vertex of the rhombus.
The perpendicular bisectors of the segments $BD, CE, AB$ and $AC$ intersect at points $O$ (the circumcenter $ABC), H, H',$ and $Q.$ Let $AD = AE = 2x, AC = 2b.$ The distance from the line $HO$ to point $C$ is $b,$ from $QH'$ to $C$ is $b - x.$ Therefore, the distance between lines $HO$ and $QH'$ is $x.$ Similarly, the distance between the lines $HQ$ and $OH'$ is $x.$
The quadrilateral $OHQH'$ is formed by the intersection of two pairs of equidistant lines $\implies$ $OHQH'$ is a parallelogram with equal heights $\implies$ $OHQH'$  is a rhombus.
Let $I$ and $I'$ be the feet of the heights of the rhombus from the vertex $O.$ In isosceles triangles $\triangle ADE$ and $\triangle OII'$ the sides are parallel, hence the bases of these triangles are parallel, $DE||II'.$
The feet of the heights $I$ and $I'$ divide the sides of the rhombus in the same ratio, which means that the diagonal $HH'$ of the rhombus is parallel to the segments $II' || DE.$
The angles between the diagonal of the rhombus and its sides are the same, so $\angle HQO = \angle H'QO.$
The lines $QF$ and $QG$ are symmetrical with respect to the diameter $OQ,$ so $QF = QG, QF' = QG'.$
The  homothety of triangles  $QHH',  QFG,$ and $QF'G'$  centered at $Q$  implies $FG || F'G' || HH'.$
Therefore, these three lines are parallel to $DE.$
Note that triangle $ABC$ may be obtuse.
vladimir.shelomovskii@gmail.com, vvsss


Let $\mathbb{Z}$ be the set of integers. Determine all functions $f : \mathbb{Z} \to \mathbb{Z}$ such that, for all
integers $a$ and $b$,

Let us substitute $0$ in for $a$ to get
Now, since the domain and range of $f$ are the same, we can let $x = f(b)$ and $f(0)$ equal some constant $c$ to get

Therefore, we have found that all solutions must be of the form $f(x) = 2x + c.$
Plugging back into the original equation, we have: $4a + c + 4b + 2c = 4a + 4b + 2c + c$ which is true. Therefore, we know that $f(x) = 2x + c$ satisfies the above for any integral constant c, and that this family of equations is unique.
(This solution does not work though because we don't know that $f$ is surjective)

We plug in $a=-b=x$ and $a=-b=x+k$ to get

respectively.
Setting them equal to each other, we have the equation  and moving "like terms" to one side of the equation yields  Seeing that this is a difference of outputs of $f,$ we can relate this to slope by dividing by $2k$ on both sides. This gives us  which means that $f$ is linear. (Functional equations don't work like that unfortunately)
Let $f(x)=mx+n.$ Plugging our expression into our original equation yields $2ma+2mb+3n=m^2a+m^2b+mn+n,$ and letting $b$ be constant, this can only be true if $2m=m^2 \implies m=0,2.$ If $m=0,$ then $n=0,$ which implies $f(x)=0.$ However, the output is then not all integers, so this doesn't work. If $m=2,$ we have $f(x)=2x+n.$ Plugging this in works, so the answer is $f(x)=2x+c$ for some integer $c.$

The only solutions are $f(x)=0, 2x+c.$ For some integer $c.$
Obviously these work. We prove these are the only linear solutions. Plug $a=0$ and $b=0$ separately to get that $f(2x)=2f(x)-f(0).$ Plug $(0, a+b)$ to see  $f(0)+f(a+b)=f(a)+f(b),$ and subtracting $2f(0)$ from both sides shows $f(a)-f(0)$ to be additive thus linear by Cauchy since this is on integers. Thus, $f(a)$ is linear, and so we are done since it can be easily shown that $0$ and $2x+c$ are the only linear solutions by plugging $mx+n$ into the equation.
~ LLL2019 (first to post correct solution on here)

We claim the only solutions are $f\equiv0$ and $f(x)=2x+c$ for some integer $c$., which obviously work. Plugging in $(0,n)$ and $(1,n-1)$ give $f(0)+2f(n)=f(f(n))=f(2)+2f(n-1)$, so $f(n)-f(n-1)=\frac{f(2)-f(0)}2$. Since this difference is constant and $f\colon\mathbb Z\rightarrow\mathbb Z$, we must have $f$ is linear (finite differences or induction). It is easy to see the only linear solutions are those specified above. $\blacksquare$
~ Ezra Guerrero


Consider the convex quadrilateral $ABCD$. The point $P$ is in the interior of $ABCD$. The following ratio equalities hold:
 Prove that the following three lines meet in a point: the internal bisectors of angles $\angle ADP$ and $\angle PCB$ and the perpendicular bisector of segment $\overline{AB}$.

The essence of the proof is the replacement of the bisectors of angles by the perpendicular bisectors of the sides of the cyclic pentagon.
Let $O$ be the circumcenter of $\triangle ABP, \angle PAD = \alpha, OE$ is the perpendicular bisector of $AP,$ and point $E$ lies on  $AD.$ Then
$\hspace{33mm} ABPE$ is cyclic.

the bisector of the $\angle ADP$ is the perpendicular bisector of the side $EP$ of the cyclic $ABPE$ that passes through the center $O.$
A similar reasoning can be done for $OF,$ the perpendicular bisector of $BP.$
vladimir.shelomovskii@gmail.com, vvsss


Let $n \geq 100$ be an integer. Ivan writes the numbers $n, n+1, \ldots, 2 n$ each on different cards. He then shuffles these $n+1$ cards, and divides them into two piles. Prove that at least one of the piles contains two cards such that the sum of their numbers is a perfect square.

If we can guarantee that there exist $3$ cards such that every pair of them sum to a perfect square, then we can guarantee that one of the piles contains $2$ cards that sum to a perfect square. Assume the perfect squares $p^2$, $q^2$, and $r^2$ satisfy the following system of equations:

where $a$, $b$, and $c$ are numbers on three of the cards. Solving for $a$, $b$, and $c$ in terms of $p$, $q$, and $r$ tells us that $a = \frac{p^2 + r^2 - q^2}{2}$, $b=\frac{p^2 + q^2 - r^2}{2}$, and $c=\frac{q^2 + r^2 - p^2}{2}$. We can then substitute $p^2 = (2e-1)^2$, $q^2 = (2e)^2$, and $r^2 = (2e+1)^2$ to cancel out the $2$s in the denominatior, and simplifying gives $a = 2e^2 + 1$, $b = 2e(e-2)$, and $c = 2e(e+2)$. Now, we have to prove that there exists three numbers in these forms between $n$ and $2n$ when $n \ge 100$. Notice that $b$ will always be the least of the three and $c$ will always be the greatest of the three. So it is sufficient to prove that there exists numbers in the form $2e(e-2)$ and $2e(e+2)$ between $n$ and $2n$.
For two numbers in the form of $2e(e-2)$ and $2e(e+2)$ to be between $n$ and $2n$, the inequalities

must be satisfied. We can then expand and simplify to get that

Then, we can complete the square on the left sides of both inequalities and isolate $e$ to get that

Notice that $e$ must be an integer, so there must be an integer between $\sqrt{1 + n} - 1$ and $\sqrt{1 + \frac{n}{2}} + 1$. If $\sqrt{1 + n} - 1$ and $\sqrt{1 + \frac{n}{2}} + 1$ differ by at least $1$, then we can guarantee that there is an integer between them (and those integers are the possible values of $e$). Setting up the inequality $\sqrt{1 + n} - \sqrt{1 + \frac{n}{2}} - 2 \ge 1$ and solving for $n$ tells us that $n \in [107, \infty)$ always works. Testing the remaining $7$ numbers ($100$ to $106$) manually tells us that there is an integer between $\sqrt{1 + n} - 1$ and $\sqrt{1 + \frac{n}{2}} + 1$ when $n \ge 100$. Therefore, there exists a triplet of integers $(a,b,c)$ with $a, b, c \in \{n, n+1, ..., 2n\}$ when $n \ge 100$ such that every pair of the numbers sum to a perfect square. By the pigeonhole principle, we know that $2$ of the numbers must be on cards in the same pile, and hence, when $n \ge 100$, there will always be a pile with $2$ numbers that sum to a perfect square. $\square$
~Mathdreams

For the statement to be true, there must be at least three pairs whose sum is each a perfect square. There must be p,q,r such that
p+q = x^2 and q+r = y^2, p+r = z^2.
by equation 1
if we add (2) and (3) to (1),
(1) + (2) + (3) =>
At this time 100 ≤ n, so let's put n = 100 to this
where
x = 16, y = 18, z = 20 fits perfectly
therefore the minimum of n fits the proposition so the proposition is true
~Mathhyhyhy


The Bank of Oslo issues two types of coin: aluminium (denoted A) and bronze (denoted B). Marianne has $n$ aluminium coins and $n$ bronze coins, arranged in a row in some arbitrary initial order. A chain is any subsequence of consecutive coins of the same type. Given a fixed positive integer $k\le 2n$, Marianne repeatedly performs the following operation: she identifies the longest chain containing the $k^{th}$ coin from the left, and moves all coins in that chain to the left end of the row. For example, if $n = 4$ and $k = 4$, the process starting from the ordering AABBBABA would be
AABBBABA → BBBAAABA → AAABBBBA → BBBBAAAA → BBBBAAAA → ...
Find all pairs $(n, k)$ with $1 \le k \le 2n$ such that for every initial ordering, at some moment during the process, the leftmost $n$ coins will all be of the same type.

Solution:
https://www.youtube.com/watch?v=nYD-qIOdi_c [Video contains solutions to all day 1 problems]
https://youtu.be/KHn3xD6wS3A
[Video contains problem 1 discussion]
We call a chain basic when it is the largest possible for the coins it consists of. Let \(A=[i,j]\) be the basic chain with the \(i\)-th and \(j\)-th coins being the first and last, respectively.
Claim:
Proof:
For \(k < n\), it is easy to see that the arrangement \(A\ldots AB\ldots BA\) remains the same.
For \(k > \lceil \frac{3n}{2} \rceil = 2n - \lfloor \frac{n}{2} \rfloor\), we obtain the arrangement \(A\ldots AB\ldots BA\ldots AB\ldots B\), where each basic chain consists of \(\lfloor \frac{n}{2} \rfloor, \lceil \frac{n}{2} \rceil, \lceil \frac{n}{2} \rceil, \lfloor \frac{n}{2} \rfloor\) coins, respectively. Since the number of coins in the last chain is \(\geq \lfloor \frac{n}{2} \rfloor\), it follows that \(k\) is greater than the number of the remaining coins, or in other words, it is always contained in the last chain.
However, we have a loop:
We will prove that in any other case, the number of basic chains decreases by a constant, which proves the claim.
For \(k \in B=[l, m]\), where \(l > 1\) and \(m < 2n\), the basic chains \(B_1=[l{'}, l-1]\) and \(B_2=[m+1, m']\) merge into one, and we are done since it is impossible to increase.
For \(k \in C=[1, l]\), where \(n + 1 > l \geq k \geq n\), it holds that \(l = n\), which is what we need to prove.
For \(k \in D=[m, 2n]\), we will prove that the basic chains are of quantity 2 or 3 (two are obtained with one move):
Indeed, if there are at least 4 basic chains, from the beginning of the pigeonhole, we have at least one chain with a number of coins < \(\lfloor \frac{n}{2} \rfloor + 1 \leq 2n - k + 1\). Therefore, \(k\) does not belong to this chain when it is the last, and then the number of basic chains decreases, which completes the proof. $\blacksquare$
In conclusion, such pairs are \((n, k)\), where \(k \in \{n, n+1, \ldots, \lceil \frac{3n}{2} \rceil\}\).


Determine all composite integers $n>1$ that satisfy the following property: if $d_1,d_2,\dots,d_k$ are all the positive divisors of $n$ with $1=d_1<d_2<\dots<d_k=n$, then $d_i$ divides $d_{i+1}+d_{i+2}$ for every $1\le i \le k-2$.

