The positive real numbers $x_0, x_1, x_2,.....x_{1994}, x_{1995}$ satisfy the relations
$x_0=x_{1995}$ and $x_{i-1}+\frac{2}{x_{i-1}}=2{x_i}+\frac{1}{x_i}$
for $i=1,2,3,....1995$
Find the maximum value that $x_0$ can have.

Solution:
First we start by solving for $x_{i}$ in the recursive relation
$x_{i-1}+\frac{2}{x_{i-1}}=2x_{i}+\frac{1}{x_{i}}$
$\frac{x_{i-1}^{2}+2}{x_{i-1}}=\frac{2x_{i}^{2}+1}{x_{i}}$
$\left( x_{i} \right)\left( x_{i-1}^{2}+2 \right)=\left( x_{i-1} \right)\left( 2x_{i}^{2}+1 \right)$
$x_{i}x_{i-1}^{2}+2x_{i}=2x_{i-1}x_{i}^{2}+x_{i-1}$
$x_{i}x_{i-1}^{2}-2x_{i-1}x_{i}^{2}+2x_{i}-x_{i-1}=0$
$x_{i}x_{i-1}\left( x_{i-1} \right)-2x_{i}x_{i-1}\left( x_{i} \right)+2x_{i}-x_{i-1}=0$
$x_{i}x_{i-1}\left( x_{i-1} -2x_{i}\right)+2x_{i}-x_{i-1}=0$
$\left( 2x_{i}-x_{i-1}\right)\left( 1-x_{i}x_{i-1} \right)=0$
$x_{i}=\frac{x_{i-1}}{2},\;$ or $x_{i}=\frac{1}{x_{i-1}}$
So we have two recursive properties to chose from.
If we want to maximize $x_{0}$ then we can use $x_{i}=\frac{x_{i-1}}{2}$ from $i=1$ to $1994$.  This will make $x_{0}$ the largest and $x_{1994}$ the smallest.
Then we can simply use $x_{i}=\frac{1}{x_{i-1}}$ to get $x_{1995}$ since the reciprocal will make it very large.
Then we use $x_{0}=x_{1995}$ and solve for $x_{0}$
This means that we can write $x_{i}$ as:
$x_{i}=\begin{cases} \frac{x_{0}}{2^{i}} & 1 \le i \le 1994 \\ \frac{1}{x_{i-1}} & i=1995\end{cases}$
Then $x_{1994}=\frac{x_0}{2^{1994}}$,
thus $x_{1995}=\frac{1}{x_{1994}}=\frac{2^{1994}}{x_0}={x_0}$
Solving for ${x_0}$ we get:
${x_0}^{2}=2^{1994}$
${x_0}=\pm\sqrt{2^{1994}}=\pm2^{997}$.  Since all ${x_i}$ are defined as positive, ${x_0}=2^{997}$.
Therefore, the maximum value that $x_0$ can have is ${x_0}=2^{997}$
~ 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.


The positive integers $a$ and $b$ are such that the numbers $15a+16b$ and $16a-15b$ are both squares of positive integers.  What is the least possible value that can be taken on by the smaller of these two squares?

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


An $n \times n$ matrix whose entries come from the set $S={1,2,...,2n-1}$ is called a $\textit{silver}$ matrix if, for each $i=1,2,...,n$, the $i$th row and the $i$th column together contain all elements of $S$.  Show that
(a) there is no $\textit{silver}$ matrix for $n=1997$;
(b) $\textit{silver}$ matrices exist for infinitely many values of $n$.


Determine all pairs $(n,p)$ of positive integers such that
$p$ is a prime, $n$ not exceeded $2p$, and $(p-1)^{n}+1$ is divisible by $n^{p-1}$


A magician has one hundred cards numbered $1$ to $100$. He puts them into three boxes,
a red one, a white one and a blue one, so that each box contains at least one card.
A member of the audience selects two of the three boxes, chooses one card from each
and announces the sum of the numbers on the chosen cards. Given this sum, the magician
identifies the box from which no card has been chosen.
How many ways are there to put all the cards into the boxes so that this trick always
works? (Two ways are considered different if at least one card is put into a different box.)

Solution:
Consider $1$, $2$ and $3$. If they are in different boxes, then $4$ must be in the
same box as $1$, $5$ in the same box as $2$ and so on. This leads to the solution where all
numbers congruent to each other mod $3$ are in the same box.
Suppose $1$ and $2$ are in box $A$ and $3$ in box $B$. Then $4$ must be in box $A$ or $B$. In
general, if $k\ge 4$ is in either box $A$ or $B$, then $k + 1$ also must be in box $A$ or $B$. Thus box
$C$ is empty which is impossible.
Similarly, it is impossible for $1$ and $3$ to be in box $A$ and $2$ in box $B$.
Thus we are left with the case where $1$ is in box $A$ and $2$ and $3$ in box $B$. Suppose box
$B$ contains $2, . . . k$, where $k \ge 3$, but does not contain $k + 1$ and $m$ is the smallest number in box $C$.
Then $m > k$.
If $m > k + 1$, then $k + 1$ must be box $A$ and we see that no box can
contain $m-1$.
Thus $m = k + 1$. If $k < 99$, we see that no box can contain $k + 2$. Thus
$k = 99$. It is easy to see that this distribution works. Thus there altogether $12$ ways - $2$ options times permutation of $3$ colors for each of $3$ boxes.
Taken from: https://sms.math.nus.edu.sg/Simo/IMO_Problems/00.pdf


Let $n_1, n_2, \dots , n_m$ be integers where $m>1$ is odd. Let $x = (x_1, \dots , x_m)$ denote a permutation of the integers $1, 2, \cdots , m$. Let $f(x) = x_1n_1 + x_2n_2 + ... + x_mn_m$. Show that for some distinct permutations $a$, $b$ the difference $f(a) - f(b)$ is a multiple of $m!$.

Solution:
Notice that if $\{0,1,2,...,m!-1\}\not=\{f(1),f(2),f(3),...,f(m!)\}\pmod{m!}$ then by the pigeon hole princible, there must be some $a,b\in\{1,2,...,m!\}$ such that $f(a)\equiv f(b)\pmod{m!}$ as desired.  Thus, we must prove that $\{0,1,2,...,m!-1\}\not=\{f(1),f(2),f(3),...,f(m!)\}\pmod{m!}$.  Suppose there was such a situation.  Then, since the two sets have the same number of elements, for every $t\in\{0,1,...,m!-1\}$, there exists a $v\in\{1,2,...,m!\}$ such that $t\equiv f(v)\pmod{m!}$.  So, let $t\equiv f(v_t)\pmod{m!}$.  Consider the sum $f(v_0)+f(v_1)+\cdots+f(v_{m!-1})\pmod{m!}$.  Using the fact that $f(v_t)\equiv t\pmod{m!}$, we find the sum is congruent to $0+1+\cdots+m!-1=\frac{m!(m!-1)}{2}$.
On the other hand, using the fact that $f(x)=x_1n_1 + x_2n_2 + ... + x_mn_m$, we can combine the terms with the same coefficient ($n_i$).  For each $n_1$, since all the $v_j$ are distinct, the coefficient in the sum would be $\sum x_i$ over all permutations $x$ of $\{1,2,...,m\}$.  Thus, the coefficient would be $\sum_{i=1}^m i(m-1)!=\frac{m!(m+1)}{2}$.  Since $m$ is odd, $\frac{m+1}{2}$ is an integer, so the coefficient is congruent to $0\pmod{m!}$.  Thus, the whole sum is congruent to $0\pmod{m!}$.  Therefore, $\frac{m!(m!-1)}{2}\equiv0\pmod{m!}$.  But, since $m>1$, we have $m!$ is even, and thus $m!-1$ is odd.  Therefore, the integer $\frac{m!(m!-1)}{2}$ has one factor of $2$ less than $m!$ does.  Contradiction!


Let $ABCD$ be a cyclic quadrilateral. Let $P$, $Q$, and $R$ be the feet of perpendiculars from $D$ to lines $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$, respectively. Show that $PQ=QR$ if and only if the bisectors of angles $ABC$ and $ADC$ meet on segment $\overline{AC}$.

Solution: Clearly $PQR$ is the Simson Line and $APDQ$, $BPDR$, $CQDR$ is cyclic. By angle chasing we have $\triangle DPQ\sim\triangle DBC$, $\triangle DQR\sim\triangle DAB$. Then by $PQ=QR$ we have $\frac{DC}{CB}=\frac{DQ}{QP}=\frac{DQ}{QR}=\frac{DA}{AB}$. Rearranging and using the angle bisector theorem we are done.


(Hojoo Lee) Let $n \geq 3$ be an integer. Let $t_1, t_2, \dots , t_n$ be positive real numbers such that

Show that $t_i$, $t_j$, $t_k$ are side lengths of a triangle for all $i$, $j$, $k$ with $1 \leq i < j < k \leq n$.

Solution:
For $n=3$, suppose (for sake of contradiction) that $t_3 = t_2 + t_1 + k$ for $k \ge 0$; then (by Cauchy-Schwarz Inequality)

so it is true for $n=3$. We now claim the result by induction; for $n \ge 4$, we have

By AM-GM, $\frac{t_n}{t_i} + \frac{t_i}{t_n} \ge 2$, so $f(n) \ge f(n-1) + 2(n-1) + 1 = f(n-1) + 2n - 1$. Then the problem is reduced to proving the statement true for $n-1$ numbers, as desired.$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\square$


Determine all positive integers relatively prime to all the terms of the infinite sequence

Solution:
Let $k$ be a positive integer that satisfies the given condition.
For all primes $p>3$, by Fermat's Little Theorem, $n^{p-1} \equiv 1\pmod p$ if $n$ and $p$ are relatively prime. This means that $n^{p-3} \equiv \frac{1}{n^2} \pmod p$. Plugging $n = p-3$ back into the equation, we see that the value$\mod p$ is simply $\frac{2}{9} + \frac{3}{4} + \frac{1}{36} - 1 = 0$. Thus, the expression is divisible by all primes $p>3.$ Since $a_2 = 48 = 2^4 \cdot 3,$ we can conclude that $k$ cannot have any prime divisors. Therefore, our answer is only $1.$


Determine all pairs $(x, y)$ of integers such that

Solution:
If $(x,y)$ is a solution then obviously $x\geq 0$ and $(x,-y)$ is a solution too.  For $x=0$ we get the two solutions $(0,2)$ and $(0,-2)$.
Now let $(x,y)$ be a solution with $x > 0$; without loss of generality confine attention to $y > 0$.  The equation rewritten as  shows that the factors $y-1$ and $y+1$ are even, exactly one of them divisible by $4$.  Hence $x\geq 3$ and one  of these factors is divisible by $2^{x-1}$ but not by $2^x$.  So

Plugging this into the original equation we obtain

or, equivalently  Therefore  For $\varepsilon = 1$ this yields $m^2 - 8 < 0$, i.e. $m=1$, which fails to satisfy $(\dagger)$. For $\varepsilon = -1$ equation $(\dagger)$ gives us  implying $2m^2 - m - 17 \leq 0$.  Hence $m\leq 3$; on the other hand $m$ cannot be $1$ by $(\dagger)$.  Because $m$ is odd, we obtain $m=3$, leading to $x=4$.  From $(*)$ we get $y=23$.  These values indeed satisfy the given equation.  Recall that then $y = -23$ is also good.  Thus we have the complete list of solutions $(x,y)$: $(0,2)$, $(0,-2)$, $(4,23)$, $(4,-23)$.


In $\triangle ABC$ the bisector of $\angle{BCA}$ intersects the circumcircle again at $R$, the perpendicular bisector of $BC$ at $P$, and the perpendicular bisector of $AC$ at $Q$. The midpoint of $BC$ is $K$ and the midpoint of $AC$ is $L$. Prove that the triangles $RPK$ and $RQL$ have the same area.

$\angle{RQL} = 90+\angle{QCL} = 90+\dfrac{C}{2}$, and similarly $\angle{RPK} = 90+\angle{PCK} = 90+\dfrac{C}{2}$. Therefore, $\angle{RQL} = \angle{RPK}$. Using the triangle area formula $A = \dfrac{1}{2}bc\sin{\angle{A}}$ yields $RQ \cdot QL = RP \cdot PK = \dfrac{PK}{QL} = \dfrac{RQ}{RP}$ after cancelling the sines and constant. Draw line $QD$ perpendicular to $BC$ that intersects $BC$ at $D$, then $QD=QL$ because the perpendicular bisectors are congruent, (or alternatively $\triangle QDC\cong\triangle QLC$). This presents us $\dfrac{PK}{QL}=\dfrac{PK}{QD}=\dfrac{PC}{QC}$ by similar triangles; now, we have only to prove $\dfrac{PC}{QC}=\dfrac{RQ}{RP}$, or $RQ \cdot QC=RP \cdot PC$.
Since $\angle{OPQ} =1 80-\angle{RPK} =1 80-\angle{RQL} = \angle{OQP}$, we have $\triangle OPQ$ is isosceles. Draw the perpendicular from $O$ to $RC$, intersecting at $E$. Then $PE = QE = x$ for a real $x$, now because the perpendicular from the center of a circle to a chord bisects that chord, $RE = CE$. Let $y = RE$, and then $RQ \cdot QC = (y+x) \cdot (y-x) = PC \cdot RP$, proving our claim.