If $n$ has at least $2$ prime divisors, WLOG let $p<q$ be the smallest two of these primes. Then the ordered tuple of divisors is of the form $(1,\,  p,\,  p^2 \dots,\,  p^a,\,  q \dots,\,  n)$ for some integer $a\geq 1$.
To prove this claim, note that $p$ is the smallest prime that divides $n$, so it is the smallest divisor not equal to $1$, meaning the first $2$ divisors are $1$ and $p$. Furthermore, the smallest divisor of $n$ that is not equal to a power of $p$ (i.e. not equal to $(1,\, p,\, p^2\dots)$ is equal to $q$. This is because all other divisors either include a prime $z$ different from both $q$ and $p$, which is larger than $q$ (since $q$ and $p$ are the smallest two prime divisors of $n$), or don’t include a different prime $z$. In the first case, since $z>q$, the divisor is larger than $q$. In the second case, all divisors divisible by $q^2$ are also larger than $q$, and otherwise are of the form $p^x \cdot q^1$ or $p^x$ for some nonnegative integer $x$. If the divisor is of the form $p^x$, then it is a power of $p$. If it is of the form $p^x \cdot q^1$, the smallest of these factors is $p^0 \cdot q^1 = q$. Therefore, (in the case where $2$ or more primes divide $n$) the ordered tuple of divisors is of the form $(1,\,  p,\,  p^2 \dots,\,  p^a,\,  q \dots,\,  n)$ for some integer $a\geq 1$, since after each divisor $p^x$, the next smallest divisor is either $p^{x+1}$ or simply $q$.
If $a\geq 2$, the condition fails. This is because $p^{a-1} \nmid p^a + q$, since $p^a$ is divisible by $p^{a-1}$, but $q$ is not since it is a prime different from $p$. If $a=1$, then $p^{a-1}=p^0=1$, which does divide $q$. Therefore $a$ must equal $1$ for the condition to be satisfied in this case. However, we know that the ordered list of divisors satisfies $d_i \cdot d_{k+1-i}=n$, meaning since the first $3$ divisors are $(1, p, q)$, then the last $3$ divisors are $(\frac{n}{q}, \frac{n}{p}, n)$, so $(\frac{n}{q})$ must divide $(\frac{n}{p} + n)$. But $\frac{n}{q}$ is divisible by $p$, so $\frac{n}{p} + n$ must also be divisible by $p$, but since $a=1$ $\frac{n}{p}$ is and $n$ isn't.
When $n=p^x$, it is easy to verify this works for all primes $p$ and all $x\geq 2$, since $p^y  \vert  p^{y+1} + p^{y+2}$, and the divisors are ordered as ${1,\, p,\, p^2…\, p^x}$.
~SpencerD.

Similar argument as above but restated.
Let $n = p_1^{e_1} p_2^{d_2} \dots p_k^{e_k}$, with $p_1 < p_2 < \dots < p_k$ primes.
If $k=1$, it's easy to verify that this property holds, since $p^l \mid p^{l+1} + p^{l+2}$.
Suppose $k \geq 2$. All divisors of $n$ which contain a factor of $p_k$ for $k \geq 2$ is at least as large as $p_2$. So, the only divisors which are possibly less than $p_2$ are of the form  $p_1^{j}$. Also, by construction $p_1$ is less than $p_2$. Putting it all together, this says that the smallest factors are $(1, p_1, \dots, p_1^j, p_2)$ for some $j \geq 1$.
If $j \geq 2$, then $p_1^{j-1} \nmid p_1^j + p_2$, since the LHS has a factor of $p_1$ and the RHS does not since.
If $j = 1$, then the largest three factors are $(\frac{n}{p_2}, \frac{n}{p_1}, n)$. But $\frac{n}{p_2}$ and $n$ have a factor of $p_1^{e_1}$, while $\frac{n}{p_1}$ only has a factor of $p_1^{e_1 - 1}$, so $\frac{n}{p_2} \nmid \frac{n}{p_1} + n$.
Hence the prime powers are the only composite numbers which satisfy the property.


For what real values of $x$ is
given (a) $A=\sqrt{2}$, (b) $A=1$, (c) $A=2$, where only non-negative real numbers are admitted for square roots?

The square roots imply that $x\ge \frac{1}{2}$.
Square both sides of the given equation:
Add the first and the last terms to get:
Multiply the middle terms, and use $(a + b)(a - b) = a^2 - b^2$ to get:
Since the term inside the square root is a perfect square, and by factoring 2 out, we get

Use the property that $\sqrt{x^2}=|x|$ to get
Case I: If $x \le 1$, then $|x-1| = 1 - x$, and the equation reduces to $A^2 = 2$. This is precisely part (a) of the question, for which the valid interval is now $x \in \left[ \frac{1}{2}, 1 \right]$
Case II: If $x > 1$, then $|x-1| = x - 1$ and we have

which simplifies to
This tells there that there is no solution for (b), since we must have $A^2 \ge 2$
For (c), we have $A = 2$, which means that $A^2 = 4$, so the only solution is $x=\frac{3}{2}$.
~flamewavelight (Expanded)
~phoenixfire (edited)

Note that the equation can be rewritten to

i.e., $\sqrt{2x-1}+1 + |\sqrt{2x-1}-1|=A\sqrt{2}$.
Case I: when $2x-1\ge 1$ (i.e., $x\ge 1$), the equation becomes $2\sqrt{2x-1}=\sqrt{2}A$. For (a), we have $x=1$; for (b) we have $x=\frac{3}{4}$; for (c) we have $x=\frac{3}{2}$. Since $x\ge 1$, (b) $x=\frac{3}{4}$ is not what we want.
Case II: when $0\le 2x-1 <1$ (i.e., $1/2\le x <1$), the equation becomes $2=\sqrt{2}A$, which only works for (a) $A=\sqrt{2}$.
In summary, any $x \in \left[\frac{1}{2}, 1\right]$ is a solution for (a); there is no solution for (b); there is one solution for (c), which is $x=\frac{3}{2}$.
~zhaoxiayu

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.


For what values of the variable $x$ does the following inequality hold:

Solution:
Set $x = -\frac{1}{2} + \frac{a^2}{2}$, where $a\ge0$.
$\frac{4\left(-\frac{1}{2}+\frac{a^2}{2}\right)^2}{\left(1-\sqrt{1+2\left(-\frac{1}{2}+\frac{a^2}{2}\right)}\right)^2}<2\left(-\frac{1}{2}+\frac{a^2}{2}\right)+9$
After simplifying, we get
$(a+1)^2<a^2+8$
So
$a^2+2a+1<a^2+8$
Which gives $a<\frac{7}{2}$ and hence $-\frac{1}{2} \le x<\frac{45}{8}$.
But $x=0$ makes the LHS indeterminate.
So, answer: $-\frac{1}{2} \le x<\frac{45}{8}$, except $x=0$.


Let $a$, $b$, and $c$ be the lengths of a triangle whose area is S.  Prove that
$a^2 + b^2 + c^2 \ge 4S\sqrt{3}$
In what case does equality hold?

Solution:
By Heron's formula, we have

This can be simplified to

Next, we can factor out all of the $2$s and use a clever difference of squares:

We can now use difference of squares again:

We must prove that the RHS of this equation is less than or equal to $a^2 + b^2 + c^2$.
Let $a^2 = A$, $b^2 = B$, $c^2 = C$. Then, our inequality can be reduced to

We now have to prove

We can simplify:

Finally, we can apply AM-GM:

Adding these all up, we have the desired inequality

and so the proof is complete.$\square$
To have $A + B + C = 4S\sqrt{3}$, we must satisfy

This is only true when $A = B = C$, and thus $a = b = c$. Therefore, equality happens when the triangle is equilateral.
~mathboy100

We firstly use the duality principle.
$a=x+y~~b=x+z~~c=y+z$
The LHS becomes $(x+y)^2+(x+z)^2+(y+z)^2$
and the RHS becomes $4\sqrt{3}\sqrt{(x+y+z)xyz}$ If we use Heron's formula.
By AM-GM $\frac{(x+y+z)^3}{27} \ge xyz$
Making this substitution $[ABC]$ becomes
$\sqrt{(x+y+z)^4\frac{1}{27}}$
and once we take the square root of the area then our RHS becomes
$\frac{4\sqrt{3}}{3\sqrt{3}}(x+y+z)^2=\frac{4}{3}(x+y+z)^2$
Multiplying the RHS and the LHS by 3 we get the LHS to be
$3((x+y)^2+(x+z)^2+(y+z)^2)=6(x^2+y^2+z^2+xy+yz+xz).$
Our RHS becomes
$4(x^2+y^2+z^2)+8(xy+yz+xz).$
Subtracting $4(x^2+y^2+z^2)+6(xy+yz+xz)$ we have the LHS equal to $(2(x^2+y^2+z^2))$ and the RHS being $2(xy+xz+yz)$
If LHS $\ge$ RHS then LHS-RHS$\ge 0$
LHS-RHS=$2(x^2+y^2+z^2)-2(xy+xz+yz)=x^2-2xy+y^2+x^2-2x+z^2+y^2-2yz+z^2=(x-y)^2+(x-z)^2+(y-z)^2.$
$(x-y)^2+(x-z)^2+(y-z)^2 \ge 0$ by the trivial inequality so therefore, $a^2 + b^2 + c^2 \ge 4S\sqrt{3}$ and we're done.
~PEKKA

Let $\theta$ be the angle between vertices $a$ and $b$. We use two formulae:

Going backwards,
$\begin{rcases}(a-b)^2=a^2-2ab+b^2\leq0\\-2\leq2\sin x\geq2\end{rcases}\forall a,b,x\in\Bbb R^+:a^2-2ab\sin x+b^2\geq0$

From $a^2-2ab\sin x+b^2\geq0$,
Equality occurs when $a=b$ and $\sin(\theta+30^\circ)=1\implies\theta=60^\circ$, which is when $a=b=c$ so that they form an equilateral triangle.
~ztilB


Determine all real numbers $x$ which satisfy the inequality:

Solution:
Obviously we need $\sqrt{3-x} \geq \sqrt{x+1}$ for the outer square root to be defined, $x\leq 3$ for the first inner square root to be defined,
and $x\geq -1$ for the second inner square root to be defined. Solving these we get that the left hand side is defined for $x\in \left[ -1,1 \right]$.
Now obviously the function $f(x)=\sqrt{\sqrt{3-x}-\sqrt{x+1}}$ is continuous on $\left[ -1,1 \right]$, with $f(-1)=\sqrt 2$ and $f(1)=0$.
Moreover, as $3-x$ is a decreasing and $x+1$ an increasing function, both $\sqrt{3-x}$ and $-\sqrt{x+1}$ are decreasing functions, and hence $f(x)$ is a decreasing function. Therefore there is exactly one solution to $f(x)=\dfrac{1}{2}$.
We can now find this solution:

(Note the little trick in the third row: placing the square roots on opposite sides of the equation. Squaring the equation in the second row would work as well, but this way is a little more pleasant, as the one remaining square root after the squaring will essentially be one of the original two, not their product.)
Solving the quadratic equation for $x$, we get
The reason why we got two roots is that while solving the original equation we squared both sides twice, and this could have created additional solutions. In this case, obviously the root that is larger than $1$ is the additional solution, and $x=1-\dfrac{\sqrt{127}}{32}$ is the root we need.
Hence the solutions to the given inequality are precisely the reals in the interval $\boxed{ \left[ ~ -1,\quad 1-\dfrac{\sqrt{127}}{32} ~ \right) }$.


Point $A$ and segment $BC$ are given. Determine the locus of points in space which are the vertices of right angles with one side passing through $A$, and the other side intersecting the segment $BC$.

Solution:
Let $\omega_1$ be the circle with diameter $AB$, and let $\omega_2$ be the circle with diameter $AC$.  Then the locus is simply the set of points inside either $\omega_1$ or $\omega_2$, but not both.
To see this, suppose the right angle's ray that does not pass through $A$ intersects segment $BC$ at $X$.  Then the right angle's vertex must lie on the circle with diameter $AX$.  So, for a particular $X$, the desired locus is a circle with diameter $AX$.  Accounting for all possible $X$, the total locus is the union of the circumferences of all circles that have a diameter $AX$, where $X$ is some point on $BC$.
As $X$ moves from $B$ to $C$, the motion of the circle with diameter $AX$ is continuous and fluid.  Any point $P$ lying within $\omega_1$ but outside $\omega_2$ will eventually be intersected by this moving circle, since it went from inside to outside the circle.  This applies similarly to all points inside $\omega_2$ but outside $\omega_1$.  Also, the points inside both $\omega_1$ and $\omega_2$ are never intersected by this moving circle, as it always stays inside.
(This proof sucks and needs some formalism)


Suppose $a, b, c$ are the sides of a triangle. Prove that

Solution:
Let $b+c-a = x$, $c+a-b = y$, and $a+b-c = z$. Then, $a = \frac{y+z}{2}$, $b = \frac{x+z}{2}$, and $c = \frac{x+y}{2}$. By AM-GM,
Multiplying these equations, we have

We can now simplify:

~mathboy100

We can use the substitution $a=x+y$, $b=x+z$, and $c=y+z$ to get

$2zx^2+2zy^2+2yx^2+2yz^2+2xy^2+2xz^2+12xyz\leq 3x^2y+3x^2z+3y^2x+3y^2z+3z^2x+3z^2y+6xyz$

This is true by AM-GM. We can work backwards to get that the original inequality is true.