The area of $\triangle{RQL}$ is given by $\dfrac{1}{2}QL*RQ\sin{\angle{RQL}}$ and the area of $\triangle{RPK}$ is $\dfrac{1}{2}RP*PK\sin{\angle{RPK}}$. Let $\angle{BCA}=C$, $\angle{BAC}=A$, and $\angle{ABC}=B$. Now $\angle{KCP}=\angle{QCL}=\dfrac{C}{2}$ and $\angle{PKC}=\angle{QLC}=90$, thus $\angle{RPK}=\angle{RQL}=90+\dfrac{C}{2}$. $\triangle{PKC} \sim \triangle{QLC}$, so $\dfrac{PK}{QL}=\dfrac{KC}{LC}$, or $\dfrac{PK}{QL}=\dfrac{BC}{AB}$. The ratio of the areas is $\dfrac{[RPK]}{[RQL]}=\dfrac{BC*RP}{AC*RQ}$. The two areas are only equal when the ratio is 1, therefore it suffices to show $\dfrac{RP}{RQ}=\dfrac{AC}{BC}$. Let $O$ be the center of the circle. Then $\angle{ROK}=A+C$, and $\angle{ROP}=180-(A+C)=B$. Using law of sines on $\triangle{RPO}$ we have: $\dfrac{RP}{\sin{B}}=\dfrac{OR}{\sin{(90+\dfrac{C}{2})}}$ so $RP*\sin{(90+\dfrac{C}{2})}=OR*\sin{B}$. $OR*\sin{B}=\dfrac{1}{2}AC$ by law of sines, and $\sin{(90+\dfrac{C}{2})}=\cos{\dfrac{C}{2}}$, thus 1) $2RP\cos{\dfrac{C}{2}}=AC$. Similarly, law of sines on $\triangle{ROQ}$ results in $\dfrac{RQ}{\sin{(180-A)}}=\dfrac{OR}{\sin{(90-\dfrac{C}{2})}}$ or $\dfrac{RQ}{\sin{A}}=\dfrac{OR}{\cos{\dfrac{C}{2}}}$. Cross multiplying we have $RQ\cos{\dfrac{C}{2}}=OR*\sin{A}$ or 2) $2RQ\cos{\dfrac{C}{2}}=BC$. Dividing 1) by 2) we have $\dfrac{RP}{RQ}=\dfrac{AC}{BC}$ $\square$
$(tkhalid)$

(Image Link) https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvYi85LzQ5OTJlZGNhYTQ0YjJjODcxMTBmZGNmMTdiZDdkMGRjZGUyOWQ5LnBuZw==&rn=U2NyZWVuIFNob3QgMjAxOS0wOC0wOCBhdCAxMi4yMC4zOCBQTS5wbmc=
WLOG, let the diameter of $(ACBR)$ be $1.$
We see that $PK = \dfrac{1}{2}a \tan \dfrac{1}{2}C$ and $QL = \dfrac{1}{2}b \tan \dfrac{1}{2}C$ from right triangles $\triangle PKC$ and $\triangle QLC.$
We now look at $AR.$ By the Extended Law of Sines on $\triangle ACR,$ we get that $AR = \sin\frac{1}{2}C.$ Similarly, $BR = \sin \frac{1}{2}C.$
We now look at $CR.$ By Ptolemy's Theorem, we have  which gives us  This means that  We now seek to relate the lengths computed with the areas.
To do this, we consider the altitude from $R$ to $PK.$ This is to find the area of $RPK.$ Finding the area of $\triangle RQL$ is similar.
We claim that $RF = \dfrac{1}{2}b.$ In order to prove this, we will prove that $\triangle RFP \cong \triangle QLC.$ In other words, we wish to prove that $PR = QC.$ This is equivalent to proving that $PC + QC = CR.$
Note that $PC = \dfrac{PK}{\sin \frac{1}{2}C}$ and $QC = \dfrac{QL}{\sin \frac{1}{2}C}.$ Therefore, we get that

Thus, $RF = \dfrac{1}{2}b.$ In this way, we get that the altidude from $R$ to $QL$ has length $\dfrac{1}{2}a.$ Therefore, we see that $[RPK] = \dfrac{1}{8}ab \tan \frac{1}{2}C$ and $[RQL] = \dfrac{1}{8}ab \tan \frac{1}{2}C,$ so the two areas are equal.
Solution by Ilikeapos

$[\triangle{RPK}]=[\triangle{RQL}], LQ*RQ*\sin\angle{LQR}=KP*PR*\sin\angle{RPK}$
Since $CR$ bisects $\angle{ACB},\angle{QCL}=\angle{PCK}$, $OL,OK$ are perpendicular to sides $AC,BC$ separately, $\angle{QLC}=\angle{PKC}=90^{\circ}, \angle{CQL}=\angle{CPK}, \angle{LQR}=\angle{RPK}$
So now, we only have to prove $RP*PK=LQ*QR$, which is $RP*(CQ+QP)*\cos\angle{CPK}=CQ*(QP+PR)*\cos\angle{CQL}$, as mentioned above, the two angles are the same, we have to prove that $RP(CQ+QP)=CQ(QP+PR)$, which is equivalent to $RP*QP=QP*CQ$, we have to prove $RP=CQ$
Now notice that $\triangle{OQP}$ is isosceles. $\angle{CQL}=\angle{OQP}=\angle{OPQ}$, construct $OJ \bot CR$, $CJ=JR,JQ=JP,CQ=PR$ as desired
~bluesoul
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.


Find all functions $f: (0, \infty) \mapsto (0, \infty)$ (so $f$ is a function from the positive real numbers) such that
for all positive real numbers $w,x,y,z,$ satisfying $wx = yz.$

Solution:
Considering $w=1$ and $z=y=\sqrt{x}$ which satisfy the constraint $wx=yz$ we get the following equation:

At once considering $x=1$ we get $(f(1))^2 = f(1)$ and knowing that $f : \mathbb{R}^+ \rightarrow \mathbb{R}^+$ the only possible solution is $f(1)=1$ since $f(1)=0$ is impossible.

So we get the quadratic equation:

Solving for $f(x)$ as a function of $x$ we get:

At once we see that for one value of $x$, $f(x)$ can only take one of 2 possible values:

.

Take into consideration that $f(2) = 2$ but $f(3) = \frac{1}{3}$ verifies the quadratic equation and thus so far we can't say that $f(x)=x \, \forall_{x \in \mathbb{R}^+}$ or alternatively $f(x)=\frac{1}{x} \, \forall_{x \in \mathbb{R}^+}$. This is indeed the case but we haven't proved it yet.

To prove the previous assertion consider 2 values $a,b \in \mathbb{R}^+$ such that $a\ne 1 \wedge b \ne 1 \wedge a \ne b$ while having $f(a) = a \wedge f(b)=\frac{1}{b}$

Consider now the original functional equation with $w=a,\ x=b,\ y=z=\sqrt{ab}$ which verifies the constraint. Substituting we have:

Now either $f(ab)=ab$ or $f(ab)=\frac{1}{ab}$. (notice that $ab \ne b \wedge ab\ne b$ by hypothesis)

If $f(ab)=ab$ then we have $ab = ab\frac{a^2 + \frac{1}{b^2}}{a^2 + b^2} \Leftrightarrow b^4=1$ and since $b>0$ the only solution is $b=1$.

If $f(ab)=\frac{1}{ab}$ then we have $\frac{1}{ab} = ab\frac{a^2 + \frac{1}{b^2}}{a^2 + b^2} \Leftrightarrow  a^2+b^2 = a^4b^2 +a^2 \Leftrightarrow a^4=1$ and since $a>0$ the only solution is $a=1$.

So the only solutions are $a=1$ or $b=1$ in which case both alternatives imply $f(1)=1$. Thus we conclude that solutions to the functional equation are a subset of $\left\{f(x)=x \ \forall_{x \in \mathbb{R}^+},\ f(x)=\frac{1}{x}\ \forall_{x \in \mathbb{R}^+} \right\}$.

Finally, plug each of these 2 functions into the functional equation and verify that they indeed are solutions.

This is trivial since $f(x)=x$ is an obvious solution and for $f(x)=\frac{1}{x}$ we have:

provided that $(wx)^2 = (yz)^2$ which verifies the original constraint.

So the functional equation has 2 solutions:


Let $ABC$ be a triangle with $AB=AC$. The angle bisectors of $\angle CAB$ and $\angle ABC$ meet the sides $BC$ and $CA$ at $D$ and $E$, respectively. Let $K$ be the incentre of triangle $ADC$. Suppose that $\angle BEK=45^\circ$. Find all possible values of $\angle CAB$.
Authors: Jan Vonk and Peter Vandendriessche, Belgium, and Hojoo Lee, South Korea


Let $P$ be a point interior to triangle $ABC$ (with $CA \neq CB$). The lines $AP$, $BP$ and $CP$ meet again its circumcircle $\Gamma$ at $K$, $L$, respectively $M$. The tangent line at $C$ to $\Gamma$ meets the line $AB$ at $S$. Show that from $SC = SP$ follows $MK = ML$.

Without loss of generality, suppose that $AS > BS$. By Power of a Point, $SP^2 = SC^2 = SB \cdot SA$, so $\overline{SP}$ is tangent to the circumcircle of $\triangle ABP$. Thus, $\angle KPS = 180 - \angle SPA = \widehat{AP}/2 = \angle ABP$. It follows that after some angle-chasing,
so $ML = MK$ as desired.

Let the tangent at $M$ to $\Gamma$ intersect $SC$ at $X$. We now have that since $\triangle{XMC}$ and $\triangle{SPC}$ are both isosceles, $\angle{SPC}=\angle{SCP}=\angle{XMC}$. This yields that $MX \| PS$.
Now consider the power of point $S$ with respect to $\Gamma$.

Hence by AA similarity, we have that $\triangle{SPA} \sim \triangle{SBP}$. Combining this with the arc angle theorem yields that $\angle{SPA}=\angle{SBP}=\angle{PKL}$. Hence $PS \| LK$.
This implies that the tangent at $M$ is parallel to $LK$ and therefore that $M$ is the midpoint of arc $LK$. Hence $MK=ML$.

Since $SP = SC$, we can construct circle $\omega$ tangent to $SP$ and $SC$ at $P$ and $C$ respectively. Then, we see that the circumcircle of $\triangle ABP$ must be tangent to $SP$ at $P$ by the converse of the radical axis theorem, since $\omega$ is internally tangent to $\Gamma$ at $C$, and $(ABP)$ meets $\Gamma$ at $A$ and $B$. Thus, $\angle SPB = \angle BAK = \angle BLK$, so lines $SP$ and $KL$ are parallel.
It then follows that $\angle SCP = \angle SPC = \angle KDC = \angle KLC + \angle LCD$. Since $SC$ is tangent to $\Gamma$, we see that $\angle KLC = \angle SCK$, and thus $\angle KCD = \angle LCD = \angle SCP - \angle KLC$. Thus since $CM$ bisects $\angle KCL$ in cyclic quad $CKML$, we have $MK = ML$. $\blacksquare$


Let $n > 0$ be an integer. We are given a balance and $n$ weights of weight $2^0,2^1, \cdots ,2^{n-1}$.  We are to place each of	the $n$ weights on the balance, one after another, in such a way that the right pan is never heavier than the left pan.  At each step we choose one of the weights that has not yet been placed on the balance, and place it on either the left pan or the right pan, until all of the weights have been placed.
Determine the number of ways in which this can be done.

Solution:
Call our answer $W(n)$.
We proceed to prove $W(n)=(2n-1)!!$.
It is evident $W(1)=1$.
Now, the key observation is that smaller weights can never add up to the weight of a larger weight, ie which side is heavier is determined completely by the heaviest weight currently placed. It follows, therefore, that the number of ways to place $n$ weights on the balance according to the rule is the same no matter which $n$ distinct powers of two are the weights, as each weight completely overpowers any smaller weight and is completely overpowered by any larger weight. That is, there is the 1st heaviest weight, the 2nd heaviest, the 3rd, ..., the n-th heaviest, and each weight is negligible compared to any heavier weight. Thus, any valid placement of $n$ weights of weight $2^0,2^1, \cdots ,2^{n-1}$ can changed by replacing $2^i$ with the $(n-i)$-th heaviest weight in the set ${2^{a_k}}$, where $a_k \in \mathbb{Z}$, and vice versa, forming a $1:1$ relation. With this in mind, we use recursion upon the last weight placement. There are $2n-1$ choices; namely, you can put any weight on either side except for the heaviest weight on the right. For the first $n-1$ weight placements, the answer reduces to $W(n-1)$. We can reduce $W(n-1)$ in the same way.
$W(n)=(2n-1)W(n-1)=(2n-1)(2n-3)W(n-2)=...=(2n-1)!!W(1)=(2n-1)!!$
$\text{QED}$


Let $ABC$ be an acute triangle with orthocenter $H$, and let $W$ be a point on the side $BC$, lying strictly between $B$ and $C$. The points $M$ and $N$ are the feet of the altitudes from $B$ and $C$, respectively. Denote by $\omega_1$ is the circumcircle of $BWN$, and let $X$ be the point on $\omega_1$ such that $WX$ is a diameter of $\omega_1$. Analogously, denote by $\omega_2$ the circumcircle of triangle $CWM$, and let $Y$ be the point such that $WY$ is a diameter of $\omega_2$. Prove that $X, Y$ and $H$ are collinear.