Rearrange to get

which is true by Schur's inequality.


Consider the system of equations

with unknowns $x_1$, $x_2$, $x_3$. The coefficients satisfy the conditions:
(a) $a_{11}$, $a_{22}$, $a_{33}$ are positive numbers;
(b) the remaining coefficients are negative numbers;
(c) in each equation, the sum of the coefficients is positive.
Prove that the given system has only the solution $x_1 = x_2 = x_3 = 0$.

Solution:
Clearly if the $x_i$ are all equal, then they are equal to 0. Now let's assume WLOG that $x_1=0$. If $x_2$ or $x_3$ is 0, then the other is clearly zero, so let's consider the case where neither are 0. $a_{12}$ and $a_{21}$ are negative, so exactly one of $x_2$ or $x_3$ is positive. Unfortunately this means that one of $a_{22}x_2 + a_{23}x_3$ or $a_{32}x_2 + a_{33}x_3 = 0$ is positive and the other is negative, so the equation couldn't possibly be satisfied if $x_2$ or $x_3$ isn't 0. We have covered the case where one of the $x_i$ is 0, now let's assume that none of them are 0.
If two are positive and one is negative, then when the negative $x_i$ is paired with one of the positive $a_i$, the corresponding equation is negative. This is bad. If two are negative and one is positive, then when the positive $x_i$ is paired with one of the positive $a_i$, the corresponding equation is positive. This is also bad. Therefore the $x_i$ all have the same sign.
Case 1: The $x_i$ are all positive. WLOG $x_1\leq x_2\leq x_3$. Now consider the third equation, $a_{31}x_1 + a_{32}x_2 + a_{33}x_3 = 0$. Therefore $x_2(a_{31} +a_{32}+a_{33})+ a_{31}(x_1-x_2)+a_{33}(x_3-x_2)= 0$, but all of the terms on the LHS are non-negative and the first one is positive, so this is impossible.
Case 2: The $x_i$ are all negative. WLOG $x_1\geq x_2\geq x_3$. Consider the third equation, $a_{31}x_1 + a_{32}x_2 + a_{33}x_3 = 0$. Therefore $x_3(a_{31}+a_{32}+a_{33})+a_{31}(x_1-x_3)+a_{32}(x_2-x_3)=0$, but all of the terms on the LHS are non-positive and the first one is negative, so this is impossible.
Therefore at least one of the $x_i$ is 0, which implies all of them are 0.


Find all natural numbers $x$ such that the product of their digits (in decimal notation) is equal to $x^2 - 10x - 22$.

Let the decimal expansion of $x$ be $\overline{d_1d_2d_3\dots d_n}$, where $d_i$ are base-10 digits. We then have that $x\geq d_1\cdot 10^{n-1}$. However, the product of the digits of $x$ is $d_1d_2d_3\dots d_n\leq d_1\cdot 10\cdot 10\dots 10=d_1\cdot 10^{n-1}$, with equality only when $x$ is a one-digit integer. Therefore the product of the digits of $x$ is always at most $x$, with equality only when $x$ is a base-10 digit. This implies that $x^2-10x-22\leq x$, so $x^2-11x-22\leq 0$. Every natural number from 1 to 12 satisfies this inequality, so we only need to check these possibilities. It is easy to rule out 1 through 11, since $x^2-10x-22<0$ for those values. However, $12^2-10\cdot 12-22=2$, which is the product of the digits of 12. Therefore $\boxed{12}$ is the only natural number with the desired properties. $\blacksquare$

It is pretty obvious that $x$ cannot be three digits or more, because then $x^2 - 10x - 22$ is way too big.
Write $x = 10a + b$ where $a$ and $b$ are digits satisfying $0 \leq a, b < 10$. Then, we can use SFFT:

We have

It is therefore clear that $a$ must be either $0$ or $1$. We can then split into two cases:
$\mathbf{a = 0:}$
We have $(b + 2)(b - 12) = -2$ or $b^2 - 10b - 22 = 0$, which is only satisfied when $b = -2$ or $12$.
$\mathbf{a = 1:}$
We have $(b + 12)(b - 2) = b - 2$. This is only satisfied when $b = 2$, or $b + 12 = 0$. Therefore, $b = 2$, and so $x = \boxed{12}.\square$
~mathboy100

Let,
$x^2-10x-22=y$
$\implies x^2-10+25-47=y$
$\implies (x-5)^2=47+y$
Now note that, if $p$ is a prime such that $p|y$ then $7\geq p$.
That means, $y=2^a*3^b*5^c*7^d$
But, $a^2 \not\equiv 2 (mod3), a^2 \not\equiv 2 (mod5), a^2 \not\equiv 5 (mod7)$ which means $3,5,7$ don't divivde $(x-5)^2-47=y.$
So, $y=2^a$ and $y+17=2^a+47=(x-5)^2$
It is easy to see that $a$ has one solution  and that is $2.$( Prove it by contradiction)
So, $(x-5)^2=47+2=49$
$\implies x=12$
$\blacksquare$


Let $a_1, a_2,\cdots, a_n$ be real constants, $x$ a real variable, and  Given that $f(x_1)=f(x_2)=0,$ prove that $x_2-x_1=m\pi$ for some integer $m.$

Solution:
Because the period of $\cos(x)$ is $2\pi$, the period of $f(x)$ is also $2\pi$.

We can get $x_2-x_1 = 2k\pi$ for $k\in N^*$. Thus, $x_2-x_1=m\pi$ for some integer $m.$


Let $a, b$, and $n$ be integers greater than 1, and let $a$ and $b$ be the bases of two number systems.  $A_{n-1}$ and $A_{n}$ are numbers in the system with base $a$ and $B_{n-1}$ and $B_{n}$ are numbers in the system with base $b$; these are related as follows:
Prove:

Solution:
Suppose $a>b$.  Then for all integers $0 \le k \le n$, $x_n x_k a^n b^k \ge x_n x_k b^n a^k$, with equality only when $k=n$ or $x_k = 0$.  (In particular, we have strict inequality for $k=n-1$.)  In summation, this becomes

or

which is equivalent to

This implies

On the other hand, if $a=b$, then evidently $A_{n-1}/A_n = B_{n-1}/B_n$, and if $a < b$, then by what we have just shown, $A_{n-1}/A_n > B_{n-1}/B_n$.  Hence $A_{n-1}/A_n < B_{n-1}/B_n$ if and only if $a>b$, as desired.  $\blacksquare$
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.


Consider a convex polyhedron $P_1$ with nine vertices $A_1, A_2, \cdots, A_9;$ let $P_i$ be the polyhedron obtained from $P_1$ by a translation that moves vertex $A_1$ to $A_i(i=2,3,\cdots, 9).$ Prove that at least two of the polyhedra $P_1, P_2,\cdots, P_9$ have an interior point in common.

Solution:
WLOG let $A_1$ be the origin $0$.
Take any point $A_i$, then $P_i=A_i+P_1$, lies in $2 P_1$, the polyhedron $P_1$ stretched by the factor $2$ on $P_1=0$.
More general: take any $p,q$ in any convex shape $S$. Then $p+q \in 2S$.
Prove: since $S$ is convex, $\frac{p+q}{2} \in S$, thus $p+q \in 2S$.
Now all these nine polyhedrons lie inside $2 P_1$. Let $V$ be the volume of $P_1$.
Then some polyhedrons with total sum of volumes $9V$ lie in a shape of volume $8V$, thus they must overlap, meaning that they have an interior point in common.
The above solution was posted by ZetaX. The original thread for this problem can be found here: [1]


Prove that if $n \geq 4$, every quadrilateral that can be inscribed in a circle can be dissected into $n$ quadrilaterals each of which is inscribable in a circle.

Solution:
Our initial quadrilateral will be $ABCD$.
For $n=4$, we do this:
Take $E\in AB,F\in CD,\ EF\|AD$ with $E,F$ sufficiently close to $A,D$ respectively. Take $U\in AD,V\in EF$ such that $AEVU$ is an isosceles trapezoid, with $V$ close enough to $F$ (or $U$ close enough to $D$) that we can find a circle passing through $U,D$ (or $F,V$) which cuts the segments $UV,DF$ in $X,Y$. Our four cyclic quadrilaterals are $BCFE,\ AEVU,\ VFYX,\ YXUD$.
For $n\ge 5$ we do the exact same thing as above, but now, since we have an isosceles trapezoid, we can add as many trapezoids as we want by dissecting the one trapezoid with lines parallel to its bases.
The above solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [1]


Determine whether or not there exists a finite set $M$ of points in space not lying in the same plane such that, for any two points $A$ and $B$ of $M$; one can select two other points $C$ and $D$ of $M$ so that lines $AB$ and $CD$ are parallel and not coincident.

Solution:
In order to solve this problem we can start by finding at least one finite set $M$ that satisfies the condition.
We start by defining our first set $M_{8}$ with the vertices of a cube of side $k$ as follows:
$M_{8} = \{ (0,0,0), (k,0,0), (k,k,0), (0,k,0), (0,0,k), (k,0,k), (k,k,k), (0,k,k) \}$
Since all the faces of this cube have a parallel face, then any two points on one face will have corresponding 2 points on the opposite face that is parallel.  However we have four diagonals on this cube that do not have two points that are parallel to any of these diagonals.
By doing a reflection of the points on the $z=k$ plane along the $xy$-plane these four diagonals will have their respective parallel diagonals on the $z \le 0$ space.
But now we have four more diagonals on the set of two cubes that do not have a parallel line.  That is, diagonal $(0,0,-k) \rightarrow (k,k,k)$ does not have a parallel line and neither do the other three.
By doing a reflection of the points on the $x=k$ plane along the $yz$-plane these new four diagonals will have their respective parallel diagonals on the $x \le 0$ space.
But now we have four more diagonals on the set of 4 cubes that do not have a parallel line.  That is, diagonal $(-k,0,-k) \rightarrow (k,k,k)$ does not have a parallel line and neither do the other three.
By doing a reflection of the points on the $y=k$ plane along the $xz$-plane these new four diagonals will have their respective parallel diagonals on the $y \le 0$ space.
The new 4 longer diagonals will cross the diagonal of two of the cubes and will have a parallel line on one of the other cubes.
So, we found a set at least one finite set $M$ that we can define as $M=\{(x,y,z)\}$ where $x,y,z \in \{-k,0,k\}$ giving a total of 27 points.  Therefore such a set exists.
Another way to define this set of points is let $M$ be:
Let $V$ be a solid cube or right angled parallelepiped
Let $M_{v}$ be the set of all 8 vertices of $V$
Let $M_{me}$ be the set of all 12 midpoints of the edges of $V$
Let $M_{mf}$ be the set of all 6 midpoints of the faces of $V$
Let $M_{c}$ be the center of $V$
$M$=$M_{v} \cup M_{me} \cup M_{mf} \cup M_{c}$
It is possible that one can construct many other sets of $M$ using regular tetrahedrons with some reflections and with less points than 27, or by translating or rotating or skewing all the the points simultaneously of the finite set $M$ that we defined here.  But that is not necessary for this problem because it asks to prove whether there exist a set with the described conditions.  By showing that at least one set exists with those conditions, the problem is proved.

~Tomas Diaz.  orders@tomasdiaz.com
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.


In the triangle $ABC$, prove that there is a point $D$ on side $AB$ such that $CD$ is the geometric mean of $AD$ and $DB$ if and only if $\sin{A}\sin{B} \leq  \sin^2 (\frac{C}{2})$.