Let $T$ be the intersection of $\omega_1$ and $\omega_2$ other than $W$.
Lemma 1: $T$ is on $XY$.
Proof: We have $\angle{XTW} = \angle{YTW} = 90^\circ$ because they intercept semicircles. Hence, $\angle{XTY} = \angle{XTW} + \angle{YTW} = 180^\circ$, so $XTY$ is a straight line.
Lemma 2: $T$ is on $AW$.
Proof: Let the circumcircles of $NBW$ and $MWC$ be $\omega_1$ and $\omega_2$, respectively, and, as $BNMC$ is cyclic (from congruent $\angle{BNC} = \angle{BMC} = 90^\circ$), let its circumcircle be $\omega_3$. Then each pair of circles' radical axises, $BN, TW,$ and $MC$, must concur at the intersection of $BN$ and $MC$, which is $A$.
Lemma 3: $YT$ is perpendicular to $AW$.
Proof: This is immediate from $\angle{YTW} = 90^\circ$.
Let $AH$ meet $BC$ at $L$, which is also the foot of the altitude to that side. Hence, $\angle{ALB} = 90^\circ.$
Lemma 4: Quadrilateral $THLW$ is cyclic.
Proof: We know that $NHLB$ is cyclic because $\angle{BNH}$ and $\angle{BLH}$, opposite and right angles, sum to $180^\circ$. Furthermore, we are given that $NTWB$ is cyclic. Hence, by Power of a Point,

The converse of Power of a Point then proves $THLW$ cyclic.
Hence, $\angle{WTH} = 180^\circ - \angle{WLH} = 90^\circ$, and so $HT$ is perpendicular to $AW$ as well. Combining this with Lemma 3's statement, we deduce that $T, H, Y$ are collinear. But, as $X$ is on $YT$ (from Lemma 1), $X, Y, H$ are collinear. This completes the proof.
$\blacksquare$
--Suli 13:51, 25 August 2014 (EDT)

Probably a simpler solution than above.
As above, let $T = \omega_1 \cap \omega_2 \neq W.$ By Miquel $MTHN$ is cyclic. Then since $\angle WCY = \angle WBX = 90^{\circ}$ we know, because $W,B,X,T \in \omega_1$ and $W,C,Y,T \in \omega_2,$ that $\angle WTY = \angle WTX = 90^{\circ},$ thus $X,T,Y$ are collinear.
There are a few ways to finish.
(a)  so $H,X,Y$ are collinear, as desired $\square$
(b)  Since $BNMC$ is cyclic we know $AN \cdot AB = AM \cdot AC$ which means $p(A,\omega_1) = p(A, \omega_2)$ so $A$ is on the radical axis, $TW,$ hence  so $H$ lies on this line as well and we may conclude $\square$
--mathguy623 03:10, 12 August 2016 (EDT)

For any point $P,$ let $p$ denote the complex number for point $P.$ First off, let $C$ be the origin. Now, since $WY$ is the diameter of the circumcircle of triangle $CMW$, we must have $\angle{YMW}=\angle{YCH}=90^{\circ}.$ Since angles inscribed in the same arc are congruent, $\angle{MYW}=\angle{MCW}=\angle{C}.$ This means that $\frac{|w-m|}{|y-m|}=\tan{\angle{C}}.$ Combining this with the fact that $\angle{YMW}$ is right, we find that  Solving, we find that  We wish to simplify $(i\tan{\angle{C}}-1)m$ first. Note that  This means that

This means that  Now, since $WX$ is a diameter of the circumcircle of triangle $NBW,$ we must have $\angle{WNX}=\angle{WBX}=90^{\circ}.$ Since angles inscribed in the same arc are congruent, $\angle{NXW}=\angle{NBW}=\angle{B}.$ This means that $\frac{|x-n|}{|w-n|}=\cot{\angle{B}}.$ Combining this with the fact that $\angle{WNX}$ is right, we find that  Solving, we find that  We wish to simplify $n(1-i\cot{\angle{B}})$ first. Note that  This means that

This means that  This means that the line through complex numbers $x$ and $y$ satisfy the equation  If there is a fixed point to the line, then the real and imaginary values of the point must not contin $w$. If the fixed point is $c,$ then we have  after comparing the coefficient of $w.$ This means that

From the Law of Sines, we find that  Substituting this, we find that

This means that

Since the $w$'s cancel out, we can just discard everything with $w$ in it. Thus,

Since  we have
Thus,  In conclusion,  which is the fixed point $XY$ always passes through. However, by inspection, $c=H=\text{the orthocenter}.$ Therefore, we conclude that $X,Y,H$ are collinear for all acute triangles $ABC.$
~pinkpig


Points $P$ and $Q$ lie on side $BC$ of acute-angled $\triangle{ABC}$ so that $\angle{PAB}=\angle{BCA}$ and $\angle{CAQ}=\angle{ABC}$. Points $M$ and $N$ lie on lines $AP$ and $AQ$, respectively, such that $P$ is the midpoint of $AM$, and $Q$ is the midpoint of $AN$. Prove that lines $BM$ and $CN$ intersect on the circumcircle of $\triangle{ABC}$.

We are trying to prove that the intersection of $BM$ and $CN$, call it point $D$, is on the circumcircle of triangle $ABC$. In other words, we are trying to prove $\angle {BDC} + \angle {BAC} = 180$.
Let the intersection of $BM$ and $AN$ be point $E$, and the intersection of $AM$ and $CN$ be point $F$.
Let us assume $\angle {BDC} + \angle {BAC} = 180$. Note: This is circular reasoning. If $\angle {BDC} + \angle {BAC} = 180$, then $\angle {BAC}$ should be equal to $\angle {BDN}$ and $\angle {CDM}$. We can quickly prove that the triangles $ABC$, $APB$, and $AQC$ are similar, so $\angle {BAC} = \angle {AQC} = \angle {APB}$. We also see that $\angle {AQC} = \angle {BQN} = \angle {APB} = \angle {CPF}$. Also because angles $BEQ$ and $NED, MFD$ and $CFP$ are equal, the triangles $BEQ$ and $NED$, $MDF$ and $FCP$ must be two pairs of similar triangles. Therefore we must prove angles $CBM$ and $ANC, AMB$ and $BCN$ are equal.
We have angles $BQA = APC = NQC = BPM$. We also have $AQ = QN$, $AP = PM$. Because the triangles $ABP$ and $ACQ$ are similar, we have $\dfrac {EC}{EN} = \dfrac {BF}{FM}$, so triangles $BFM$ and $NEC$ are similar. So the angles $CBM$ and $ANC, BCN$ and $AMB$ are equal and we are done.

Let $L$ be the midpoint of $BC$. Easy angle chasing gives $\angle{AQP} = \angle{APQ} = \angle{BAC}$. Because $P$ is the midpoint of $AM$, the cotangent rule applied on triangle $MBA$ gives us

Hence, by the cotangent rule on $ABC$, we have

Because the period of cotangent is $180^\circ$, but angles are less than $180^\circ$, we have $\angle{BAL} = \angle{MBC}.$
Similarly, we have $\angle{LAC} = \angle{NCB}.$ Hence, if $BM$ and $CN$ intersect at $Z$, then $\angle{BZC} = 180^\circ - \angle{BAC}$ by the Angle Sum in a Triangle Theorem. Hence, $BACZ$ is cyclic, which is equivalent to the desired result.
--Suli 23:27, 7 February 2015 (EST)

Let $L$ be the midpoint of $BC$. By AA Similarity, triangles $BAP$ and $BCA$ are similar, so $\dfrac{BA}{AP} = \dfrac{BC}{CA}$ and $\angle{BPA} = \angle{BAC}$. Similarly, $\angle{CQA} = \angle{BAC}$, and so triangle $AQP$ is isosceles. Thus, $AQ = AP$, and so $\dfrac{BA}{AQ} = \dfrac{BC}{CA}$. Dividing both sides by 2, we have $\dfrac{BA}{AN} = \dfrac{BL}{AC}$, or

But we also have $\angle{ABL} = \angle{CAQ}$, so triangles $ABL$ and $NAC$ are similar by $SAS$ similarity. In particular, $\angle{ANC} = \angle{BAL}$. Similarly, $\angle{BMA} = \angle{CAL}$, so $\angle{ANC} + \angle{BMA} = \angle{BAC}$. In addition, angle sum in triangle $AQP$ gives $\angle{QAP} = 180^\circ - 2\angle{A}$. Therefore, if we let lines $BM$ and $CN$ intersect at $T$, by Angle Sum in quadrilateral $AMTN$ concave $\angle{NTM} = 180^\circ + \angle{A}$, and so convex $\angle{BTC} = 180^\circ - \angle{A}$, which is enough to prove that $BACT$ is cyclic. This completes the proof.

--Suli 10:38, 8 February 2015 (EST)

Let $D_1$ be the second intersection of $NC$ with the circumcircle of $\triangle ABC,$ and $D_2$ the second intersection of $MB$ with the circumcircle of $\triangle ABC.$ By inscribed angles, the tangent at $C$ is parallel to $AN.$ Let $P_{\infty}$ denote the point at infinity along line $AN.$ Note that  So, $ABD_1C$ is harmonic. Similarly, we can find $ABD_2C$ is harmonic. Therefore, $D_1=D_2,$ which means that $BM$ and $CN$ intersect on the circumcircle. $\blacksquare$

We use barycentric coordinates. Due to the equal angles, $AC$ is tangent to the circumcircle of $ABQ$ and $AB$ is tangent to the circumcircle of the $APC.$ Therefore, we can use power of a point to solve for side ratios. We have
Therefore, $D=(-a^2:2b^2:2c^2),$ as $BM$ and $CN$ are cevians. Note that $(x,y,z)$ lies on the circumcircle iff $a^2yz+b^2xz+c^2xy=0.$ Substituting the values in, we have  so we are done. $\blacksquare$

Note that the givens immediately imply that $\triangle{ABC} \sim \triangle{QAC} \sim \triangle{PBA}$, hence $\angle{AQP}=\angle{APQ}=\angle{A}$. Let $D$ be the midpoint of BC, $E$ be the midpoint of $AC$, and $F$ the midpoint of $AB$. By the similar triangles, we have $\angle{BAD}=\angle{AQE}=\angle{AMC}$. We also have $\angle{BAD}=\angle{BPF}=\angle{MNB}$, so we find $\angle{AMC}=\angle{MNB}$. We note that $\angle{AMC}+\angle{CMN}=\angle{AMN}=\angle{AQP}=\angle{A}$, so $\angle{CMN}+\angle{MNB}=A$, which gives that $\angle{BKC}=180-\angle{A}$ and we are done.
As an addition, $AK$ is the A-symmedian in $\triangle{ABC}$.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.


Triangle $ABC$ has circumcircle $\Omega$ and circumcenter $O$. A circle $\Gamma$ with center $A$ intersects the segment $BC$ at points $D$ and $E$, such that $B$, $D$, $E$, and $C$ are all different and lie on line $BC$ in this order. Let $F$ and $G$ be the points of intersection of $\Gamma$ and $\Omega$, such that $A$, $F$, $B$, $C$, and $G$ lie on $\Omega$ in this order. Let $K$ be the second point of intersection of the circumcircle of triangle $BDF$ and the segment $AB$. Let $L$ be the second point of intersection of the circumcircle of triangle $CGE$ and the segment $CA$.
Suppose that the lines $FK$ and $GL$ are different and intersect at the point $X$. Prove that $X$ lies on the line $AO$.
Proposed by Silouanos Brazitikos and Evangelos Psychas, Greece

Solution:
Lemma (On three chords). If two lines pass through different endpoints of two circles' common chord, then the other two chords cut by these lines on the circles are parallel.
Proof The second and the third chords are anti-parallel to the first (common) chord with respect to the given lines, so they are parallel to each other. $\Box$
To solve this problem, it is sufficient to apply the lemma 5 times. Indeed, let the lines $FD, GE, FK, GL$ meet $\Omega$ second time at $H, I, M, N$ respectively. One of the circles that figure in lemma is always $\Omega$, while the other is one of three other circles from the problem statement. Applying the lemma to the lines $FDH$ and $GEI$, $FKM$ and $BDC$, $FDH$ and $BKA$, $GLN$ and $CEB$, $GEI$ and $CLA$, we get $DE \parallel IH$, $KD \parallel MC$, $KD \parallel AH$, $LE \parallel NB$, $LE \parallel AI$, respectively. From this, $BC \parallel IH$, $MC \parallel AH$, $NB \parallel AI$. Therefore, $AN=IB=HC=AM$. This means that $N$ and $M$ are symmetric wrt $AO$, a diameter of $\Omega$ through $A$. So are $F$ and $G$, as $AF=AG$. Therefore, the lines $FM$ and $GN$ are symmetric wrt $AO$ and meet on it.


A set of positive integers is called fragrant if it contains at least two elements and each of its elements has a prime factor in common with at least one of the other elements.  Let $P(n)=n^2+n+1$.  What is the least possible positive integer value of $b$ such that there exists a non-negative integer $a$ for which the set $\{P(a+1),P(a+2),\ldots,P(a+b)\}$ is fragrant?

Solution:
Consider $P(x)$ and $P(x+y)$. We note that in order to $p \mid P(x)$ and $p \mid P(x+y)-P(x)$ we must have $p \mid x^2+x+1$ and $p \mid y(2x+y+1)$. It is obvious that $p \equiv 1 \pmod{6}$ since $p \mid n^2+n+1 \mid (2n+1)^2+3$ or $\left( \tfrac{-3}{p} \right)=1$.
If $y=1$ then $p \mid 2x+2$ implies $p \mid x+1$. WLOG, we can let $x=p-1$ and see that $p \nmid x^2+x+1$. So there doesn't exists $x$ so that $\gcd \left( P(x),P(x+1) \right)>1$.
If $y=2$, we find $p \mid 2x+3$ and we let $x=\tfrac{p-3}{2}$. Hence, from $p \mid x^2+x+1$ we get $p=7$. So there is one prime $p=7$ such that $\gcd \left( P(x),P(x+2) \right)>1$.
If $y=3$, it is obvious that $p=3$ satisfy and is the only answer.
If $y=4$, we do the similar thing and get $p \mid 2x+5$ and $p \mid x^2+x+1$ so $p=19$.
___________________________
Now, back to the problem, since there doesn't exists any prime $p$ for $y=1$ so $b \ge 3$. The only prime for $y=2$ is $p=7$ so if we choose $7 \mid P(x+1), 7 \mid P(x+3)$ then there will be a number $7 \nmid P(x+2)$ remains. This means $b \ge 5$ since we need a prime $p \mid P(x+2)$ and $p \mid P(x+y)$ but $y \ge 5$ since $p \ne 7$.
If $b=5$, we consider two cases, where there are two numbers that are divisible by $19$ (which means $19 \mid P(x+1), 19 \mid P(x+5)$), the middle-three numbers $P(x+2),P(x+3),P(x+4)$ we can't find a way make each two of them have common prime factor. If no two are divisible by $19$ then they can only be divisible by $7,3$, but it can't cover all $5$ "consecutive" numbers.
If $b=6$ then we can pick $19 \mid P(x+2),P(x+6), 3 \mid P(x+1), P(x+4), 7 \mid P(x+3), 7 \mid P(x+5)$.
So the final answer is $b=6$.


Let $R$ and $S$ be different points on a circle $\Omega$ such that $RS$ is not a diameter. Let $\ell$ be the tangent line to $\Omega$ at $R$. Point $T$ is such that $S$ is the midpoint of the line segment $RT$. Point $J$ is chosen on the shorter arc $RS$ of $\Omega$ so that the circumcircle $\Gamma$ of triangle $JST$ intersects $\ell$ at two distinct points. Let $A$ be the common point of $\Gamma$ and $\ell$ that is closer to $R$. Line $AJ$ meets $\Omega$ again at $K$. Prove that the line $KT$ is tangent to $\Gamma$.

Solution:
We construct inversion which maps $KT$ into the circle $\omega_1$ and  $\Gamma$ into  $\Gamma.$ Than we prove that $\omega_1$ is tangent to $\Gamma.$
Quadrangle $RJSK$ is cyclic $\implies \angle RSJ = \angle RKJ.$
Quadrangle $AJST$ is cyclic $\implies \angle RSJ = \angle TAJ \implies AT||RK.$
We construct circle $\omega$ centered at $R$ which maps  $\Gamma$ into  $\Gamma.$
Let $C = \omega \cap RT \implies RC^2 = RS \cdot RT.$ Inversion with respect to $\omega$ swap $T$ and $S \implies  \Gamma$ maps into  $\Gamma (\Gamma = \Gamma').$
Let $O$ be the center of $\Gamma.$
Inversion with respect to $\omega$  maps $K$ into $K'$.
$K$ belong $KT \implies$ circle  $K'SR = \omega_1$ is the image of $KT$. Let $Q$ be the center of $\omega_1.$
$K'T$ is the image of $\Omega$ at this inversion, $l = AR$ is tangent line to $\Omega$ at $R,$ so $K'T||AR.$
$K'$ is image K at this inversion $\implies K \in RK' \implies RK'||AT \implies ARK'T$ is parallelogram.
$S$ is the midpoint of $RT \implies S$ is the center of symmetry of $ATK'R \implies$
$\triangle RSK'$ is symmetrical to $\triangle TSA$ with respect to $S \implies$
$\omega_1$  is symmetrical to $\Gamma$ with respect to $S \implies$
$O$ is symmetrycal $Q$ with respect to $S.$
$S$ lies on $\Gamma$ and on $\omega_1 \implies \Gamma$ is tangent to $\omega_1 \implies$ line $KT$ is tangent to $\Gamma.$
vladimir.shelomovskii@gmail.com, vvsss

We use the tangent-chord theorem: the angle formed between a chord and a tangent line to a circle is equal to the inscribed angle on the other side of the chord.
Quadrangle $RJSK$ is cyclic $\implies \angle RSJ = \angle RKJ.$
Quadrangle $AJST$ is cyclic $\implies \angle RSJ = \angle TAJ$
(One can use Reim’s theorem – it is shorter way.)
Let $B$ be symmetric to $A$ with respect to $S \implies$
$ATBR$ is parallelogram.

$\angle RBT = \angle RAT \implies \angle KST + \angle KBT = 180^\circ$
$\implies SKBT$ is cyclic.
Inscribed angle of $\Gamma (\angle SAT)$ is equal to angle between $KT$ and chord $ST \implies$
$KT$ is tangent to $\Gamma$ by the inverse of tangent-chord theorem.
vladimir.shelomovskii@gmail.com, vvsss


A site is any point $(x, y)$ in the plane such that $x$ and $y$ are both positive integers less
than or equal to 20.
Initially, each of the 400 sites is unoccupied. Amy and Ben take turns placing stones with Amy
going first. On her turn, Amy places a new red stone on an unoccupied site such that the distance
between any two sites occupied by red stones is not equal to $\sqrt{5}$. On his turn, Ben places a new blue
stone on any unoccupied site. (A site occupied by a blue stone is allowed to be at any distance from
any other occupied site.) They stop as soon as a player cannot place a stone.
Find the greatest $K$ such that Amy can ensure that she places at least $K$ red stones, no matter
how Ben places his blue stones.

Solution:
The maximal K is 100.
Amy can reach at least 100 by playing only on sites for which x+y is even. There are 200 such sites, none are of distance $\sqrt{5}$ from each other and Ben can occupy at most half of them.
On the other hand Ben can prevent Amy from reaching more than 100 using the following strategy:
Picture the sites as a 20 by 20 board and divide it into 25 non overlapping 4-by-4 squares. We label each site in the square as follows:
Whenever Amy plays in a square Ben plays in the same square and in the site with the same label. In each square Amy can place at most 2 stones in sites labeled 1,4,6,7 (no three sites with labels from this set are free from distance $\sqrt{5}$ and Amy can play one stone on each label since Ben plays the other). Likewise for the sites labeled 2,3,5,8. So in total Amy can place at most 4 stones in each of the 25 squares for a total of 100 stones.


Find all pairs $(k,n)$ of positive integers such that

$LHS$
$k! = 1$ (when $k = 1$), $2$ (when $k = 2$), $6$ (when $k = 3$)
$RHS = 1$(when $n = 1$), $6$ (when $n = 2$)
Hence, $(1,1)$, $(3,2)$ satisfy
For $k = 2: RHS$ is strictly increasing, and will never satisfy $k$ = 2 for integer n since $RHS = 6$ when $n = 2$.

In all solutions, for any prime $p$ and positive integer $N$, we will denote by  the exponent of the largest power of $p$ that divides $N$. The right-hand side of $(1)$ will be denoted by  that is,
=
On the other hand,  is expressed by the $Legendre$ $formula$ as
Thus, $k! = L_n$ implies the inequality
In order to obtain an opposite estimate, observe that
We claim that  for all $n \geq 6$
For $n \geq 6$ the estimate (3) is true because
and
~flamewavelight and ~phoenixfire


Let $\Gamma$ be a circle with centre $I$, and $ABCD$ a convex quadrilateral such that each of
the segments $AB, BC, CD$ and $DA$ is tangent to $\Gamma$. Let $\Omega$ be the circumcircle of the triangle $AIC$.
The extension of $BA$ beyond $A$ meets $\Omega$ at $X$, and the extension of $BC$ beyond $C$ meets $\Omega$ at $Z$.
The extensions of $AD$ and $CD$ beyond $D$ meet $\Omega$ at $Y$ and $T$, respectively. Prove that

Let $O$ be the centre of $\Omega$.
For $AB=BC$ the result follows simply. By Pitot's Theorem we have  so that, $AD = CD.$ The configuration becomes symmetric about $OI$ and the result follows immediately.
Now assume WLOG $AB < BC$. Then $T$ lies between $A$ and $X$ in the minor arc $AX$ and $Z$ lies between $Y$ and $C$ in the minor arc $YC$.
Consider the cyclic quadrilateral $ACZX$.
We have $\angle CZX = \angle CAB$ and $\angle IAC = \angle IZC$. So that,   Since $I$ is the incenter of quadrilateral $ABCD$, $AI$ is the angular bisector of $\angle DBA$. This gives us,  Hence the chords  $IX$ and $IY$ are equal.
So $Y$ is the reflection of $X$ about $OI$.
Hence,  and now it suffices to prove
Let $P, Q, N$ and $M$ be the tangency points of $\Gamma$ with $AB, BC, CD$ and $DA$ respectively. Then by tangents we have, $AD = AM + MD = AP + ND$. So $AD + DT + XA = AP + ND + DT + XA = XP + NT$.
Similarly we get, $CD + DY + ZC = ZQ + YM$. So it suffices to prove,
Consider the tangent $XJ$ to $\Gamma$ with $J \ne P$. Since $X$ and $Y$ are reflections about $OI$ and $\Gamma$ is a circle centred at $I$ the tangents $XJ$ and $YM$ are reflections of each other. Hence  By a similar argument on the reflection of $T$ and $Z$ we get $NT = ZQ$ and finally,
 as required.
$QED$
~BUMSTAKA

Denote $AD$ tangents to the circle $I$ at $N$, $CD$ tangents to the same circle at $M$; $XB$ tangents at $F$ and $ZB$ tangents at $J$. We can get that $AD=AM+MD;CD=DN+CN$.Since $AM=AF,XA=XF-AM;ZC=ZJ-CN$ Same reason, we can get that $DT=TN-DM;DY=YM-DM$
We can find that $AD+XA+DT=XF+TN;CD+DY+ZC=ZJ+YM$. Connect $IM,IT,IT,IZ,IX,IN,IJ,IF$ separately, we can create two pairs of congruent triangles. In $\triangle{XIF},\triangle{YIM}$, since $\widehat{AI}=\widehat{AI},\angle{FXI}=\angle{MYI}$ After getting that $\angle{IFX}=\angle{IMY};\angle{FXI}=\angle{MYI};IF=IM$, we can find that $\triangle{IFX}\cong \triangle{IMY}$. Getting that $YM=XF$, same reason, we can get that $ZJ=TN$.
Now the only thing left is that we have to prove $TX=YZ$. Since $\widehat{IX}=\widehat{IY};\widehat{IT}=\widehat{IZ}$ we can subtract and get that $\widehat{XT}=\widehat{YZ}$,means $XT=YZ$ and we are done
~bluesoul

We use the equality of the tangent segments and symmetry.
Using Claim 1 we get $\overset{\Large\frown} {TX}$ symmetric to $\overset{\Large\frown} {ZY}$ with respect $IO.$
Therefore $\hspace{10mm} TX = ZY.$
Let $P, Q, N$ and $M$ be the tangency points of $\Gamma$ with $AB, BC, CD,$ and $DA,$ respectively.
Using Claim 2 we get $TM = QZ, PX = NY.$
Claim 1
Let $O$ be the center of $\Omega.$ Then point $T$ is symmetric to $Z$ with respect $IO,$ point $X$ is symmetric to $Y$ with respect $IO.$
Proof
Let $\angle BAD =2\alpha, \angle ABC =2\beta, \angle BCD =2\gamma, \angle ADC =2\delta.$
We find measure of some arcs:

$\overset{\Large\frown} {IX}= 2\pi - \overset{\Large\frown} {IY} - \overset{\Large\frown} {XY} = 2\alpha =\overset{\Large\frown} {IY}\implies$ symmetry $X$ and $Y.$

$\overset{\Large\frown} {IZ}= 2\pi - \overset{\Large\frown} {IT} - \overset{\Large\frown} {TZ}= 2\gamma= \overset{\Large\frown} {IT}\implies$ symmetry $T$ and $Z.$
Claim 2
Let circles $\omega$ centered at $I$ and $\Omega$ centered at $O$ be given. Let points $A$ and $A'$ lies on  $\Omega$ and $A$ be symmetric to $A'$ with respect $OI.$ Let $AC$ and $A'B$ be tangents to  $\omega$. Then $AC = A'B.$
Proof
vladimir.shelomovskii@gmail.com, vvsss


Let $ABCDE$ be a convex pentagon such that $BC = DE$. Assume that there is a
point $T$ inside $ABCDE$ with $TB = TD$, $TC = TE$ and $\angle ABT = \angle TEA$. Let line $AB$ intersect
lines $CD$ and $CT$ at points $P$ and $Q$, respectively. Assume that the points $P, B, A, Q$ occur on their
line in that order. Let line $AE$ intersect lines $CD$ and $DT$ at points $R$ and $S$, respectively. Assume
that the points $R, E, A, S$ occur on their line in that order. Prove that the points $P, S, Q, R$ lie on
a circle.

Solution:
$\hspace{28mm}CDQS$ is cyclic $\implies \angle QCD = \angle QSD.$

$\hspace{43mm}PRQS$ is cyclic.
vladimir.shelomovskii@gmail.com, vvsss


Let $x_1, x_2, \cdots , x_{2023}$ be pairwise different positive real numbers such that

is an integer for every $n = 1,2,\cdots,2023$. Prove that $a_{2023} \ge 3034$.

Solution:
We solve for $a_{n+2}$ in terms of $a_n$ and $x.$
$a_{n+2}^2 \\  = (\sum^{n+2}_{k=1}x_k)(\sum^{n+2}_{k=1}\frac1{x_k}) \\ = (x_{n+1}+x_{n+2}+\sum^{n}_{k=1}x_k)(\frac{1}{x_{n+1}}+\frac{1}{x_{n+2}}+\sum^{n}_{k=1}\frac1{x_k}) \\  = \frac{x_{n+1}}{x_{n+1}} + \frac{x_{n+1}}{x_{n+2}} + \frac{x_{n+2}}{x_{n+1}} + \frac{x_{n+2}}{x_{n+2}} +  \frac{1}{x_{n+1}}\sum^{n}_{k=1}x_k + x_{n+1}\sum^{n}_{k=1}\frac1{x_k} + \frac{1}{x_{n+2}}\sum^{n}_{k=1}x_k + x_{n+2}\sum^{n}_{k=1}\frac1{x_k} + (\sum^{n}_{k=1}x_k)(\sum^{n}_{k=1}\frac1{x_k}) \\ = \frac{x_{n+1}}{x_{n+1}} + \frac{x_{n+1}}{x_{n+2}} + \frac{x_{n+2}}{x_{n+1}} + \frac{x_{n+2}}{x_{n+2}} +  \frac{1}{x_{n+1}}\sum^{n}_{k=1}x_k + x_{n+1}\sum^{n}_{k=1}\frac1{x_k} + \frac{1}{x_{n+2}}\sum^{n}_{k=1}x_k + x_{n+2}\sum^{n}_{k=1}\frac1{x_k} + a_n^2 \\ \text{}$
Again, by AM-GM, the above equation becomes
$a_{n+2}^2 \ge 4 \sqrt[4]{(\frac{x_{n+1}}{x_{n+1}})(\frac{x_{n+1}}{x_{n+2}})(\frac{x_{n+2}}{x_{n+1}})(\frac{x_{n+2}}{x_{n+2}})} +  4\sqrt[4]{ (\frac{1}{x_{n+1}}\sum^{n}_{k=1}x_k)(x_{n+1}\sum^{n}_{k=1}\frac1{x_k})(\frac{1}{x_{n+2}}\sum^{n}_{k=1}x_k)(x_{n+2}\sum^{n}_{k=1}\frac1{x_k}) } + a_n^2  = a_n^2+4a_n+4 = (a_n+2)^2 \\ \text{}$
Hence, $a_{n+2} \ge a_{n} + 2,$ but equality is achieved only when $\frac{x_{n+1}}{x_{n+1}},\frac{x_{n+1}}{x_{n+2}},\frac{x_{n+2}}{x_{n+1}},$ and $\frac{x_{n+2}}{x_{n+2}}$ are equal. They can never be equal because there are no two equal $x_k.$So $a_{2023} \ge a_1 + 3\times \frac{2023-1}{2} = 1 + 3033 = 3034$


Consider the cube $ABCDA'B'C'D'$ (with face $ABCD$ directly above face $A'B'C'D'$).
a) Find the locus of the midpoints of the segments $XY$, where $X$ is any point of $AC$ and $Y$ is any point of $B'D'$;
b) Find the locus of points $Z$ which lie on the segment $XY$ of part a) with $ZY = 2XZ$.