Solution:
Let a point $D$ on the side $AB$.
Let $CF$ the altitude of the triangle $\triangle ABC$, and $C'$ the symmetric point of $C$ through $F$.
We bring a parallel line $L$ from $C'$ to $AB$. This line intersects the ray $CD$ at the point $E$, and we know that $DE=DC$.
The distance $d(L,AB)$ between the parallel lines $L$ and $AB$ is $CF$.
Let $w = (O,R)$ the circumscribed circle of $\triangle ABC$, and $MM'$ the perpendicular diameter to $AB$, such that $M,C$ are on difererent sides of the line $AB$.
In fact, the problem asks when the line $L$ intersects the circumcircle. Indeed:
Suppose that $DC$ is the geometric mean of $DA,DB$.
$DA \cdot DB = DC^{2}\Rightarrow DA \cdot DB = DC \cdot DE$
Then, from the power of $D$ we can see that $E$ is also a point of the circle $w$.
Or else, the line $L$ intersects $w \Leftrightarrow$
$d(L,AB)\leq d(M,AB) \Leftrightarrow$
$CF \leq MN,$ where $MN$ is the altitude of the isosceles $\triangle MAB$.
$\Leftrightarrow \frac{1}{2}CF \cdot AB \leq \frac{1}{2}MN \cdot AB \Leftrightarrow$ $(ABC) \leq (MAB) \Leftrightarrow$ $\frac{AB \cdot BC \cdot AC}{4R}\leq \frac{AB \cdot MA^{2}}{4R}\Leftrightarrow$
$BC \cdot AC \leq MA^{2}$
We use the formulas:
$BC = 2R \cdot \sin A$
$AC = 2R \cdot \sin B$
and $\angle CMA = \frac{C}{2}\Rightarrow MA = 2R \cdot \sin\frac{C}{2}$
So we have
$(2R \cdot \sin A)(2R \cdot \sin B) \leq (2R \cdot \sin\frac{C}{2})^{2}\Leftrightarrow$
$\sin A \cdot \sin B \leq \sin^{2}\frac{C}{2}$

For $(\Leftarrow)$
Suppose that $\sin A \cdot \sin B \leq \sin^{2}\frac{C}{2}$
Then we can go inversely and we find that $d(L,AB)\leq d(M,AB) \Leftrightarrow$
the line $L$ intersects the circle $w$ (without loss of generality; if $d(L,AB)=d(M,AB)$ then $L$ is tangent to $w$ at $M$)
So, if $E \in L \cap w$ then for the point $D = CE \cap AB$ we have $DC=DE$ and $AD \cdot AB = CD \cdot DE \Rightarrow$
$AD \cdot AB = CD^{2}$
The above solution was posted and copyrighted by pontios. The original thread for this problem can be found here: [1]


Let $a_1, a_2, a_3, \cdots$ be an infinite increasing sequence of positive integers. Prove that for every $p \geq 1$ there are infinitely many $a_m$ which can be written in the formwith $x, y$ positive integers and $q > p$.

Solution:
If we can find $p\ne q$ such that $(a_p,a_q)=1$, we're done: every sufficiently large positive integer $n$ can be written in the form $xa_p+ya_q,\ x,y\in\mathbb N$. We can thus assume there are no two such $p\ne q$. We now prove the assertion by induction on the first term of the sequence, $a_1$. The base step is basically proven, since if $a_1=1$ we can take $p=1$ and any $q>1$ we want. There must be a prime divisor $u|a_1$ which divides infinitely many terms of the sequence, which form some subsequence $(a_{k_n})_{n\ge 1},\ k_1=1$. Now apply the induction hypothesis to the sequence $\left(\frac{a_{k_n}}u\right)_{n\ge 1}$.
The above solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [1]


Let $P_{1}(x) = x^{2} - 2$ and $P_{j}(x) = P_{1}(P_{j - 1}(x))$ for $j= 2,\ldots$ Prove that for any positive integer n the roots of the equation $P_{n}(x) = x$ are all real and distinct.

Solution:
I shall prove by induction that $P_n(x)$ has $2^n$ distinct real solutions, where $2^{n-1}$ are positive and $2^{n-1}$ are negative. Also, for ever root $r$, $|r|<2$.
Clearly, $P_1(x)$ has 2 real solutions, where 1 is positive and 1 is negative. The absolute values of these two solutions are also both less than 2. This proves the base case.
Now assume that for some positive integer $k$, $P_k(x)$ has $2^k$ distinct real solutions with absolute values less than 2, where $2^{k-1}$ are positive and $2^{k-1}$ are negative.
Choose a root $r$ of $P_{k+1}(x)$. Let $P_1(r)=s$, where $s$ is a real root of $P_k(x)$. We have that $-2<s<2$, so $0<r^2<4$, so $r$ is real and $|r|<2$. Therefore all of the roots of $P_{k+1}$ are real and have absolute values less than 2.
Note that the function $P_{k+1}(x)$ is an even function, since $P_1(x)$ is an even function. Therefore half of the roots of $P_{k+1}$ are positive, and half are negative.
Now assume for the sake of contradiction that $P_{k+1}(x)$ has a double root $r$. Let $P_1(r)=s$. Then there exists exactly one real number $r$ such that $r^2-2=s$. The only way that this could happen is when $s+2=0$, or $s=-2$. However, $|s|<2$ from our inductive hypothesis, so this is a contradiction. Therefore $P_{k+1}(x)$ has no double roots. This proves that that the roots of $P_{k+1}(x)$ are distinct.
This completes the inductive step, which completes the inductive proof.


In a finite sequence of real numbers the sum of any seven successive terms is negative and the sum of any eleven successive terms is positive. Determine the maximum number of terms in the sequence.

Solution:
Let $x_1,x_2,\ldots$ be the given sequence and let $s_n=x_1+x_2+\ldots+x_n$.
The conditions from the hypothesis can be now written as $s_{n+7}<s_n$ and $s_{n+11}>s_n$ for all $n\ge 1$.
We then have:
$0<s_{11}<s_4<s_{15}<s_8<s_1<s_{12}<s_5<s_{16}<s_9<s_2<s_{13}<s_6<s_{17}<s_{10}<s_3<s_{14}<s_7<0,$
a contradiction. Therefore, the sequence cannot have $17$ terms.
In order to show that $16$ is the answer, just take 16 real numbers satisfying
$s_{10}<s_3<s_{14}<s_7<0<s_{11}<s_4<s_{15}<s_8<s_1<s_{12}<s_5<s_{16}<s_9<s_2<s_{13}<s_6$.
We have $x_1=s_1$ and $x_n=s_n-s_{n-1}$ for $n\ge 2$. Thus we found all sequences with the given properties.
The above solution was posted and copyrighted by enescu. The original thread for this problem can be found here: [1]


We consider a fixed point $P$ in the interior of a fixed sphere$.$ We construct three segments $PA, PB,PC$, perpendicular two by two$,$ with the vertexes $A, B, C$ on the sphere$.$ We consider the vertex $Q$ which is opposite to $P$ in the parallelepiped (with right angles) with $PA, PB, PC$ as edges$.$ Find the locus of the point $Q$ when $A, B, C$ take all the positions compatible with our problem.

Solution:
Let $R$ be the radius of the given fixed sphere.
Let point $O$ be the center of the sphere.
Let point $D$ be the 4th vertex of the face of the parallelepiped that contains points $P$, $A$, and $B$.
Let point $E$ be the point where the line that passes through $OP$ intersects the circle on the side nearest to point $A$
Let $\alpha=\angle AOP,\;\beta=\angle BPD,\;\theta=\angle APE$
We start the calculations as follows:
$\left| AB \right|= \left| PD \right|$
$\left| AB \right|^{2}=\left| PA \right|^{2}+\left| PB \right|^{2}$
Therefore, $\left| PD \right|^{2}=\left| PA \right|^{2}+\left| PB \right|^{2}$ [Equation 1]
Using law of cosines:
$R^{2}=\left| OP \right|^{2} + \left| PB \right|^{2} - 2 \left| OP \right| \left| PB \right| cos (\angle OPB)$
$R^{2}=\left| OP \right|^{2} + \left| PB \right|^{2} - 2 \left| OP \right| \left| PB \right| cos \left( \frac{\pi}{2}-\theta \right)$
$R^{2}=\left| OP \right|^{2} + \left| PB \right|^{2} - 2 \left| OP \right| \left| PB \right| sin (\theta)$
$\left| PB \right|^{2} =R^{2}-\left| OP \right|^{2} + 2 \left| OP \right| \left| PB \right| sin (\theta)$ [Equation 2]
Using law of cosines again we also get:
$\left| PA \right|^{2} =R^{2}+\left| OP \right|^{2} - 2 \left| OP \right| R cos(\alpha)$
Since $R cos(\alpha) = \left| PA \right| cos(\theta) + \left| OP\right|$, then
$\left| PA \right|^{2} =R^{2}+\left| OP \right|^{2} - 2 \left| OP \right| \left[ \left| PA \right| cos(\theta) + \left| OP\right| \right]$
$\left| PA \right|^{2} =R^{2}-\left| OP \right|^{2} - 2 \left| OP \right| \left| PA \right| cos(\theta)$ [Equation 3]
Substituting [Equation 2] and [Equation 3] into [Equation 1] we get:
$\left| PD \right|^{2}=2R^{2}-2\left| OP \right|^{2}+2\left| OP \right| \left[ \left| PB \right| sin(\theta) - \left| PA \right| cos(\theta) \right]$ [Equation 4]
Now we apply the law of cosines again:
$\left| OD \right|^{2}=\left| OP \right|^{2}+\left| PD \right|^{2}-2\left| OP \right| \left| PD \right|cos(\angle OPD)$
$\left| OD \right|^{2}=\left| OP \right|^{2}+\left| PD \right|^{2}-2\left| OP \right| \left| PD \right| cos(\angle OPB+\angle BPD)$
$\left| OD \right|^{2}=\left| OP \right|^{2}+\left| PD \right|^{2}-2\left| OP \right| \left| PD \right| cos \left(\frac{\pi}{2}-\theta+\beta \right)$
$\left| OD \right|^{2}=\left| OP \right|^{2}+\left| PD \right|^{2}-2\left| OP \right| \left| PD \right| sin(\theta-\beta)$
$\left| OD \right|^{2}=\left| OP \right|^{2}+\left| PD \right|^{2}-2\left| OP \right| \left| PD \right| \left[sin(\theta)cos(\beta)-sin(\beta)cos(\theta)  \right]$
Since, $sin(\beta)=\frac{\left| PA \right|}{\left| PD \right|}$ and $cos(\beta)=\frac{\left| PB \right|}{\left| PD \right|}$ then,
$\left| OD \right|^{2}=\left| OP \right|^{2}+\left| PD \right|^{2}-2\left| OP \right| \left| PD \right| \left[\frac{\left| PB \right|}{\left| PD \right|}sin(\theta)-\frac{\left| PA \right|}{\left| PD \right|}cos(\theta)  \right]$
$\left| OD \right|^{2}=\left| OP \right|^{2}+\left| PD \right|^{2}-2\left| OP \right| \left[\left| PB \right|sin(\theta)-\left| PA \right|cos(\theta)  \right]$ [Equation 5]
Substituting [Equation 4] into [Equation 5] we get:
$\left| OD \right|^{2}=\left| OP \right|^{2}+2R^{2}-2\left| OP \right|^{2}+2\left| OP \right| \left[ \left| PB \right| sin(\theta) - \left| PA \right| cos(\theta) \right]-2\left| OP \right| \left[\left| PB \right|sin(\theta)-\left| PA \right|cos(\theta)  \right]$
Notice that all of the terms with $\theta$ cancel and thus we're left with:
$\left| OD \right|^{2}=2R^{2}-\left| OP \right|^{2}$ regardless of $\theta$. [Equation 6]
Now we need to find $\left| PC \right|$
Since points $O$, $P$, and $C$ are on the plane perpendicular to the plane with points $O$, $P$, and $A$, then these points lie on the big circle of the sphere.  Therefore the distance $\left| PC \right|$ can be found using the formula:
$R^{2}=\left| OP \right|^{2}+\left| PC \right|^2$
Solving for $\left| PC \right|^2$ we get:
$\left| PC \right|^2=R^{2}-\left| OP \right|^{2}$ [Equation 7]
Now we need to get $\left| OQ \right|^{2}$ which will be using the formula:
$\left| OQ \right|^{2}=\left| OD \right|^{2}+\left| PC \right|^2$ [Equation 8]
Substituting [Equation 6] and [Equation 7] into [Equation 8] we get:
$\left| OQ \right|^{2}=2R^{2}-\left| OP \right|^{2}+R^{2}-\left| OP \right|^{2}$
This results in:
$\left| OQ \right|^{2}=3R^{2}-2\left| OP \right|^{2}$
which is constant regardless of $\theta$ and constant regardless of where points $A$, $B$, and $C$ are located as long as they're still perpendicular to each other.
In space, this is a sphere with radius $\left| OQ \right|$ which is equal to $\sqrt{3R^{2}-2\left| OP \right|^{2}}$
Therefore, the locus of vertex $Q$ is a sphere of radius $\sqrt{3R^{2}-2\left| OP \right|^{2}}$ with center at $O$, where $R$ is the radius of the given sphere and $\left| OP \right|$ the distance from the center of the given sphere to point $P$
~ Tomas Diaz. orders@tomasdiaz.com
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.