Solution:
Let $A=(0,0,2)$, $B=(2,0,2)$, $C=(2,0,0)$, $D=(0,0,0)$, $A'=(0,2,2)$, $B'=(2,2,2)$, $C'=(2,2,0)$, and $D'=(0,2,0)$. Then there exist real $x$ and $y$ in the closed interval $[0,2]$ such that $X=(x,0,2-x)$ and $Y=(y,2,y)$.
The midpoint of $XY$ has coordinates $((x+y)/2, 1, (2-x+y)/2)$. Let $a$ and $b$ be the $x$- and $z$-coordinates of the midpoint of $XY$, respectively. We then have that $a+b=y+1$ and $a-b=x-1$, so $a+b\in [1,3]$ and $a-b\in [-1,1]$. The region of points that satisfy these inequalities is the closed square with vertices at $(1,1,2)$, $(2,1,1)$, $(1,1,0)$, and $(0,1,1)$. For every point $P$ in this region, there exist unique points $X$ and $Y$ such that $P$ is the midpoint of $XY$.
If $Z\in XY$ and $ZY=2XZ$, then $Z$ has coordinates $((2x+y)/3, 2/3, (4-2x+y)/3)$. Let $a$ and $b$ be the $x$- and $z$- coordinates of $Z$. We then have that $a+b=(4/3)+(2/3)y$ and $a-b=(4x-4)/3$, and $a\in (4/3,8/3)$ and $b\in (-4/3,4/3)$. The region of points that satisfy these inequalities is the closed rectangle with vertices at $(0,2/3,4/3)$, $(2/3,2/3,1)$, $(1,2/3,2/3)$, and $(4/3,2/3,0)$. For every point $Z$ in this region, there exist unique points $X$ and $Y$ such that $Z\in XY$ and $ZY=2XZ$.


Construct a triangle ABC if the following elements are given: $AC = b, AB = c$, and $\angle AMB = \omega \left(\omega < 90^{\circ}\right)$ where M is the midpoint of BC.  Prove that the construction has a solution if and only if
$b \tan{\frac{\omega}{2}} \le c < b$
In what case does equality hold?

Solution:
Prolong BA to a point D such that $BD = 2AB$. Take circle through B and D such that the minor arc BD is equal to $2*\omega$ so that for points P on the major arc BD we have $\angle BPD = \omega$. Draw a circle with center A and radius AC, and the point of intersection of this circle and the major arc BD will be C. In general there are two possibilities for C.
Let X be the intersection of the arc BN and the perpendicular to the segment BN through A. For the construction to be possible we require $AX \geqslant AC > AB$. But $\frac{AB}{AX} =  \tan{\frac{\omega}{2}}$, so we get the condition in the question


On the circle $K$ there are given three distinct points $A,B,C$. Construct (using only straightedge and compass) a fourth point $D$ on $K$ such that a circle can be inscribed in the quadrilateral thus obtained.

Solution:
Aviso: La siguiente solución está en español
generalidad ya que es simétrico para todas las combinaciones posibles)
Circunferencia K en 2 puntos, uno de ello se nombrará con E (va a estar del mismo lado que B con
respecto a AC) y el otro se nombrará F
iguales, abrir con el compás la distancia BE y se traza la Circunferencia de Centro en F y Radio BE,
así esta nueva circunferencia corta a la circunferencia K en 2 puntos, llámese D al punto que está
del mismo lado de A con respecto a la mediatriz de AC, ese es el punto pedido que cumple la


Solution:
Because the sum of the $x$-coordinates of the seventh roots of unity is $0$, we have

Now, we can apply $\cos{x} = \cos{\left(-x\right)}$ to obtain

Finally, since $\cos{x} = -\cos{(\pi-x)}$,

~mathboy100

Let $\cos{\frac{\pi}{7}}-\cos{\frac{2\pi}{7}}+\cos{\frac{3\pi}{7}}=S$. We have

Then, by product-sum formulae, we have

Thus $S = 1/2$. $\blacksquare$

Let $a=\sin{\frac{\pi}{7}}$ and $b=\cos{\frac{\pi}{7}}$. From the addition formulae, we have

From the Trigonometric Identity, $a^2=1-b^2$, so

We must prove that $S=1/2$. It suffices to show that $8b^3-4b^2-4b+1=0$.
Now note that $\cos{\frac{4\pi}{7}}=-\cos{\frac{3\pi}{7}}$. We can find these in terms of $a$ and $b$:

Therefore $8b^4-8b^2+1=-(3b-4b^3)\Rightarrow 8b^4+4b^3-8b^2-3b+1=0$. Note that this can be factored:

Clearly $b\neq -1$, so $8b^3-4b^2-4b+1=0$. This proves the result. $\blacksquare$

Let $\omega=\mathrm{cis}\left(\frac{\pi}{14}\right)$. Thus it suffices to show that $\omega+\omega^{-1}-\omega^2-\omega^{-2}+\omega^3+\omega^{-3}=1$. Now using the fact that $\omega^k=\omega^{14+k}$ and $-\omega^2=\omega^9$, this is equivalent to

But since $\omega$ is a $14$th root of unity, $\omega^{14}=1$. The answer is then $-\omega^{7}=1$, as desired.
~yofro

We let $\omega = \mathrm{cis} \ \left(\dfrac{2\pi}{7} \right)$. We therefore have $w^i$, where $0 \leq i \leq 6$, are the $7^{\text{th}}$ roots of unity. Since $\sum_{i = 0}^6 \omega^i = 0$, then $\sum_{i = 1}^6 \omega^i = -1$, so $\sum_{i = 1}^3 \mathrm{Re}(\omega^i) = -\dfrac{1}{2}$. Therefore, because $\mathrm{cis} \ \alpha = \cos \alpha + i \sin \alpha, \mathrm{Re}(\mathrm{cis} \ \alpha) = \cos \alpha$, so
Since $\cos \alpha = - \cos(\pi - \alpha)$, we have $\cos \left(\dfrac{\pi}{7} \right) - \cos \left(\dfrac{2\pi}{7} \right) + \cos \left(\dfrac{3\pi}{7} \right) = \dfrac{1}{2}$ and we are done $\blacksquare$
~Yiyj1


Suppose five points in a plane are situated so that no two of the straight lines
joining them are parallel, perpendicular, or coincident. From each point perpendiculars
are drawn to all the lines joining the other four points. Determine
the maximum number of intersections that these perpendiculars can have.

Solution:
Suppose, those five points are $(A, B, C, D, E)$. Now, we want to create some special structure. Let, we take the line $BC$ and draw a perpendicular from $A$ on $BC$, andd call it $P_1$. We can do this set up in $\binom{5}{1}\binom{4}{2}=30$ ways. There will $30$ such $P_i$ s.
Now, we will find how many other perpendiculars intersect the line. We can do this in total $20$ ways. Why? See, can draw perpendiculars from $B$ and $C$ to other lines( we haven't counted the perpendicular from $B$ to $AC$ and perpendicular from $C$ on $AB$ , as they intersect $P_1$ at the same point) in $5$ ways for each. So, total $10$ ways.
Now, $5$ perpendiculars from each $D$ and $E$ on the other lines except on $BC$( because in this case teh perpendiculars from $D$ and $E$ will be parallel to $P_1$ , and so shall not intersect). So,total $10$ cases.
From, these two cases we get $P_1$ will be intersected at at most $5*6*(5+5+5+5)=600$ points.
But, as we have passed this algorithm over all the five points, we have counted each intersection points twice. So, there are total $\frac{600}{2}=300$ ways.
Now, as we had excluded  the orthocentres, we have to add now. There are total $\binom{5}{3}=10$ orthocentres. Also we should add those vertices as these are also point of intersection of silimar perpendiculars, there are $5$ such.
So, total ways $300+10+5=315$.


Consider $\triangle OAB$ with acute angle $AOB$. Through a point $M \neq O$ perpendiculars are drawn to $OA$ and $OB$, the feet of which are $P$ and $Q$ respectively. The point of intersection of the altitudes of $\triangle OPQ$ is $H$. What is the locus of $H$ if $M$ is permitted to range over (a) the side $AB$, (b) the interior of $\triangle OAB$?

Solution:
Let $O(0,0),A(a,0),B(b,c)$.
Equation of the line $AB: y=\frac{c}{b-a}(x-a)$.
Point $M \in AB : M(\lambda,\frac{c}{b-a}(\lambda-a))$.
Easy, point $P(\lambda,0)$.
Point $Q = OB \cap MQ$, $MQ \bot OB$.
Equation of $OB : y=\frac{c}{b}x$, equation of $MQ : y=-\frac{b}{c}(x-\lambda)+\frac{c}{b-a}(\lambda-a)$.
Solving: $x_{Q}=\frac{1}{b^{2}+c^{2}}\left[b^{2}\lambda+\frac{c^{2}(\lambda-a)b}{b-a}\right]$.
Equation of the first altitude: $x=\frac{1}{b^{2}+c^{2}}\left[b^{2}\lambda+\frac{c^{2}(\lambda-a)b}{b-a}\right] \quad (1)$.
Equation of the second altitude: $y=-\frac{b}{c}(x-\lambda)\quad\quad (2)$.
Eliminating $\lambda$ from (1) and (2):

a line segment $MN , M \in OA , N \in OB$.
Second question: the locus consists in the $\triangle OMN$.


Solution:
Take a1 > a2 > a3 > a4. Subtracting the equation for i=2 from that for i=1 and dividing by (a1 - a2) we get:

Subtracting the equation for i=4 from that for i=3 and dividing by (a3 - a4) we get:

Hence x1 = x4. Subtracting the equation for i=3 from that for i=2 and dividing by (a2 - a3) we get:

Hence $x2 = x3 = 0$, and $x1 = x4 = 1/(a1 - a4)$.


Let $f$ be a real-valued function defined for all real numbers $x$ such that, for some positive constant $a$, the equation

holds for all $x$.
(a) Prove that the function $f$ is periodic (i.e., there exists a positive number $b$ such that $f(x + b) = f(x)$ for all $x$).
(b) For $a = 1$, give an example of a non-constant function with the required properties.

Solution:
(a) Since

is true for any $x$, and
We have:

Therefore $f$ is periodic, with $2a>0$ as a period.
(b) $f(x) = 1$ when $2n\le x < 2n+1$ for some integer $n$, and $f(x)=\frac{1}{2}$ when $2n+1\le x < 2n+2$ for some integer $n$.


Given $n > 4$ points in the plane such that no three are collinear, prove that there are at least $(n-3)/2$ convex quadrilaterals whose vertices are four of the given points.

Solution:
Orient the points so that one of them corresponds to the origin (A), another lies on the y-axis (B), and all the others are in either quandrant I or IV. (In other words, you are creating a boundary line.) Select one of the points (C) which minimizes the angle BAC. You cannot have two of them because then they would be collinear. Also no points lie inside the area of ABC because if such a point D did exist, then the angle DAC would be less than BAC - contradiction. So there are none of the other n-3 points inside the area of ABC.
Now we will show that for any two of the remaining n-3 points, you can create a convex quadrilateral containing two of them and two of A,B,C.
Now select two arbitrary points from the remaining n-3. Call them E and F, and draw a line though them; call this line l. If l does not intersect ABC, then ABCEF is convex and you choose the quadrilateral ABEF.
The other case is that it separates ABC such that one of A,B,C is one one side of l and the other two are on the other side. Without loss of generality, A is on one side of l while B and C are on the other side. Consider the quadrilateral BCEF (or CBEF, depending on orientation.) We know that BC are on the same side of EF. Also, since no points are in the third quadrant, we know that EF are on the same side of BC. Now we can choose either BCEF or CBEF to be our convex quadrilateral.
So we now have that for the given ABC, all pairs of the other n points give a distinct convex quadrilateral. So we have n-3 points to choose from, and we must choose two of them. So this gives us at least C(n-3,2) = $\frac{(n-3)(n-4)}{2}$ convex quadrilaterals.
The above solution was posted and copyrighted by Philip_Leszczynski. The original thread for this problem can be found here: [1]


In the tetrahedron $ABCD$, angle $BDC$ is a right angle.  Suppose that the foot $H$ of the perpendicular from $D$ to the plane $ABC$ in the tetrahedron is the intersection of the altitudes of $\triangle ABC$.  Prove that
For what tetrahedra does equality hold?

Solution:
Let us show first that angles $ADB$ and $ADC$ are also right. Let $H$ be the intersection of the altitudes
of $ABC$ and let $CH$ meet $AB$ at $X$. Planes $CED$ and $ABC$ are perpendicular and $AB$ is perpendicular to
the line of intersection $CE$. Hence $AB$ is perpendicular to the plane $CDE$ and hence to $ED$. So $BD^2 = DE^2 + BE^2.$ Also $CB^2 = CE^2 + BE^2.$ Therefore $CB^2 - BD^2 = CE^2 - DE^2.$ But $CB^2 - BD^2 = CD^2,$ so $CE^2 = CD^2 + DE^2$, so angle $CDE = 90^{\circ}$. But angle $CDB = 90^{\circ}$, so $CD$ is perpendicular to
the plane $DAB$, and hence angle $CDA$ = $90^{\circ}$. Similarly, angle $ADB = 90^{\circ}$.
Hence $AB^2 + BC^2 + CA^2 = 2(DA^2 + DB^2 + DC^2)$. But now we are done, because Cauchy's
inequality gives $(AB + BC + CA)^2 = 3(AB^2 + BC^2 + CA^2).$ We have equality if and only if
we have equality in Cauchy's inequality, which means $AB = BC = CA.$


Prove that for every natural number $m$; there exists a finite set $S$ of points in a plane with the following property: For every point $A$ in $S$; there are exactly $m$ points in $S$ which are at unit distance from $A$.

I shall prove a more general statement about the unit distance graph($V=\mathbb{R}^2$, adjacency iff the Euclidean distance between the points is $1$) and this will follow as a consequence.
if $G,H$ occur as unit distance graphs, then so does $G\times H$( here $G\times H$ is described as $V(G\times H)=V(G)\times V(H), (v_1,w_1)\leftrightarrow (v_2,w_2) \Leftrightarrow v_1=v_2,w_1\leftrightarrow w_2$ or $v_1\leftrightarrow v_2,w_1=w_2$).
this is seen by describing the vertices by complex numbers. suppose there is an embedding of $G$ by the complex numbers $v_1,v_2\ldots,v_n$ and for $H$ by the numbers $w_1,w_2,\ldots, w_m$. we claim that for some choice of $0\leq\theta<2\pi$, $v_i+e^{i\theta}w_j$ will do the job(a suitable rotation).
what we need is that $(v_i+e^{i\theta}w_j)\leftrightarrow (v_k+e^{i\theta}w_l)$ iff either $v_i=v_k,|w_j-w_l|=1$ or $w_j=w_l,|v_i-v_k|=1$. clearly if the condition holds then the adjacency is satisfied. suppose $i\neq k,j\neq l$ and that the corresponding complex numbers are at a distance $1$ from one another. Then this gives a quadratic in $e^{i\theta}$ and hence $\theta$ can take only finitely many values here.ruling this out for each such set of $i,j,k,l$ hence rules out finitely many values of $\theta$ and therefore a suitable choice exists.
for the given problem we need a unit distance graph which is regular of degree $m$ for any given $m$. since we can form the graph $K_2$, we can form $Q_n=Q_{n-1}\times K_2$(the unit cube) and that solves the problem.
This solution was posted and copyrighted by seshadri. The original thread for this problem can be found here: [1]

Suppose $S$ has the formwhere $V_m=\{\mathbf v_1,\mathbf v_2,\dots,\mathbf v_m\}$ is unknown set of distinct unit vectors in $\mathbb R^2$. We can build $V_m$ inductively, starting from the empty set and adding vectors to it, one by one. We just need to make sure that two thing are respected:
1. All $2^m$ vectors in $S_m$ are distinct;
2. Two vector sums are at unit distance from one another if and only if they differ in presence of exactly one summand (i.e. one and only one coefficient $b_i$ in the sum changes from $0$ to $1$).
If these two conditions are kept, then each of $2^m$ points at $S_m$ will be at unit distance from exactly $m$ points corresponding to sums at which one and only one of $m$ coefficients differs from coefficients of this point. However, respecting these conditions is not hard because $\left\|\sum_{\mathbf v \in U_1} \mathbf v - \sum_{\mathbf v \in U_2} \mathbf v\right\|=\left\|\sum_{\mathbf v \in U_1 \setminus U_2} \mathbf v \,-\! \sum_{\mathbf v \in U_2 \setminus U_1} \mathbf v\right\|$ and for each new vector being added to $V_m$ there is at most some finite set of forbidden endpoints given by sums/differences of already determined vectors but the rest of the (infinite) unit circle is permissible.
This solution was posted and copyrighted by Bandera. The original thread for this problem can be found here: [2]


Determine all possible values of  where $a, b, c, d,$ are arbitrary positive numbers.

Solution:
Note that  We will now prove that $S$ can reach any range in between $1$ and $2$.
Choose any positive number $a$. For some variables such that $k, m, l > 0$ and $k + m + l = 1$, let $b = ak$, $c = am$, and $d = al$. Plugging this back into the original fraction, we get

The above equation can be further simplified to

Note that $S$ is a continuous function and that $f(m) = m + \frac{1}{2-m}$ is a strictly increasing function. We can now decrease $k$ and $l$ to make $m$ tend arbitrarily close to $1$. We see $\lim_{m\to1} m + \frac{1}{2-m} = 2$, meaning $S$ can be brought arbitrarily close to $2$.
Now, set $a = d = x$ and $b = c = y$ for some positive real numbers $x, y$. Then

Notice that if we treat the numerator and denominator each as a quadratic in $y$, we will get $1 + \frac{g(x)}{2y^2 + 5xy + 2x^2}$, where $g(x)$ has a degree lower than $2$. This means taking $\lim_{y\to\infty} 1 + \frac{g(x)}{2y^2 + 5xy + 2x^2} = 1$, which means $S$ can be brought arbitrarily close to $1$. Therefore, we are done.

~Imajinary


Determine, with proof, whether or not one can find $1975$ points on the circumference of a circle with unit radius such that the distance between any two of them is a rational number.

Solution:
Since there are infinitely many primitive Pythagorean triples, there are infinitely many angles $\theta$ s.t. $\sin\theta, \cos\theta$ are both rational. Call such angles good. By angle-sum formulas, if $a,b$ are good, then $a+b,a-b$ are also good.
For points $A,B$ on the circle $\omega$, let $\angle AB$ be the angle subtended by $AB$. Now inductively construct points on $\omega$ s.t. all angles formed by them are good; for 1,2 take any good angle. If there are $n$ points chosen, pick a good angle $\theta$ and a marked point $A$ s.t. the point $B$ on $\omega$ with $\angle AB=\theta$ is distinct from the $n$ points. Since there are infinitely many good angles but finitely many marked points, such $\theta$ exists. For a previously marked point $P$ we have $\angle BP=\pm \angle AP\pm \angle AB$ for suitable choices for the two $\pm$. Since $\angle AP ,\angle AB$ are both good, it follows that $\angle BP$ is good, which finishes induction by adding $B$.
Observe that these points for $n=1975$ work: since $AB=27sin\angle AB$ for $A,B$ on the circle, it follows that $AB$ is rational, and so we're done.
The above solution was posted and copyrighted by tobash_co. The original thread for this problem can be found here: [1]


We consider the following system
with $q = 2p$:
$\begin{matrix} a_{11}x_{1} + \ldots + a_{1q}x_{q} = 0, \\ a_{21}x_{1} + \ldots + a_{2q}x_{q} = 0, \\ \ldots , \\ a_{p1}x_{1} + \ldots + a_{pq}x_{q} = 0, \\ \end{matrix}$
in which every coefficient is an element from the set $\{ - 1,0,1\}$$.$ Prove that there exists a solution $x_{1}, \ldots,x_{q}$ for the system with the properties:
a.) all $x_{j}, j = 1,\ldots,q$ are integers$;$
b.) there exists at least one j for which $x_{j} \neq 0;$
c.) $|x_{j}| \leq q$ for any $j = 1, \ldots ,q.$

Solution:
First of all note that we have $(q + 1)^q - 1$ possible nonzero vectors $(x_1,\cdots,x_q)$ such that $0\leq x_i\leq q$ are integers.
But $a_{j1}x_1 + \cdots + a_{jq}x_q$ can only assume $q^2 + 1$ different values, because if it is maximized/minimized by $(M_1,M_2,\cdots,M_q)/(m_1,m_2,\cdots,m_q)$, we have that $\sum_{i = 1}^{q}a_{ji}(M_i - m_i)\leq q\times q = q^2$ (if $a_{ji} = 0$, it doesn't affect the sum, if it is $1$, $M_i = q,m_i = 0$, and if it is $- 1$, $M_i = 0,m_i = q$).
From this we conclude that there are at most $(q^2 + 1)^p = (q^2 + 1)^{q/2}$ possible values for the vector $(a_{11}x_{1} + \ldots + a_{1q}x_{q},\cdots,a_{p1}x_{1} + \ldots + a_{pq}x_{q})$.
But we have that:
$(q + 1)^q - 1 = (q^2 + 1 + 2q)^{q/2} - 1 =$
$= (q^2 + 1)^{q/2} + \left( - 1 + \sum_{j = 1}^{q/2}{{q/2}\choose j}(q^2 + 1)^{q/2 - j}(2q)^j\right) > (q^2 + 1)^{q/2}$
We conclude that by the pigeonhole principle there are two distinct vectors being mapped to the same vector. Taking their difference we have a vector with the desired properties.
The above solution was posted and copyrighted by Jorge Miranda. The original thread for this problem can be found here: [1]


Let $a,b$ be two natural numbers. When we divide $a^2+b^2$ by $a+b$, we the the remainder $r$ and the quotient $q.$ Determine all pairs $(a, b)$ for which $q^2 + r = 1977.$

Solution:
Using $r=1977-q^2$, we have $a^2+b^2=(a+b)q+1977-q^2$, or $q^2-(a+b)q+a^2+b^2-1977=0$, which implies $\Delta=7908+2ab-2(a^2+b^2)\ge 0$. If we now assume Wlog that $a\ge b$, it follows $a+b\le 88$. If $q\le 43$, then $r=1977-q^2\ge 128$, contradicting $r<a+b\le 88$. But $q\le 44$ from $q^2+r=1977$, thus $q=44$. It follows $r=41$, and we get $a^2+b^2=44(a+b)+41\Leftrightarrow (a-22)^2+(b-22)^2=1009\in \mathbb{P}$. By Jacobi's two squares theorem, we infer that $15^2+28^2=1009$ is the only representation of $1009$ as a sum of squares. This forces $\boxed{(a,b)=(37,50) , (7, 50)}$, and permutations. $\blacksquare$
The above solution was posted and copyrighted by cobbler. The original thread for this problem can be found here: [1]


Let $f$ be an injective function from ${1,2,3,\ldots}$ in itself. Prove that for any $n$ we have: $\sum_{k=1}^{n} f(k)k^{-2} \geq \sum_{k=1}^{n} k^{-1}.$

Solution:
We know that all the unknowns are integers, so the smallest one must greater or equal to 1.
Let me denote the permutations of $(k_1,k_2,...,k_n)$ with $(y_1,y_2,...,y_n)=y_i (*)$.
From the rearrangement's inequality we know that $\text{Random Sum} \geq \text{Reversed Sum}$.
We will denote we permutations of $y_i$ in this form $y_n \geq ...\geq y_1$.
So we have $\frac{k_1}{1^2}+\frac{k_2}{2^2}+...+\frac{k_n}{n^2} \geq \frac{y_1}{1^2}+ \frac{y_2}{2^2}+...+ \frac{y_n}{n^2} \geq 1+\frac{1}{2}+...+\frac{1}{n}$.
Let's denote $\frac{y_1}{1^2}+ \frac{y_2}{2^2}+...+ \frac{y_n}{n^2}=T$ and $1+\frac{1}{2}+...+\frac{1}{n}=S$.
We have $T \geq S$. Which comes from $y_1 \geq1, y_2 \geq2, ...,y_n \geq n$.
So we are done.
The above solution was posted and copyrighted by Davron. The original thread for this problem can be found here: [1]


Determine all real numbers a for which there exists non-negative reals $x_{1}, \ldots, x_{5}$ which satisfy the relations $\sum_{k=1}^{5} kx_{k}=a,$ $\sum_{k=1}^{5} k^{3}x_{k}=a^{2},$ $\sum_{k=1}^{5} k^{5}x_{k}=a^{3}.$

Solution:
Let $\Sigma_1= \sum_{k=1}^{5} kx_{k}$, $\Sigma_2=\sum_{k=1}^{5} k^{3}x_{k}$ and $\Sigma_3=\sum_{k=1}^{5} k^{5}x_{k}$. For all pairs $i,j\in \mathbb{Z}$, letThen we have on one handTherefore \\(1)and on the other hand \\ (2)Then from (1) we haveand from (2)so $a\in [0,25]$ Besides we also have from (1)and from (2)and for $n=1,2,3,4$where in the right hand we have that, so,and, sofor $n=1,2,3,4$ From the latter and (2) we also haveSo we have that
If $a=k^2$, $k=0,1,2,3,4,5$ take $x_k=k$, $x_j=0$ for $j\neq k$. Then $\Sigma_1=k^2=a$, $\Sigma_2=k^3k=k^4=a^2$, and $\Sigma_3=k^5k=k^6=a^3$


Three congruent circles have a common point $O$ and lie inside a given triangle. Each circle touches a pair of sides of the triangle. Prove that the incenter and the circumcenter of the triangle and the point $O$ are collinear.

Let the triangle have vertices $A,B,C$, and sides $a,b,c$, respectively, and let the centers of the circles inscribed in the angles $A,B,C$ be denoted $O_A, O_B, O_C$, respectively.
The triangles $O_A O_B O_C$ and $ABC$ are homothetic, as their corresponding sides are parallel.  Furthermore, since $O_A$ lies on the  bisector of angle $A$ and similar relations hold for the triangles' other corresponding points, the center of homothety is the incenter of both the triangles.  Since $O$ is clearly the circumcenter of $O_A O_B O_C$, $O$ is collinear with the incenter and circumcenter of $ABC$, as desired.

Suppose 3 congruent circles with centres P,Q,R lie inside ABC and are such that the circle with centre P touches AB & AC and the circle with centre Q touches CA & BC.an R with remaining 2.
Since O lies in all 3 circles, PO=QO=RO. Therefore, O is circumcentre of PQR. let O' be circumcentre of ABC.
Since BC is tangent to the circles with centers Q & R, the lengths of perpendiculars from Q & R, the lengths are equal. therefore,   QR//BC,RP//CA,PQ//AB.
Again, since AB and AC both touch circle with centre P. Therefore P is equidistant from AB & AC. Therefore P lies on the internal bisector of angle A. Similarly Q & R lie internal bisectors of angle B and angle C respectively. Therefore, AP,BQ,CR produced meet at incenter I. Since, QR//BC,RP//CA,PQ//AB, it follows that I is also incentre of PQR, I being the centre of homothety. By the property of enlargements, O and O' must be co-linear with I , the centre of enlargement.


The diagonals $AC$ and $CE$ of the regular hexagon $ABCDEF$ are divided by inner points $M$ and $N$ respectively, so thatDetermine $r$ if $B,M$ and $N$ are collinear.