We consider a prism which has the upper and inferior basis the pentagons: $A_{1}A_{2}A_{3}A_{4}A_{5}$ and $B_{1}B_{2}B_{3}B_{4}B_{5}$. Each of the sides of the two pentagons and the segments $A_{i}B_{j}$ with $i,j=1,\ldots$,5 is colored in red or blue. In every triangle which has all sides colored there exists one red side and one blue side. Prove that all the 10 sides of the two basis are colored in the same color.

Solution:
Let us prove first that the edges $A_1A_2,A_2A_3,\cdots ,A_5A_1$ are of the same color. Assume the contrary, and let w.l.o.g. $A_1A_2$ be red and $A_2A_3$ be green. Three of the segments $A_2B_l, (l = 1, 2, 3, 4, 5)$, say $A_2B_i,A_2B_j,A_2B_k$,have to be of the same color, let it w.l.o.g. be red. Then $A_1B_i,A_1B_j,A_1B_k$ must be green. At least one of the sides of triangle $B_iB_jB_k$, say $B_iB_j$ ,must be an edge of the prism. Then looking at the triangles $A_1B_iB_j$ and $A_2B_iB_j$ we deduce that $B_iB_j$ can be neither green nor red, which is acontradiction. Hence all five edges of the pentagon $A_1A_2A_3A_4A_5$ have thesame color. Similarly, all five edges of $B_1B_2B_3B_4B_5$ have the same color.We now show that the two colors are the same. Assume otherwise, i.e.,that w.l.o.g. the $A$ edges are painted red and the $B$ edges green. Let us call segments of the form $A_iB_j$ diagonal ($i$ and $j$ may be equal). We now count the diagonal segments by grouping the red segments based on their $A$ point, and the green segments based on their $B$ point. As above, the assumption that three of $A_iB_j$ for fixed $i$ are red leads to a contradiction. Hence at most two diagonal segments out of each $A_i$ may be red, which counts up to at most $10$ red segments. Similarly, at most $10$ diagonal segments can be green. But then we can paint at most $20$ diagonal segments out of $25$, which is a contradiction. Hence all edges in the pentagons $A_1A_2A_3A_4A_5$ and $B_1B_2B_3B_4B_5$ have the same color.
This solution was posted and copyrighted by Learner94. The original thread for this problem can be found here: [1]


Let $1 \le r \le n$ and consider all subsets of $r$ elements of the set $\{ 1, 2, \ldots , n \}$.  Each of these subsets has a smallest member.  Let $F(n,r)$ denote the arithmetic mean of these smallest numbers; prove that

Clearly, the sum of the desired least elements is $\sum_{k=1}^{n} k {n-k \choose r-1} = \sum_{k=1}^{n} {k \choose 1}{n-k \choose r-1}$, which we claim to be equal to ${n+1 \choose r+1}$ by virtue of the following argument.
Consider a binary string of length $n+1$ which contains $r+1$ 1s.  For some value of $k$ between 1 and $n$, inclusive, we say that the second 1 will occur in the $(k+1)$th place.  Clearly, there are ${k \choose 1}$ ways to arrange the bits coming before the second 1, and ${n-k \choose r-1}$ ways to arrange the bits after the second 1.  Our identity follows.
Since the sum of the least elements of the sets is ${n+1 \choose r+1}$, the mean of the least elements is $\frac{{n+1 \choose r+1}}{{n \choose r}} = \frac{n+1}{r+1}$, Q.E.D.

We proceed as in the previous solution, but we prove our identity using the following manipulations:
Q.E.D.

We proceed by strong induction.
We define $F(k, k-1)$ to be zero (the empty sum).
We consider $r$ to be fixed.  The assertion obviously holds for $r = n$.  We now assume the problem to hold for values of $n$ less than or equal to $k$.  By considering subsets containing $k+1$ and not containing $k+1$, respectively, we conclude that
This completes our induction, Q.E.D.

Consider a bipartite graph $G$ with bipartition $\{A,B\}$. The vertices in $A$ are the $(r+1)$-element subsets of $\{0, \dots , n\}$, and the vertices in $B$ are the $r$-element subsets of $\{1, \dots , n\}$, and we draw an edge $\overline{ab}$ iff the subset $b \in B$ may be obtained from $a \in A$ by deleting the smallest element in $a$.
Note that
The degree of a vertex in $B$ is the value of the least element of its corresponding subset. Hence

We will count how many times each element of the set $\{1, 2, \ldots, n\}$ appear as the smallest element in a set, which will lead us to the result.
To count how many times $1$ appears as the smallest element of a subset, we need to choose an $r-1$ element subset from the remaining $n - 1$ numbers, as all of them are greater than $1.$ There are simply $\dbinom{n - 1}{r - 1}$ ways to do this.
Similarly, to count how many times $2$ appears as the smallest element, we need to choose an $r - 1$ element subset from the remaining $n - 2$ numbers greater than $2.$ There are $\dbinom{n - 2}{r - 1}$ ways to do this.
As there are $\dbinom{n}{r}$ subsets, and thus smallest elements, in total, it now becomes clear that we need to evaluate the sum  We can rearrange this sum as  Using the Hockey Stick Identity on each of the smaller sums on the numerator gives us  Using Hockey stick again gives us  which simplifies to  as desired.
Solution by Ilikeapos
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.


A non-isosceles triangle $A_{1}A_{2}A_{3}$ has sides $a_{1}$, $a_{2}$, $a_{3}$ with the side $a_{i}$ lying opposite to the vertex $A_{i}$. Let $M_{i}$ be the midpoint of the side $a_{i}$, and let $T_{i}$ be the point where the inscribed circle of triangle $A_{1}A_{2}A_{3}$ touches the side $a_{i}$. Denote by $S_{i}$ the reflection of the point $T_{i}$ in the interior angle bisector of the angle $A_{i}$. Prove that the lines $M_{1}S_{1}$, $M_{2}S_{2}$ and $M_{3}S_{3}$ are concurrent.

Solution:
Rename the vertices so that $A_1=A,A_2=B$ and $A_3=C$. Let $I$ be the incenter of $\triangle ABC$.
Claim 1: $S_1, S_2, S_3$ all lie on the incircle.
Proof: By construction, $IS_1$ is the reflection of $IT_1$ with respect to the $A$ angle bisector since $I$ lies on the angle bisector of $A$. So, $IS_1=IT_1$ so $S_1$ lies on the incircle. Similarly, $S_2$ and $S_3$ lie on the incircle. $\Box$
Note that $I$ is the circumcenter of $\triangle S_1S_2S_3$ because of equal radii due to Claim 1.
Claim 2: $\angle S_1IS_2=2\angle C.$
Proof: In cyclic quadrilateral $T_1IT_2C$, $\angle T_1IT_2=180^{\circ}-\angle C$. Let $D$ be the point where the $A$ angle bisector intersects $BC$. Due to the construction of $S_1$, $\angle S_1IT_1=2\angle DIT_1.$ But, $\angle ADC=180^{\circ}-\angle C-\frac{1}{2}\angle A$ so $\angle DIT_1=\frac{1}{2}\angle A+\angle C-90^{\circ}.$ Thus,Similarly, $\angle S_2IT_2=\angle B+2\angle C-180^{\circ}.$ Therefore, $\angle S_1IS_2=\angle S_1IT_1+\angle T_1IT_2+\angle T_2IS_2=2\angle C.$$\Box$
Hence, we use Claim 2 to get $\angle S_1S_3S_2=\frac{1}{2}\angle S_1IS_2=\angle C$ and we similarly get $\angle S_3S_1S_2=\angle A$ and $\angle S_1S_2S_3=\angle B.$
Claim 3: Let $X_1$ be the midpoint of $S_2S_3.$ Then, $T_1IX_1$ is collinear.
Proof: We already derived that $\angle T_1IT_2=180^{\circ}-\angle C$ and $\angle S_2IT_2=\angle B+2\angle C-180^{\circ}.$ Recall that a radius is perpendicular to the midpoint of a chord so $S_2S_3\perp IX_1.$ Then, $\angle X_1IS_2=90^{\circ}-\angle X_1S_2I=90^{\circ}-(90^{\circ}-\angle A)=\angle A.$ Thus,as desired.$\Box$
Then, $T_1IX_1\perp S_2S_3$ because a diameter is perpendicular to a chord at its midpoint and because $T_1IX_1$ is part of a diameter, by Claim 3. Hence, $S_2S_3\parallel BC.$ But, $M_2M_3\parallel BC$ so $S_2S_3\parallel M_2M_3$. Similarly, we get $S_1S_2\parallel M_1M_2$ and $S_1S_3\parallel M_1M_3$ so $\triangle M_1M_2M_3\sim \triangle S_1S_2S_3$ with corresponding parallel sides. Thus, there exists a homothety that maps $M_1M_2M_3$ to $S_1S_2S_3$ and $M_1S_1, M_2S_2,$ and $M_3S_3$ intersect at the center of the homothety. $\blacksquare$
This solution was posted and copyrighted by Jzhang21. The original thread for this problem can be found here: [1]


Let $A$ be one of the two distinct points of intersection of two unequal coplanar circles $C_1$ and $C_2$ with centers $O_1$ and $O_2$ respectively. One of the common tangents to the circles touches $C_1$ at $P_1$ and $C_2$ at $P_2$, while the other touches $C_1$ at $Q_1$ and $C_2$ at $Q_2$. Let $M_1$ be the midpoint of $P_1Q_1$ and $M_2$ the midpoint of $P_2Q_2$. Prove that $\angle O_1AO_2=\angle M_1AM_2$.

Let $A$ be one of the two distinct points of intersection of two unequal coplanar circles $C_1$ and $C_2$ with centers $O_1$ and $O_2$ respectively. Let $S$ be such point on line $O_1O_2$ so that tangents on $C_1$ touches it at $P_1$ and $Q_1$ and tangents on $C_2$ touches it at $P_2$ and $Q_2$. Let $M_1$ be the midpoint of $P_1Q_1$ and $M_2$ the midpoint of $P_2Q_2$. Prove that $\angle O_1AO_2 = \angle M_1AM_2$.
Proof: Since $S$ is image of $M_1$ under inversion wrt circle $C_1$ we have:Since $S$ is image of $M_2$ under inversion wrt circle $C_2$ we have:Image of $A$ is in both cases $A$ itself, since it lies on both circles.
Since $\angle O_1SA = \angle O_2SA$ we have:Now:
This solution was posted and copyrighted by Number1. The original thread for this problem can be found here: [1]

Let $P_1P_2$ and $Q_1Q_2$ meet at $R$. Let $RA$ meet $C_2$ at $B$. Now, it is well-known that $O_1O_2$, $P_1P_2$, and $Q_1Q_2$ are concurrent at $R$, the center of homothety between $C_1$ and $C_2$. Now, it is well-known that $O_1O_2$ bisects $\angle P_1RQ_1$. Since $P_1R = Q_1R$, we have that $O_1O_2R$ meets $P_1Q_1$ at its midpoint, $M_1$, and $P_1Q_1$ is perpendicular to $O_1O_2R$. Similarly, $O_1O_2R$ passes through $M_2$ and is perpendicular to $P_2Q_2$. Since $O_2P_2\perp P_2R$, we have that $RA\cdot RB = RP_2^2 = RM_2\cdot RO_2$, which implies that $ABM_2O_2$ is cyclic. Yet, since $A$ and $B$ lie on $C_1$ and $C_2$ respectively and are collinear with $R$, we see that the homothety that maps $C_1$ to $C_2$ about $R$ maps $A$ to $B$. Also, $O_1$ is mapped to $O_2$ by this homothety, and since $M_1$ and $M_2$ are corresponding parts in these circles, $M_1$ is mapped to $M_2$ by this homothety, so $\angle O_1AM_1 = \angle O_2BM_2 = \angle O_2AM_2$, from which we conclude that $\angle O_2AO_1 = \angle M_2AOM_1$.
This solution was posted and copyrighted by The QuattoMaster 6000. The original thread for this problem can be found here: [2]

Let $B$ be the other intersection point of these two circles. Let $P_1P_2$, $Q_1Q_2$, $O_1O_2$ meet at $Q$. Let $AB$ meet $P_1P_2$ at $P$.
Clearly, $P_1Q_1$ and $O_1O_2$ are perpendicular at $M_1$; $P_2Q_2$ and $O_1O_2$ are perpendicular at $M_2$.
Since $\triangle O_1P_1M_1 \sim \triangle O_2P_2M_2$,Since $P$ is on the radical axis, $PP_1 = PP_2$, so in the right trapezoid $P_1P_2M_2M_1$, $AB$ is the midsegment. So we have $M_1A = AM_2$ and $\angle AM_1O_1 = \angle AM_2O_1$.
Let $M$ be a point on $O_1O_2$ such that $\triangle AM_1O_1 \cong \triangle AM_2M$ (which means $AM=AO_1$ ve $MM_2 = M_1O_1$). So, from $(*)$, we getThis means, $AM_2$ is the angle bisector of $\angle MAO_2$. So,
This solution was posted and copyrighted by matematikolimpiyati. The original thread for this problem can be found here: [3]

Let, $P_1P_2,Q_1Q_2,O_1O_2$ concur at $Q$.Then a homothety with centre $Q$ that sends $C_1$ to $C_2$.Let $QA \cap C_1 =B$.Under the homothety $A$ is the image of $B$.So, $\angle M_1BO_1 =\angle M_2AO_2$ and $\triangle QP_1O_1 \sim \triangle QM_1P_1 \implies \frac{QO_{1}}{QP_{1}} = \frac{QP_{1}}{QM_{1}}$.so $QO_1.QM_1 = Q.P^2_1 = QA .QB$ Hence $A,M_1,B,O_1$ are concyclic.so $\angle O_1BM_1=\angle O_1AM_1 \implies \angle O_1AM_1= \angle O_2AM_2 \implies \angle O_1AO_2=\angle M_1AM_2$
This solution was posted and copyrighted by spikerboy. The original thread for this problem can be found here: [4]

It can be easily solved with this lemma:
Let $A$-symmedian of $\triangle ABC$ intersect its circumscribed circle $u$ at $D, P$ on $AD$ satisfying $AP = PD$. Let us define center of $u$ as $O$. Then $(B, C, P, O)$ are concyclic.
Now let $S$ be intersection of tangents to circles from the problem, $AS$ intersects $C_1$ at $B$. One can prove that $\square AP_1BQ_1$ is harmonic qudrilateral, so $P_1Q_1$ is symmedian of $\triangle AP_1B$. By lemma we get that $(B, A, O_1, M_1)$ are concyclic, thus $\angle O_1AM_1$ is equivalent to $\angle O_1BM_1$. By homothety we get $\angle O_1BM_1 = \angle O_2AM_2$.
$Q. E. D.$
This solution was posted and copyrighted by LeoLeon. The original thread for this problem can be found here: [5]


Find one pair of positive integers $a,b$ such that $ab(a+b)$ is not divisible by $7$, but $(a+b)^7-a^7-b^7$ is divisible by $7^7$.

Solution:
So we want $7 \nmid ab(a+b)$ and $7^7 | (a+b)^7-a^7-b^7 = 7ab(a+b)(a^2+ab+b^2)^2$, so we want $7^3 | a^2+ab+b^2$.
Now take e.g. $a=2,b=1$ and get $7|a^2+ab+b^2$. Now by some standard methods like Hensels Lemma (used to the polynomial $x^2+x+1$, so $b$ seen as constant from now) we get also some $\overline{a}$ with $7^3 | \overline{a}^2+\overline{a}b+b^2$ and $\overline{a} \equiv a \equiv 2 \mod 7$, so $7\nmid \overline{a}b(\overline{a}+b)$ and we are done. (in this case it gives $\overline{a}=325$)
This solution was posted and copyrighted by ZetaX. The original thread for this problem can be found here: [1]


Let $n$ and $k$ be given relatively prime natural numbers, $k < n$.  Each number in the set $M = \{ 1,2, \ldots , n-1 \}$ is colored either blue or white.  It is given that
(i) for each $i \in M$, both $i$ and $n-i$ have the same color;
(ii) for each $i \in M, i \neq k$, both $i$ and $|i-k|$ have the same color.
Prove that all the numbers in $M$ have the same color.

Solution:
We may consider the elements of $M$ as residues mod $n$.  To these we may add the residue 0, since (i) may only imply that 0 has the same color as itself, and (ii) may only imply that 0 has the same color as $k$, which put no restrictions on the colors of the other residues.
We note that (i) is equivalent to saying that $i$ has the same color as $-i$, and given this, (ii) implies that $i$ and $(-i + k)$ have the same color.  But this means that $i, -i$, and $i+k$ have the same color, which is to say that all residues of the form $i + mk \; (m \in \mathbb{N}_0)$ have the same color.  But these are all the residues mod $n$, since $k$ and $n$ are relatively prime.  Q.E.D.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.


In an acute-angled triangle $ABC$ the interior bisector of the angle $A$ intersects $BC$ at $L$ and intersects the circumcircle of $ABC$ again at $N$. From point $L$ perpendiculars are drawn to $AB$ and $AC$, the feet of these perpendiculars being $K$ and $M$ respectively.  Prove that the quadrilateral $AKNM$ and the triangle $ABC$ have equal areas.

Solution:
We are to prove that $[AKNM]=[ABC]$ or equivilently, $[ABC]+[BNC]-[KNC]-[BMN]=[ABC]$.  Thus, we are to prove that $[BNC]=[KNC]+[BMN]$.  It is clear that since $\angle BAN=\angle NAC$, the segments $BN$ and $NC$ are equal.  Thus, we have $[BNC]=\frac{1}{2}BN^2\sin BNC=\frac{1}{2}BN^2\sin A$ since cyclic quadrilateral $ABNC$ gives $\angle BNC=180-\angle A$.  Thus, we are to prove that
$\frac{1}{2}BN^2\sin A=[KNC]+[BMN]$
$\Leftrightarrow \frac{1}{2}BN^2\sin A=\frac{1}{2}CN\cdot CK\sin NCA+\frac{1}{2}BN\cdot BM\sin NBA$
$\Leftrightarrow BN\sin A=CK\sin NCA+BM\sin NBA$
From the fact that $\angle BNC=180-\angle A$ and that $BNC$ is iscoceles, we find that $\angle NBC=\angle NCB=\frac{1}{2}A$.  So, we have $BN\cos\frac{1}{2}A=\frac{1}{2}BC\Rightarrow BN=\frac{BC}{2\cos \frac{1}{2}A}$.  So we are to prove that
$\frac{BC\sin A}{2\cos \frac{1}{2}A}=CK\sin NCA+BM\sin NBA$
$\Leftrightarrow BC\sin \frac{1}{2}A=CK\sin (C+ \frac{1}{2}A)+BM\sin (C+ \frac{1}{2}A)$
$\Leftrightarrow BC=CK(\sin C\cot\frac{1}{2}A+\cos C)+BM(\sin B\cot\frac{1}{2}A+\cos B)$
We have $\sin C=\frac{KL}{CL}$,$\cos C=\frac{CK}{CL}$, $\cot\frac{1}{2}A=\frac{AK}{KL}=\frac{AM}{LM}$, $\sin B=\frac{LM}{BL}$,$\cos B=\frac{BM}{ML}$, and so we are to prove that
$BC=CK(\frac{KL}{CL}\frac{AK}{KL})+\frac{CK}{CL})+BM(\frac{LM}{BL}\frac{AM}{LM}+\frac{BM}{ML})$
$\Leftrightarrow BC=CK(\frac{AK}{CL}+\frac{CK}{CL})+BM(\frac{AM}{BL}+\frac{BM}{ML})$
$\Leftrightarrow BC=\frac{CK\cdot AC}{CL}+\frac{BM\cdot AB}{BL}$
$\Leftrightarrow BC=AC\cos C+AB\cos B$
We shall show that this is true:  Let the altitude from $A$ touch $BC$ at $A^\prime$.  Then it is obvious that $AC\cos C=CA^\prime$ and $AB\cos B=A^\prime B$ and thus $AC\cos C+AB\cos B=BC$.
Thus we have proven that $[AKNM]=[ABC]$.

Clearly, $AKLM$ is a kite, so its diagonals are perpendicular. Furthermore, we have triangles $ABN$ and $ALC$ similar because two corresponding angles are equal.
Hence, we have $[AKNM] = \frac{1}{2} AN \cdot KM = \frac{1}{2} \frac{AB \cdot AC}{AL} \cdot KM.$ Notice that we used the fact that a quadrilateral's area is equal to half the product of its perpendicular diagonals (if they are, in fact, perpendicular).
But in (right) triangle $AKL$, we have $<LAB = <A/2$. Furthermore, if $Q$ is the intersection of diagonals $AL$ and $KM$ we have $Q$ the midpoint of $KM$ and $KQ$ an altitude of $AKL$, so

so $\frac{KM}{AL} = \sin <A$. Hence  as desired.

Proceed as in Solution 2. To prove that $\frac{KM}{AL} = \sin A$, consider that

via usage of definition of sine, equal angles in a right triangle if its altitude is drawn, and the Law of Sines.


Let $n$ be a positive integer and let $A_1, A_2, \cdots, A_{2n+1}$ be subsets of a set $B$.
Suppose that
(a) Each $A_i$ has exactly $2n$ elements,
(b) Each $A_i\cap A_j$ $(1\le i<j\le 2n+1)$ contains exactly one element, and
(c) Every element of $B$ belongs to at least two of the $A_i$.
For which values of $n$ can one assign to every element of $B$ one of the numbers $0$ and $1$ in such a way that $A_i$ has $0$ assigned to exactly $n$ of its elements?

Solution:
Answer: All $n$ such that $4|n$
We first make the following $2$ claims:
Claim $1$: Each element of union belongs to exactly $2$ subsets.
Proof:
Consider a subset $A_i$. Assume that some element $x \in\ A_i$ also $\in A_k, A_l$. There are $n-1$ elements remaining in $A_i$ and there are $n-2$ subsets to choose from. By pigeon hole principle, at least $1$ of the remaining elements in $A_i$ must $\in A_k$ or $\in A_l$. This contradicts the assumption that any $2$ subsets have only $1$ element in common.
Claim $2$: $4|n$
Proof:
Now, since each element in $A_i \in$ exactly $1$ other subset, total number of elements present in the union = $n*(n+1)/2$.
If each subset must have $n/2$ elements assigned a value of $1$, the total number of elements assigned value of $1$ = $n/2*(n+1)/2 = n*(n+1)/4$.
Thus $4$ must divide $n$.
Now we make our final claim:
Claim $3$: $4|n$ is a sufficient condition to assign every element of the union one of the numbers 0 and 1 in such a manner that each of the sets has exactly $\frac {n}{2}$ zeros.
Proof:
Consider a regular polygon consisting of $n+1$ vertices where each line joining two vertices $A_i, A_j$ represents the element which $\in A_i, A_j$. Clearly there are a total of $n*(n+1)/2$ such lines representing the total number of elements of the union where each vertex is connected to $n$ vertices, meaning each of the $n$ elements of $A_i$ is part of $1$ other subset.
Starting with $i = 1$, let us start coloring all lines joining vertices $A_i$, $A_{i+1}$ with color Red, all lines joining $A_i$, $A_{i+2}$ with color White, $A_i$, $A_{i+3}$ with color Red, $A_i$, $A_{i+4}$ with color White and so on ... $A_i$, $A_{i+n/2}$ with color White.
Clearly each line from vertex $A_i$ alternates Red, White for first $n/2$ lines and then alternates White, Red for remaining $n/2$ lines implying that we could have exactly $n/2$ red lines emanating from each vertex $A_i$. But these $n/2$ lines represent $n/2$ elements of each subset $A_i$ which could each be assigned a value of $0$.  This completes the proof.
- Kris17
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.