O is the center of the regular hexagon. Then we clearly have $ABC\cong COA\cong EOC$. And therefore we have also obviously $ABM\cong AOM\cong CON$, as $\frac{AM}{AC} =\frac{CN}{CE}$.
So we have $\angle{BMA} =\angle{AMO} =\angle{CNO}$ and $\angle{NOC} =\angle{ABM}$. Because of $\angle{AMO} =\angle{CNO}$ the quadrilateral $ONCM$ is cyclic. $\Rightarrow \angle{NOC} =\angle{NMC} =\angle{BMA}$. And as we also have $\angle{NOC} =\angle{ABM}$ we get $\angle{ABM} =\angle{BMA}$. $\Rightarrow AB=AM$. And as $AC=\sqrt{3} \cdot AB$ we get $r=\frac{AM}{AC} =\frac{AB}{\sqrt{3} \cdot AB} =\frac{1}{\sqrt{3}}$.
This solution was posted and copyrighted by Kathi. The original thread for this problem can be found here: [1]

Let $X$ be the intersection of $AC$ and $BE$. $X$ is the mid-point of $AC$.
Since $B$, $M$, and $N$ are collinear, then by Menelaus Theorem,
$\frac{CN}{NE}\cdot\frac{EB}{BX}\cdot\frac{XM}{MC}=1$.
Let the sidelength of the hexagon be $1$. Then $AC=CE=\sqrt{3}$.
$\frac{CN}{NE}=\frac{CN}{CE-CN}=\frac{\frac{CN}{CE}}{1-\frac{CN}{CE}}=\frac{r}{1-r}$
$\frac{EB}{BX}=\frac{2}{\frac{1}{2}}=4$
$\frac{XM}{MC}=\frac{AM-AX}{AC-AM}=\frac{\frac{AM}{AC}-\frac{AX}{AC}}{1-\frac{AM}{AC}}=\frac{r-\frac{1}{2}}{1-r}$
Substituting them into the first equation yields
$\frac{r}{1-r}\cdot\frac{4}{1}\cdot\frac{r-\frac{1}{2}}{1-r}=1$
$3r^2=1$
$\therefore r=\frac{\sqrt{3}}{3}$
This solution was posted and copyrighted by leepakhin. The original thread for this problem can be found here: [2]

Note $x=m(\widehat {EBM})$. From the relation $r=\frac{AM}{AC}=\frac{CN}{CE}$ results $\frac{MA}{MC}=\frac{NC}{NE}$, i.e.
$\frac{BA}{BC}\cdot \frac{\sin (60+x)}{\sin (60-x)}=\frac{BC}{BE}\cdot \frac{\sin (60-x)}{\sin x}$. Thus, $2\sin x\sin (60+x)=\sin^2(60-x)\Longrightarrow$
$2[\cos 60-\cos (60+2x)]=1-\cos (120-2x)\Longrightarrow \cos (60+2x)=0\Longrightarrow x=15^{\circ}.$
Therefore, $\frac{MA}{MC}=\frac{\sin 75}{\sin 45}=\frac{1+\sqrt 3}{2}$, i.e. $r=\frac{MA}{AC}=\frac{1+\sqrt 3}{3+\sqrt 3}=\frac{1}{\sqrt 3}\Longrightarrow r=\frac{\sqrt 3}{3}.$
This solution was posted and copyrighted by Virgil Nicula. The original thread for this problem can be found here: [3]


Is it possible to choose $1983$ distinct positive integers, all less than or equal to $10^5$, no three of which are consecutive terms of an arithmetic progression? Justify your answer.

Solution:
The answer is yes. We will construct a sequence of integers that satisfies the requirements.
Take the following definition of $a_n$: $a_n$ in base 3 has the same digits as $n$ in base 2. For example, since $6=110_2$, $a_6=110_3=12$.
First, we will prove that no three $a_n$'s are in an arithmetic progression. Assume for sake of contradiction, that $i < j < k$ are numbers such that $2a_j=a_i + a_k$. Consider the base 3 representations of both sides of the equation. Since $a_j$ consists of just 0's and 1's in base three, $2a_j$ consists of just 0's and 2's base 3. No whenever $2a_j$ has a 0, both $a_i$ and $a_k$ must have a 0, since both could only have a 0 or 1 in that place. Similarly, whenever $2a_j$ has a 2, both $a_i$ and $a_k$ must have a 1 in that place. This means that $a_i=a_k$, which is a contradiction.
Now since $1983=11110111111_2$, $a_{1983} = 11110111111_3 < \frac{3^{11}}{2} < 10^5$. Hence, the set $\{a_1,a_2,...a_{1983}\}$ is a set of 1983 integers less than $10^5$ with no three in arithmetic progression.


Let $d$ be the sum of the lengths of all the diagonals of a plane convex polygon with $n$ vertices (where $n>3$). Let $p$ be its perimeter. Prove that:

where $[x]$ denotes the greatest integer not exceeding $x$.

Solution:
Consider all the pairs of non-adjacent sides in the polygon. Each pair uniquely determines two diagonals which intersect in an interior point. The sum of these two diagonals is greater than the sum of the two sides (triangle inequality). Each side will appear in a pair $n-3$ times. And each time the side appears it also includes two diagonals. Thus, $(n-3)p < 2d$.
Note that for any diagonal, its length is less than the sum of the sides on either side of it (triangle inequality). Consider all the diagonals from a fixed vertex and using the least number of sides in each application of the triangle inequality. We can sum everything up over all the diagonals considering the two cases (even or odd) to get the result.
This solution was posted and copyrighted by aznlord1337. The original thread for this problem can be found here: [1]


A circle with center $O$ passes through the vertices $A$ and $C$ of the triangle $ABC$ and intersects the segments $AB$ and $BC$ again at distinct points $K$ and $N$ respectively. Let $M$ be the point of intersection of the circumcircles of triangles $ABC$ and $KBN$ (apart from $B$). Prove that $\angle OMB = 90^{\circ}$.

Solution: $M$ is the Miquel Point of quadrilateral $ACNK$, so there is a spiral similarity centered at $M$ that takes $KN$ to $AC$. Let $M_1$ be the midpoint of $KA$ and $M_2$ be the midpoint of $NC$. Thus the spiral similarity must also send $M_1$ to $M_2$ and so $BMM_1 M_2$ is cyclic. $OM_1 B M_2$ is also cyclic with diameter $BO$ and thus $M$ must lie on the same circumcircle as $B$, $M_1$, and $M_2$ so $\angle OMB = 90^{\circ}$.

Let $\Omega, \Omega', \omega$ and $O,O',O''$ be the circumcircles and circumcenters of $AKNC, ABC, BNKM,$ respectively.
Let $\angle ACB = \gamma, AKNC$ is cyclic $\implies \angle BKN = \gamma.$
The radius of $\omega$ is $MO'' = BO'' = \frac {BN}{2 \sin \gamma}.$
Let $D$ and $E$ be midpoints of $BC$ and  $NC$ respectively.
$OE \perp BC, OD \perp BC, OO' \perp AC, DE = \frac {BC}{2} - \frac {NC}{2} = \frac {BN}{2}$
$\implies OO' = \frac {DE}{\sin \gamma} = \frac {BN}{2 \sin \gamma} = MO''.$
$M$ is the Miquel Point of quadrilateral $ACNK,$ so  $MO''O'O$ is cyclic.
$MO''O'O$ is trapezium $\implies O''O' || MO.$ $O''O' \perp BM \implies MO\perp BM$ as desired.
vladimir.shelomovskii@gmail.com, vvsss

Consider $\triangle MKA$ and $\triangle MNC$, they are similar because $\angle MAK$ = $\angle MCN$, and also $\angle MKA = \angle MNC$.
Now draw $OP \perp AB$, and intersecting $AB$ at $P$; $OQ \perp BC$, at $Q$. Naturally $OP$ bisects $AK$, and $OQ$ bisects $CN$. We claim $\triangle MAP \sim \triangle MCQ$, because
$\frac {AP}{CQ} = \frac {AK}{CN} = \frac {AM}{CB}.$
Thus $\angle AMP = \angle CMQ$, this implies $\angle PMQ = \angle AMC = \angle ABC = \angle PBQ$. Obviously BMPQ is cyclic, and so is BPOQ. Finally, we have $OM \perp MB$. (by gougutheorem)


Find all (if any) functions $f$ taking the non-negative reals onto the non-negative reals, such that
(a) $f(xf(y))f(y) = f(x+y)$ for all non-negative $x$, $y$;
(b) $f(2) = 0$;
(c) $f(x) \neq 0$ for every $0 \leq x < 2$.

For all $x+y\ge2$, there exists a nonnegative real $t$ such that $t+2=x+y$ and $f(x+y)=f(t+2)=f(tf(2))f(2)=0$. Hence, $f(x)=0$ for all $x\ge 2$.
For $x<2$, take $0=f(2)=f((2-x)+(x))=f((2-x)f(x))f(x)$.
Since $f(x)\neq0$ for $x<2$, then $f((2-x)f(x))=0$.
Thus, $(2-x)f(x)\ge2\Leftrightarrow f(x)\ge\frac{2}{2-x}$ for all $x<2$.
Suppose $f(a)>\frac{2}{2-a}$ so that $f(a)=\frac{2}{2-a-\epsilon}$ for some $\epsilon>0$ and $0\le a<2$.
Then $f\left((2-a-\epsilon)f(a)\right)f(a)=f(2-\epsilon)$. The left hand side is equivalent to $f\left((2-a-\epsilon)\left(\frac{2}{2-a-\epsilon}\right)\right)f(a)=f(2)=0$, but the right hand side is nonzero, since $2-\epsilon<2$ and $f(x)\neq0$ for all $x<2$.
Hence, $f(x)=\frac{2}{2-x}$ for $0\le a<2$.
For $x+y<2$,\begin{eqnarray*}f(xf(y))f(y) &=& f\left((x)\left(\frac{2}{2-y}\right)\right)\left(\frac{2}{2-y}\right)\\ &=&\left(\frac{2}{2-\frac{2x}{2-y}}\right)\left(\frac{2}{2-y}\right)\\ &=&\frac{4}{4-2y-2x}\\ &=&\frac{2}{2-y-x}\\ f(x+y)&=&\frac{2}{2-y-x}.\end{eqnarray*}
Indeed, $f(xf(y))f(y)=f(x+y)$, so the desired function is
This solution was posted and copyrighted by towersfreak2006. The original thread for this problem can be found here: [1]

Taking $y=2$ gives $f(x+2)=0 \forall x\ge 0$, or equivalently $f(x)=0  \forall x\ge 2$. Taking $y=2-x$ for $0\le x<2$ yieldsSince $0<2-x\le 2$, we must have $f(xf(2-x))=0$, which forcesNow taking $x=2/f(y)$, for $0\le y<2$ gives,Thus,Hence, this forces $f(x)=2/(2-x)$ for $0\le x<2$, giving a final answer of
This solution was posted and copyrighted by winnertakeover. The original thread for this problem can be found here: [2]

Let $P(x,y)$ be the assertion. Note that $P(x,2)\implies f(x+2)=0.$
Since $x$ is defined for non-negative reals, we have $x+2\geq 2.$ So, $\boxed{f(x)=0\qquad \forall x\geq2}.$
Now, $P(2-y,y)\implies f((2-y)f(y))f(y)=0.$
Here, we are working with $0\leq y<2.$ Since we may not have $f(y)=0$ due to condition (iii), we must have $f((2-y)f(y))=0.$
From our first solution, it is imperative that $(2-y)f(y)\geq 2.$ So, $f(y)\geq \dfrac{2}{2-y} \qquad (1).$
Now consider $P(\dfrac{2}{f(y),y}$. We get,Similar to the previous step, we have $\dfrac{2+yf(y)}{f(y)}\geq 2.$ Re arranging this, we get $f(y)\leq \dfrac{2}{2-y}\qquad (2).$
Since $f(y)$ satisfies both $(1)$ and $(2),$ we must have $\boxed{f(x)=\dfrac{2}{2-x}\qquad \forall x\in [0,2)}.$
Of course, we plug both solution back to see if they satisfy the three conditions. Sure enough, they do!
This solution was posted and copyrighted by proshi. The original thread for this problem can be found here: [3]


Let $n$ be an integer greater than or equal to 3.  Prove that there is a set of $n$ points in the plane such that the distance between any two points is irrational and each set of three points determines a non-degenerate triangle with rational area.

Solution:
Consider the set of points $S = \{ (x,x^2) \mid 1 \le x \le n , x \in \mathbb{N} \}$ in the $xy$-plane.
The distance between any two distinct points $(x_1,x^2_1)$ and $(x_2,x^2_2)$ in $S$ (with $x_1 \neq x_2$) is:
$d = \sqrt{(x_1-x_2)^2+\left(x^2_1-x^2_2\right)^2} =$ $\sqrt{(x_1-x_2)^2+(x_1-x_2)^2(x_1+x_2)^2} = |x_1-x_2|\sqrt{1+(x_1+x_2)^2}$.
Since $1+(x_1+x_2)^2$ is an integer and not a perfect square, $\sqrt{1+(x_1+x_2)^2}$ is irrational.
Since $|x_1-x_2|$ is a non-zero integer, $d$ is irrational as desired.
All the points in $S$ lie on the parabola $y = x^2$. Thus, it is impossible of any set of three points to be collinear, since no line can intersect a parabola at more than two points. Therefore, any triangle with all vertices in $S$ must be non-degenerate as desired.
Since all the points in $S$ are lattice points, by Pick's Theorem, the area of any triangle with all vertices in $S$ must be in the form $A = I + \dfrac{B}{2} - 1$ where $I$ and $B$ are integers. Thus, the area of the triangle must be rational as desired.
This completes the proof.


In a right-angled triangle $ABC$ let $AD$ be the altitude drawn to the hypotenuse and let the straight line joining the incentres of the triangles $ABD, ACD$ intersect the sides $AB, AC$ at the points $K,L$ respectively. If $E$ and $E_1$ dnote the areas of triangles $ABC$ and $AKL$ respectively, show that

Solution:
Lemma: Through the incenter $I$ of $\triangle{ABC}$ draw a line that meets the sides $AB$ and $AC$ at $P$ and $Q$, then:

Proof of the lemma:
Consider the general case: $M$ is any point on side $BC$ and $PQ$ is a line cutting AB, AM, AC at P, N, Q. Then:
$\frac{AM}{AN}=\frac{S_{APMQ}}{\triangle{APQ}}=\frac{\triangle{APM}+\triangle{AQM}}{\triangle{PQA}}=\frac{\frac{AP}{AB}\triangle{ABM}+\frac{AQ}{AC}\triangle{ACM}}{\frac{AP\cdot AQ}{AB \cdot AC}}=$
$=\frac{AC}{AQ}\cdot \frac{BM}{BC}+\frac{AB}{AP}\cdot \frac{CM}{BC}$
If $N$ is the incentre then $\frac{AM}{AN}=\frac{AB+BC+CA}{AB+AC}$, $\frac{BM}{BC}=\frac{AB}{AB+AC}$ and $\frac{CM}{BC}=\frac{AC}{AC+AB}$. Plug them in we get:
Back to the problem
Let $I_1$ and $I_2$ be the areas of $\triangle{ABD}$ and $\triangle{ACD}$ and $E$ be the intersection of $KL$ and $AD$. Thus apply our formula in the two triangles we get:

and

Cancel out the term $\frac{AD}{AE}$, we get:

So we conclude $AK=AL$.
Hence $\angle{AKI_1}=45^o=\angle{ADI_1}$ and $\angle{ALI_2}=45^o=\angle{ADI_2}$, thus $\triangle{AK_1} \cong \triangle{ADI_1}$ and $\triangle{ALI_2} \cong \triangle{ADI_2}$. Thus $AK=AD=AL$. So the area ratio is:
This solution was posted and copyrighted by shobber. The original thread for this problem can be found here: [1]


Prove that for each positive integer $n$ there exist $n$ consecutive positive integers none of which is an integral power of a prime number.