Let $n\ge3$ and consider a set $E$ of $2n-1$ distinct points on a circle. Suppose that exactly $k$ of these points are to be colored black. Such a coloring is “good” if there is at least one pair of black points such that the interior of one of the arcs between them contains exactly $n$ points from $E$. Find the smallest value of $k$ so that every such coloring of $k$ points of $E$ is good.

Solution:
We form a graph on the $2n-1$ points by connecting two of these points iff one of the arcs they determine has exactly $n$ points in its interior. If $3\not|2n-1$, then this graph is a cycle, and the question becomes: what's the minimal $k$ s.t. whenever we color $k$ vertices of a cycle of length $2n-1$, there are two adjacent colored vertices? Obviously, the answer is $k=n$. On the other hand, when $3|2n-1$, the graph is a disjoint union of three cycles of length $\frac{2n+1}3$, and the answer will be, in this case, $3\cdot\left\lfloor\frac{2n-1}6\right\rfloor+1$.
This solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [1]


Let $\,n > 6\,$ be an integer and $\,a_{1},a_{2},\cdots ,a_{k}\,$ be all the natural numbers less than $n$ and relatively prime to $n$. If

prove that $\,n\,$ must be either a prime number or a power of $\,2$.

We use Bertrand's Postulate: for $u\ge 2$ is a positive integer, there is a prime in the interval $(u,2u)$.
Clearly, $a_2$ must be equal to the smallest prime $p$ which does not divide $n$. If $p=2$, then $n$ is a prime since the common difference $a_{i+1}-a_i$ is equal to $1$, i.e. all positive integers less than $n$ are coprime to $n$. If, on the ther hand, $p=3$, we find $n$ to be a power of $2$: the positive integers less than $n$ and coprime to it are precisely the odd ones. We may thus assume that $p\ge 5$. Furthermore, since $n>6$, the positive integers less than $n$ and coprime to it cannot be $a_1,a_2$ alone, i.e. $k=\varphi(n)\ge 3$.
By Bertrand's Postulate, the largest prime less than $p$ is strictly larger than $\frac{p-1}2$, so it cannot divide $p-1$. We will denote this prime by $q$. We know that $q|n$. It's easy to check that for $p\le 31$ there is a prime $r$ strictly between $a_2=p,\ a_3=2p-1$. $rq|n$, so, in particular, $pq<rq\le n$. On the other hand, if $p>32$, the two largest primes $r_1<r_2$ which are smaller than $q$ satisfy $r_2\ge\frac p4,\ r_1\ge\frac{r_2}2\ge\frac p8$ (again, Bertrand's Postulate), so $pq<\frac{p^2}{32}q\le r_1r_2q\le n$ so, once again, we have $pq<n$ (this is what I wanted to prove here).
Now, $q\not|p-1$, and all the numbers $1,1+p-1,\ldots,1+(q-1)(p-1)$ are smaller than $n$ (they are smaller than $pq$, which is smaller than $n$, according to the paragraph above). One of these numbers, however, must be divisible by $q|n$. Since our hypothesis tells us that all of them must be coprime to $n$, we have a contradiction.
This solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [1]

Let $d=a_{i+1}-a_i$. Then we have $a_i=1_1+d(i-1)=1+d(i-1)$. Since $\gcd(n-1,n)=1$, so $n-1$ is the largest positive integer smaller than $n$ and co-prime to $n$. Thus, $a_k=n-1$ and $1+d(k-1)=n-1$ which shows $d(k-1)=n-2$ and so $d$ divides $n-2$. We make two cases.
Case 1: $n$ is odd. Then $\gcd(n-2,n)=1$. So, every divisor of $n-2$ is co-prime to $n$ too, so is $d$. Thus, there is a $j$ such that $d=a_j=1+d(j-1)$ which implies $d|1$. Therefore, in this case, $d=1$ and we easily find that all numbers less than $n$ are co-prime to $n$, so $n$ must be a prime.
Case 2: $n$ is even. But then $\gcd(n-2,n)=2$ and also $d=1$ is ruled out. So $d>1$. Say, $n=2l$. Then, $d$divides $2(l-1)$. If $d$ is odd, $d|l$ and $d|l-1$ forcing $d|1$, contradiction! So, $d=2s$ with $s|l-1$. If $l=2r+1$ for some $r$, then $\gcd(r,n)=1$. So, again $r=1+d(j-1$ for a $j$. If $s$ is odd, then $s|r$ and thus, $s|1$. Similarly, we find that actually $d|2$ must occur. We finally have, $d=2$. But then all odd numbers less than $n$ are co-prime to $n$. So, $n$ does not have any odd factor i.e. $n=2^k$ for some $k\in\mathbb N$.
This solution was posted and copyrighted by ngv. The original thread for this problem can be found here: [2]

We use Bonse's Inequality: $p_{m+1}^2 < p_1p_2\cdots p_m$, for all $m \ge 4$, $p_1 = 2$.
Let $p$ denote the smallest prime which does not divide $n$.
Case 1) $p = 2$. This implies $n$ is a prime.
Case 2) $p = 3$. This implies $n = 2^k$ for some $k \in \mathbb{N}$.
Case 3) $p = 5$. Since $n > 6$, we have that $n > 9$. Therefore $a_3 = 9$, but $\gcd (n, 9) > 1$.
Case 4) $p \ge 7$. Write $n = p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}$. We have $(p - 2)^2 < p_1p_2\cdots p_k \le n = p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}$. Hence $(p-2)^2 = a_i$, a contradiction since $\gcd(p-2, n) > 1$.
This solution was posted and copyrighted by SFScoreLow. The original thread for this problem can be found here: [3]

Clearly, $a_1 = 1$. Now let $p$ be the smallest prime number which is not a divisor of $n$. Hence, $a_2 = p$, and the common difference $d = p-1$. Observe that since $a_k = n-1$, we have that, or. This means that all prime factors of $p-1$ are prime factors of $n-2$. However, due to the minimality of $p$, all primes factors of $p-1$ are prime factors of $n$. By the Euclidean algorithm, the only prime factors of $p-1$ are $2$, so $p = 2^{2^k} + 1$ or $p = 2$, by the theory of Fermat primes. We will now do casework on $p$.
Case 1: $p = 2$
This case is relatively easy. $a_2 = 2$, so all numbers less than $n$ must be relatively prime to $n$. However, if $n = pq$, where $p > 1$ and $q > 1$, observe that $p$ is not relatively prime to $n$, so $n$ must be prime for this case to work.
Case 2: $k = 0$ and $p = 3$.
This case is somewhat tricky. Observe that $n$ is even. Since $a_2 = 3$, we have that all numbers less than $n$ which are odd are relatively prime to $n$. However, if $n$ has a odd divisor $p > 1$, clearly $p$ is not relatively prime to $n$, so $n$ must be a power of two.
Case 3: $k \ge 1$.
Note that $3$ must be a divisor of $n$, so $a_k$ cannot be a multiple of $3$ for any $k$. I will contradict this statement. First, I claim that $k = \phi(n) > 2$ for all integers $n > 6$. $\phi{n} = 1$ is trivial, so let me show that $\phi(n) = 2$ has only solutions less than or equal to $6$. Since $\phi$ is a multiplicative function, observe that $n$ cannot be a multiple of a prime greater than $3$, because for such primes $p$ we have that $\phi(p^k) = (p-1)p^{k-1} \ge p-1 > 2$, which is contradiction. Hence, $n$ is just composed of factors of $2$ and $3$. If it divides both, and is of the form $2^m 3^n$ for $m$ and $n$ positive, we get that we need $2^{m} 3^{n-1} = 2$, so $n = 1$ and $m = 1$ in this case. If it is of the form $n = 2^k$, we get $k = 2$. And if it is of the form $3^k$, we get $k = 1$. $3$, $4$, and $6$ are all less than or equal to $6$, so $\phi(n) > 2$. Now we're almost done. Since $k > 2$, the number $a_3 = 2p-1$ must be in this sequence. However, $a_3 = 2p-1$ is of the form $2^{2^n+1} + 1$, which is a multiple of $3$, and we have contradiction, so we are done.
This solution was posted and copyrighted by va2010. The original thread for this problem can be found here: [4]

Clearly that $a_1=1$. Let $a_{i+1}-a_i = d \; (1 \le i \le \varphi(n)-1)$ thenWe can see that if $n$ is a prime then $n$ satisfies the condition. We consider the case when $n$ is a composite number. Let $p$ be the smallest prime divisor of $n$ then $p \le \sqrt{n}$. Note that $\varphi (n) > \sqrt{n} \ge p$ for all prime $n>6$ since $p_i^{k_i}-p_i^{k_i-1} \ge p_i^{k_i/2}$ for all $p \ge 3$. Therefore, if $p \nmid d$ then there will exists $1 \le i \le \varphi(n)$ such that $(i-1)d \equiv -1 \pmod{p}$ or $p \mid a_i$, a contradiction since $\gcd (a_i,n)=1$. Thus, $p \mid d$. We consider two cases:
If $d \ge 2p$ then $p+1 \ne a_i \; (1 \le i \le \varphi(n))$. Hence, $\gcd (p+1,n)>1$. This means there exists a prime $q>p \ge 2$ such that $q \mid p+1$. Since $q>p$ so we get $q=p+1$ or $p=2,q=3$, a contradiction since $n>6$. Thus, $d <2p$ or $d=p$.
If $p=d=2$ then all odd numbers $a_i \le n$ are relatively prime to $n$. This can only happen when $n=2^x \; (x \in \mathbb{N})$.
If $p=d >3$ then $p+3 \ne a_i \; \forall 1 \le i \le \varphi(n)$ since $a_i \equiv 1 \pmod{p} \; \forall 1 \le i \le \varphi(n)$. Thus, $\gcd (p+3,n)>1$ or there exist a prime $q>p>3$ such that $q \mid p+3$. Note that $2 \mid p+3$ so $q \le \tfrac{p+3}{2}<p$, a contradiction.
Thus, $n$ must be either a prime number or a power of $2$. $\blacksquare$
This solution was posted and copyrighted by shinichiman. The original thread for this problem can be found here: [5]


Let $\mathbb{R}$ denote the set of all real numbers.  Find all functions $f:\mathbb{R} \to \mathbb{R}$ such that

Solution:
[quote=probability1.01]Set x = 0 to get $f(f(y)) = y+f(0)^{2}$. We'll let $c = f(0)^{2}$, so $f(f(y)) = y+c$. Then
$f(a^{2}+f(f(b))) = f(b)+f(a)^{2}$
$f(a^{2}+b+c) = f(b)+f(a)^{2}$
$f(f(a^{2}+b+c)) = f(f(b)+f(a)^{2}) = b+f(f(a))^{2}$
$a^{2}+b+2c = b+(a+c)^{2}$
$2c = c^{2}+2ac$
Since this holds for all $a$, it follows that $c = 0$. Now we have $f(0) = 0 \implies f(f(y)) = y$. Note that f must be surjective since we may let y vary among all reals, and f must be injective since if $f(a) = f(b)$, then $a+f(x)^{2}= f(x^{2}+f(a)) = f(x^{2}+f(b)) = b+f(x)^{2}$. Finally, if $u > v$, then there is some $t$ s.t. $u = t^{2}+v$, and so $f(u) = f(t^{2}+v) = f^{-1}(v)+f(t)^{2}> f^{-1}(v) = f(f(f^{-1}(v))) = f(v)$. Hence f is strictly increasing. It is now clear that since $f(f(y)) = y$, we must have $f(x) = x$ for all x.[/quote]


Let $D$ be a point inside acute triangle $ABC$ such that $\angle ADB = \angle ACB+\frac{\pi}{2}$ and $AC\cdot BD=AD\cdot BC$.
(a) Calculate the ratio $\frac{AC\cdot CD}{AB\cdot BD}$.
(b) Prove that the tangents at $C$ to the circumcircles of $\Delta ACD$ and $\Delta BCD$ are perpendicular.

Solution: This problem needs a solution. If you have a solution for it, please help us out by adding it.


(Nazar Agakhanov, Russia)
Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that

We make the substitution $x= 1/a$, $y=1/b$, $z=1/c$.  Then

Since $(x^2,y^2,z^2)$ and $\bigl( 1/(y+z), 1/(z+x), 1/(x+y) \bigr)$ are similarly sorted sequences, it follows from the Rearrangement Inequality that

By the Power Mean Inequality,

Symmetric application of this argument yields

Finally, AM-GM gives us

as desired.  $\blacksquare$

We make the same substitution as in the first solution.  We note that in general,

It follows that $(x,y,z)$ and $\bigl(x/(y+z), y/(z+x), z/(x+y)\bigr)$ are similarly sorted sequences.  Then by Chebyshev's Inequality,

By AM-GM, $\frac{x+y+z}{3} \ge \sqrt[3]{xyz}=1$, and by Nesbitt's Inequality,

The desired conclusion follows.  $\blacksquare$

Without clever substitutions:
By Cauchy-Schwarz,  Dividing by $2(ab+bc+ac)$ gives  by AM-GM.

Without clever notation:
By Cauchy-Schwarz,
Dividing by $2(ab + bc + ac)$ and noting that $ab + bc + ac \ge 3(a^2b^2c^2)^{\frac{1}{3}} = 3$ by AM-GM gives

as desired.

After the setting $a=\frac{1}{x}, b=\frac{1}{y}, c=\frac{1}{z},$ and as $abc=1$ so $\left(\frac{1}{a}\cdot\frac{1}{b}\cdot\frac{1} {c}=1\right)$ concluding $x y z=1 .$
$\textsf{Claim}:$

By Titu Lemma,

Now by AM-GM we know that and $xyz=1$ which concludes to $\implies  (x+y+z)\geq3\sqrt[3]{1}$
Therefore we get
Hence our claim is proved ~~ Aritra12

Proceed as in Solution 1, to arrive at the equivalent inequality

But we know that  by AM-GM. Furthermore,

by Cauchy-Schwarz, and so dividing by $2(x + y + z)$ gives

as desired.

Without clever substitutions, and only AM-GM!
Note that $abc = 1 \implies a = \frac{1}{bc}$. The cyclic sum becomes $\sum_{cyc}\frac{(bc)^3}{b + c}$. Note that by AM-GM, the cyclic sum is greater than or equal to $3\left(\frac{1}{(a+b)(b+c)(a+c)}\right)^{\frac13}$. We now see that we have the three so we must be on the right path. We now only need to show that $\frac32 \geq 3\left(\frac{1}{(a+b)(b+c)(a+c)}\right)^\frac13$. Notice that by AM-GM, $a + b \geq 2\sqrt{ab}$, $b + c \geq 2\sqrt{bc}$, and $a + c \geq 2\sqrt{ac}$. Thus, we see that $(a+b)(b+c)(a+c) \geq 8$, concluding that $\sum_{cyc} \frac{(bc)^3}{b+c} \geq \frac32 \geq 3\left(\frac{1}{(a+b)(b+c)(a+c)}\right)^{\frac13}$

Rewrite $\frac{1}{a^3(b+c)} + \frac{1}{b^3(a+c)} + \frac{1}{c^3(a+b)}$ as $\frac{(1/a)^2}{a(b+c)} + \frac{(1/b)^2}{b(a+c)} + \frac{(1/c)^2}{c(a+b)}$.
Now applying Titu's lemma yields $\frac{(1/a)^2}{a(b+c)} + \frac{(1/b)^2}{b(a+c)} + \frac{(1/c)^2}{c(a+b)} \geq \frac{(\frac{1}{a} + \frac{1}{b} + \frac{1}{c})^2}{a(b+c) + b(a+c) + c(a+b)} = \frac{(ab + bc + ca)^2}{2(ab + bc + ca)} = \frac{ab + bc + ca}{2}$.
Now applying the AM-GM inequality on $ab + bc +ca \geq 3((abc)^2)^{\frac{1}{3}} = 3$. The result now follows.
Note: $ab + bc + ca = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$, because $abc = 1$. (Why? Because $a = \frac{1}{bc}$, and hence $\frac{1}{a} = bc$).
~th1nq3r
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Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.


Let $P$ be a point inside triangle $ABC$ such that

Let $D$, $E$m be the incenters of triangles $APB$, $APC$, respectively.  Show that $AP$, $BD$, $CE$ meet at a point.


The angle at $A$ is the smallest angle of triangle $ABD$. The points $B$ and $C$ divide the circumcircle of the triangle into two arcs.  Let $U$ be an interior point of the arc between $B$ and $C$ which does not contain $A$.  The perpendicular bisectors of $AB$ and $AC$ meet the line $AU$ and $V$ and $W$, respectively.  The lines $BV$ and $CW$ meet at $T$.  Show that.
$AU=TB+TC$


In a competition, there are a contestants and b judges, where b ≥ 3 is an odd
integer. Each judge rates each contestant as either “pass” or “fail”. Suppose k
is a number such that, for any two judges, their ratings coincide for at most k
contestants. Prove that k/a ≥ (b − 1)/(2b).


(Marcin Kuczma, Poland)
Let $n \geq 2$ be a fixed integer.

Solution:
The answer is $C=1/8$, and equality holds exactly when two of the $x_i$ are equal to each other and all the other $x_i$ are zero.  We prove this by induction on the number of nonzero $x_i$.
First, suppose that at most two of the $x_i$, say $x_a$ and $x_b$, are nonzero.  Then the left-hand side of the desired inequality becomes $x_a x_b (x_a^2 + x_b^2)$ and the right-hand side becomes $(x_a + x_b)^4/8$.  By AM-GM,

with equality exactly when $x_a = x_b$, as desired.
Now, suppose that our statement holds when at most $k$ of the $x_i$ are equal to zero.  Suppose now that $k+1$ of the $x_i$ are equal to zero, for $k\ge 2$.  Without loss of generality, let these be $x_1 \ge \dotsb \ge x_k \ge x_{k+1}$.
We define

and for convenience, we will denote $S= \sum_{i=1}^{k-1} x_i$.  We wish to show that by replacing the $x_i$ with the $y_i$, we increase the left-hand side of the desired inequality without changing the right-hand side; and then to use the inductive hypothesis.
We note that

If we replace $x$ with $y$, then $S(x_k^3 + x_{k+1}^3) + x_k^3 x_{k+1} + x_k x_{k+1}^3$ becomes

but none of the other terms change.  Since $3S > S \ge a_1 \ge a_k, a_{k+1}$, it follows that we have strictly increased the right-hand side of the equation, i.e.,

By inductive hypothesis,

and by our choice of $y_i$,

Hence the problem's inequality holds by induction, and is strict when there are more than two nonzero $x_i$, as desired.  $\blacksquare$

I claim that $C=\frac{1}{8}$. Let $P_2=\sum_{i=1}^nx_i^2$ and $S_2=\sum_{1\leq i<j\leq n}x_ix_j$. Note that  Equality holds in the first ineq when all but two of the $x_i$ are zero, and this reduces to the $n=2$ case which we can easily show to be equivalent to $\left(x_1-x_2\right)^4\geq0$, so $x_1=x_2$. That is, equality holds when two $x_i$ are the same and the rest are zero.
Q.E.D.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.


Let $a, b, c$ be positive real numbers with $abc=1$. Show that

Solution:
There exist positive reals $x$, $y$, $z$ such that $a = \frac{x}{y}$, $b = \frac{y}{z}$, $c = \frac{z}{x}$. The inequality then rewrites as
or  Set $p=x-y+z$,$q=y-z+x$,$r=z-x+y$, we get
Since at most one of $p,q,r$ can be negative (if 2 or more are negative, then one of $a,b,c$ will become negative), for all positive we apply AM-GM, for one negative we have $LHS<0<RHS$.


Let $a,b,c$ be positive real numbers. Prove that $\frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ca}}+\frac{c}{\sqrt{c^{2}+8ab}}\ge 1$.

Solution:
Firstly, $a^{2}+8bc=(a^{2}+2bc)+6bc \leq^{AG} (a^{2}+b^{2}+c^{2})+6bc=S+6bc$ (where $S=a^{2}+b^{2}+c^{2}$) and its cyclic variations.
Next note that $(a,b,c)$ and $\left( \frac{1}{\sqrt{S+6bc}}, \frac{1}{\sqrt{S+6ca}}, \frac{1}{\sqrt{S+6ab}} \right)$ are similarly oriented sequences. Thus

Hence the inequality has been established.
Equality holds if $a=b=c=1$.
Notation: $AG$: AM-GM inequality, $AH$: AM-HM inequality, $Cheff$: Chebyshev's inequality, $QA$: QM-AM inequality / RMS inequality


(Aleksander Ivanov, Bulgaria)
Determine all pairs of positive integers $(a,b)$ such that

is a positive integer.

Solution:
The only solutions are of the form $(a,b) = (2n,1)$, $(a,b) = (n,2n)$, and $(8n^4-n,2n)$ for any positive integer $n$.
First, we note that when $b=1$, the given expression is equivalent to $a/2$, which is an integer if and only if $a$ is even.
Now, suppose that $(a,b)$ is a solution not of the form $(2n,1)$.  We have already given all solutions for $b=1$; then for this new solution, we must have $b>1$.  Let us denote

Denote

Since $k(b^3-1) >0$, and $a$ is a positive integer root of $P$, there must be some other root $a'$ of $P$.
Without loss of generality, let $a' \ge a$.  Then $a^2 \le aa' = k(b^3-1)$, so

or

which reduces to

It follows that

or

Since $a$ and $b$ are integers, this can only happen when $2a-b=0$, so $(a,b)$ can be written as $(n,2n)$, and $k = n^2$.  It follows that

Since $a'$ is the other root of $P$, it follows that $(a',b)$ also satisfies the problem's condition.  Therefore the solutions are exactly the ones given at the solution's start.  $\blacksquare$
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
In this problem we can do it by an alternative method
a^2/2ab^2-b^3+1>=1
a^2>=2ab^2-b^3+1
a^2-2ab+b^2>=1/b
(a-b)^2>=1/b
The solutions are a>=2 and b>=1 are all the solutions


Find all polynomials $f$ with real coefficients such that for all reals $a,b,c$ such that $ab + bc + ca = 0$ we have the following relations

From $b=c=0$, we have $f(a) + f(-a) = 2f(a) \Longrightarrow f(a) = f(-a)$, so $f$ is even, and all the degrees all of its terms are even. Let $\text{deg}\, f(x) = n$
Let $(a,b,c) = (6x, 3x, -2x)$*; then we have $f(3x) + f(5x) + f(8x) = 2f(7x)$. Comparing lead coefficients, we have $3^n + 5^n + 8^n = 2 \cdot 7^n$, which cannot be true for $n \ge 5$. Hence, we have $f(x) = c_1 x^4 + c_2 x^2$. We can easily verify by expanding that all such polynomials work.

Let $a = (1 - \sqrt {3})x$, $b = x$, and $c = (1 + \sqrt {3})x$.  Then it is easy to check that $ab + bc + ca = 0$, so

for all $x$.  Hence, for the coefficient of $x^n$ to be nonzero, we must have $2( - \sqrt {3})^n + (2 \sqrt {3})^n = 2 \cdot 3^n$.
This does not hold for $n = 1$, and if $n$ is odd and $n \ge 3$, then the LHS is irrational and the RHS is a positive integer, so $n$ must be even.
Let $n = 2m$.  Then $2 \cdot 3^m + 12^m = 2 \cdot 3^{2m}$, so $2 + 4^m = 2 \cdot 3^m$.  This holds for $m = 1$ and $m = 2$, and $(4/3)^3 = 64/27 > 2$, so $2 + 4^m > 4^m > 2 \cdot 3^m$ for $m \ge 3$.  Therefore, $P(x)$ must be of the form $a_2 x^2 + a_4 x^4$.


Let $a_1, a_2, \dots$ be a sequence of integers with infinitely many positive and negative terms. Suppose that for every positive integer $n$ the numbers $a_1, a_2, \dots, a_n$ leave $n$ different remainders upon division by $n$. Prove that every integer occurs exactly once in the sequence.

Solution:
${a_n}$ satisfies the conditions if and only if ${a_n-a_1}$ does.
Therefore we can assume that $a_1=0$
First of all, $|a_n|<n$
Otherwise, $a_n = 0$ mod $a_{a_n}$
Claim: $a_{k+1}$ is either the smallest positive number not in ${a_1, a_2, ..., a_k}$ or the largest negative number not in this set.
This proves that if ${a_n}$ contains an infinite number of positive and negative numbers, it must contain each integer exactly once.
~Kscv