There are at most $1+\sqrt[2]{n}+\sqrt[3]{n}+\sqrt[4]{n}+...+\sqrt[\left\lfloor \log_2(n)\right\rfloor]{n} \leq 1+ \sqrt n log_2(n)$ 'true' powers $m^k , k\geq 2$ in the set $\{1,2,...,n\}$. So when $p(n)$ gives the amount of 'true' powers $\leq n$ we get that $\lim_{n \to \infty} \frac{p(n)}{n} = 0$.
Since also $\lim_{n \to \infty} \frac{\pi(n)}{n} = 0$, we get that $\lim_{n \to \infty} \frac{p(n)+\pi(n)}{n} = 0$. Now assume that there is no 'gap' of lenght at least $k$ into the set of 'true' powers and the primes. Then this would give that $\frac{p(n)+\pi(n)}{n} \geq \frac{1}{k}$ for all $n$ in contrary to the above (at east this proves a bit more).
Edit: to elementarize the $\lim_{n \to \infty} \frac{\pi(n)}{n} = 0$ part:
Look $\mod (k+1)!$. Then all numbers in the residue classes $2,3,4,...,k+1$ are not primes (except the smallest representants sometimes). So when there wouldn't exist a gap of length $k$, there has to be a 'true' power in each of these gaps of the prime numbers, so at least one power each $(k+1)!$ numbers, again contradicting $\lim_{n \to \infty} \frac{p(n)}{n} = 0$.
This solution was posted and copyrighted by ZetaX. The original thread for this problem can be found here: [1]

By Chinese Remainder theorem, there exists $x$ such that:
$x \equiv -1\;mod\;p_1 q_1\\  x \equiv -2\;mod\;p_2 q_2\\ x \equiv -3\;mod\;p_3 q_3\\ ...\newline x \equiv -n\;mod\;p_n q_n$
Where $p_1, p_2, ..., p_n, q_1, q_2, ..., q_n$ are distinct primes.
The n consecutive numbers $x+1, x+2, ..., x+n$ each have at least two prime factors, so none of them can be expressed as an integral power of a prime.


Given an initial integer $n_{0}\textgreater1$, two players, $\mathbb{A}$ and $\mathbb{B}$, choose integers $n_{1}, n_{2},n_{3}$, . . . alternately according to the following rules:
Knowing $n_{2k}$, $\mathbb{A}$ chooses any integer $n_{2k+1}$ such that
.
Knowing $n_{2k+1}$, $\mathbb{B}$ chooses any integer $n_{2k+2}$ such that

is a prime raised to a positive integer power.
Player $\mathbb{A}$ wins the game by choosing the number 1990; player $\mathbb{B}$ wins by choosing
the number 1. For which $n_{0}$ does:
(a) $\mathbb{A}$ have a winning strategy?
(b) $\mathbb{B}$ have a winning strategy?
(c) Neither player have a winning strategy?

Solution:
If $n_0<6$ it is clear that B wins, because A can only choose numbers with at most 2 prime factors (30 is the smallest with three and $30>5^2$), which B can either make smaller (choose the lesser of the two) or the number contains one prime factor in which case B wins. Once $n_{2k}=2$, A can only choose prime powers, so B wins.
If $n_0=6,7$ it's a draw, because the only numbers that give A a chance of winning are those with three prime factors (for reasons discussed above). Less than 49, this gives 30 and 42 as the only choices which won't cause A to lose. However, if A chooses 30, B may choose 6, and we have a stalemate. If A chooses 42, B may choose 6, and we still have a stalemate.
If $n_0 \geq 8$, if $n_0$ is less than 45 by choosing (perhaps starting later) 3*4*5, then 3*5*7, then 2*3*5*7, and then 3*4*5*7 A can force B to make $n_{2k} \geq 60$ for some k.
If $45 \leq n_{2k} \leq 1990$ for some k, then A wins (he may pick 1990). If $n_{2k}>1990$ pick the least natural value of m such that $2^m 47\geq n_{2k}$. The best B can do is half it, taking it to $n_{2k+2} = 2^{m-1} 47 < n_{2k}$. However, B cannot make $n_{2k+2}$ drop below 45 because $47*32<1990 \Rightarrow 2^m \geq 64$. Thus we will get a descending sequence of integers until we end up with $45 \leq n_{2k} \leq 1990$, at which point A wins.
This solution was posted and copyrighted by Ilthigore. The original thread for this problem can be found here: [1]


Let $\,ABC\,$ be a triangle and $\,P\,$ an interior point of $\,ABC\,$. Show that at least one of the angles $\,\angle PAB,\;\angle PBC,\;\angle PCA\,$ is less than or equal to $30^{\circ }$.

Let $A_{1}$ , $A_{2}$, and $A_{3}$ be $\angle CAB$, $\angle ABC$, $\angle BCA$, respectively.
Let $\alpha_{1}$ , $\alpha_{2}$, and $\alpha_{3}$ be $\angle PAB$, $\angle PBC$, $\angle PCA$, respcetively.
Using law of sines on $\Delta PAB$ we get: $\frac{\left| PA \right|}{sin(A_{2}-\alpha_{2})}=\frac{\left| PB \right|}{sin(\alpha_{1})}$, therefore, $\frac{\left| PA \right|}{\left| PB \right|}=\frac{sin(A_{2}-\alpha_{2})}{sin(\alpha_{1})}$
Using law of sines on $\Delta PBC$ we get: $\frac{\left| PB \right|}{sin(A_{3}-\alpha_{3})}=\frac{\left| PC \right|}{sin(\alpha_{2})}$, therefore, $\frac{\left| PB \right|}{\left| PC \right|}=\frac{sin(A_{3}-\alpha_{3})}{sin(\alpha_{2})}$
Using law of sines on $\Delta PCA$ we get: $\frac{\left| PC \right|}{sin(A_{1}-\alpha_{1})}=\frac{\left| PA \right|}{sin(\alpha_{3})}$, therefore, $\frac{\left| PC \right|}{\left| PA \right|}=\frac{sin(A_{1}-\alpha_{1})}{sin(\alpha_{3})}$
Multiply all three equations we get:
$\frac{\left| PA \right|}{\left| PB \right|}\frac{\left| PB \right|}{\left| PC \right|}\frac{\left| PC \right|}{\left| PA \right|}=\frac{sin(A_{2}-\alpha_{2})}{sin(\alpha_{1})}\frac{sin(A_{3}-\alpha_{3})}{sin(\alpha_{2})}\frac{sin(A_{1}-\alpha_{1})}{sin(\alpha_{3})}$
$1=\frac{sin(A_{2}-\alpha_{2})}{sin(\alpha_{1})}\frac{sin(A_{3}-\alpha_{3})}{sin(\alpha_{2})}\frac{sin(A_{1}-\alpha_{1})}{sin(\alpha_{3})}$
$\prod_{i=1}^{3}\frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}=1$
Using AM-GM we get:
$\frac{1}{3}\sum_{i=1}^{3}\frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}\ge \sqrt[3]{\prod_{i=1}^{3}\frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}}$
$\frac{1}{3}\sum_{i=1}^{3}\frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}\ge 1$
$\sum_{i=1}^{3}\frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}\ge 3$.  [Inequality 1]
$\sum_{i=1}^{3}\frac{sin(A_{i})cos(\alpha_{i})-cos(A_{i})sin(\alpha_{i})}{sin(\alpha_{i})}\ge 3$
$\sum_{i=1}^{3}\left[ sin(A_{i})cot(\alpha_{i})-cos(A_{i})\right]\ge 3$
Note that for $0<\alpha_{i}<180^{\circ}$, $cot(\alpha_{i})$ decreases with increasing $\alpha_{i}$ and fixed $A_{i}$
Therefore, $\left[ sin(A_{i})cot(\alpha_{i})-cos(A_{i})\right]$ decreases with increasing $\alpha_{i}$ and fixed $A_{i}$
From trigonometric identity:
$sin(x)+sin(y)=2sin\left( \frac{x+y}{2} \right)cos\left( \frac{x-y}{2} \right)$,
since $-1\le cos\left( \frac{x-y}{2} \right) \le 1$, then:
$sin(x)+sin(y) \le 2sin\left( \frac{x+y}{2} \right)$
Therefore,
$sin(A_{1}-30^{\circ})+sin(A_{2}-30^{\circ}) \le 2sin\left( \frac{A_{1}+A_{2}-60^{\circ}}{2} \right)$
and also,
$sin(A_{3}-30^{\circ})+sin(30^{\circ}) \le 2sin\left( \frac{A_{3}}{2} \right)$
Adding these two inequalities we get:
$sin(30^{\circ})+\sum_{i=1}^{3}sin(A_{i}-30^{\circ})\le 2\left[ sin\left( \frac{A_{1}+A_{2}-60^{\circ}}{2} \right)+sin\left( \frac{A_{3}}{2} \right) \right]$
$\frac{1}{2}+\sum_{i=1}^{3}sin(A_{i}-30^{\circ})\le 2\left[ 2sin\left( \frac{A_{1}+A_{2}+A_{3}-60^{\circ}}{4} \right) \right]$
$\frac{1}{2}+\sum_{i=1}^{3}sin(A_{i}-30^{\circ})\le 2\left[ 2sin\left( \frac{180^{\circ}-60^{\circ}}{4} \right) \right]$
$\frac{1}{2}+\sum_{i=1}^{3}sin(A_{i}-30^{\circ})\le 4sin\left( 30^{\circ} \right)$
$\sum_{i=1}^{3}sin(A_{i}-30^{\circ})\le \frac{3}{2}$
$2\sum_{i=1}^{3}sin(A_{i}-30^{\circ})\le 3$.
$\sum_{i=1}^{3}\frac{sin(A_{i}-30^{\circ})}{sin(30^{\circ})}\le 3$. [Inequality 2]
Combining [Inequality 1] and [Inequality 2] we see the following:
$\sum_{i=1}^{3}\frac{sin(A_{i}-30^{\circ})}{sin(30^{\circ})}\le \sum_{i=1}^{3}\frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}$
This implies that for at least one of the values of $i=1$,$2$,or $3$, the following is true:
$\frac{sin(A_{i}-30^{\circ})}{sin(30^{\circ})}\le \frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}$
or
$\frac{sin(\alpha_{i})}{sin(A_{i}-\alpha_{i})}\le \frac{sin(30^{\circ})}{sin(A_{i}-30^{\circ})}$
Which means that for at least one of the values of $i=1$,$2$,or $3$, the following is true:
$\alpha_{i} \le 30^{\circ}$
Therefore, at least one of the angles $\,\angle PAB,\;\angle PBC,\;\angle PCA\,$ is less than or equal to $30^{\circ }$.
~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.

At least one of $\angle ABC, \angle BCA, \angle CAB \ge 60^\circ$. Without loss of generality, assume that $\angle BCA \ge 60^\circ$
If $\angle PAB > 30^\circ$ and $\angle PBC > 30^\circ$
Draw a circle $R$ centered at $O$ and passing through $A, P, B$. Since $P$ is an interior point of $\triangle ABC$, thus $C$ is outside the circle $R$
Draw two lines $CD, CE$ passing through $C$ and tangent to $R$. Line $CD$ intersect $R$ at $D$, and line $CE$ intersect $R$ at $E$. Choose $D$ near $A$, and choose $E$ near $B$
Extends line $BC$, and intersect $R$ at $F$ other than $B$ when $BC$ is not tangent to $R$. If $BC$ is tangent to $R$, we have $B = E$ be the tangent point, and simply let $F = B = E$
Draw the segment $OE$, and choose a point $G$ on $R$ such that $\angle GOE = 60^\circ$. There are two possible points, we choose $G$ near point $P$. Draw segments $OG, GE$, thus $\triangle GOE$ is an equilateral triangle
Draw segments $OP, OC, OB, OF, PB, GC$
$\angle OCE = \dfrac{1}{2} \angle DCE \ge \dfrac{1}{2} \angle BCA \ge 30^\circ$. Then we have $\angle COE = 90^\circ - \angle OCE \le 60^\circ = \angle GOE$
$\angle POB = 2 \angle PAB > 60^\circ, \angle POF = 2 \angle PBC > 60^\circ$, since we have either $\angle POE \ge \angle POB$ or $\angle POE \ge \angle POF$, thus $\angle POE > 60^\circ = \angle GOE$
Thus we have $\angle COE \le \angle GOE < \angle POE$, then $\angle OCE \le \angle GCE < \angle PCE$
Because $\angle GCE \ge \angle OCE \ge 30^\circ = \angle GEC$, thus $GC \le GE = OG$, and $\angle GCO \ge \angle GOC$
Finally, $\angle PCA = \angle ACE - \angle PCE < \angle ACE - \angle GCE = \angle ACO - \angle GCO$
Since $\angle ACO \le \angle DCO$, and $\angle GCO \ge \angle GOC$, thus we have $\angle PCA < \angle ACO - \angle GCO \le \angle DCO - \angle GOC = 90^\circ - \angle COE - \angle GOC = 90^\circ - \angle GOE = 30^\circ$
We have proved that when $\angle PAB > 30^\circ$ and $\angle PBC > 30^\circ$, the angle $\angle PCA$ must be less than $30^\circ$.  Thus at least one of $\angle PAB, \angle PBC, \angle PCA$ should less than or equal to $30^\circ$
~Joseph Tsai, mgtsai@gmail.com


Let $S$ be a finite set of points in three-dimensional space.  Let $S_{x}$,$S_{y}$,$S_{z}$, be the sets consisting of the orthogonal projections of the points of $S$ onto the $yz$-plane, $zx$-plane, $xy$-plane, respectively.  Prove that

where $|A|$ denotes the number of elements in the finite set $|A|$. (Note: The orthogonal projection of a point onto a plane is the foot of the perpendicular from that point to the plane)

Solution:
Let $Z_{i}$ be planes with index $i$ such that $1 \le i \le n$ that are parallel to the $xy$-plane that contain multiple points of $S$ on those planes such that all points of $S$ are distributed throughout all planes $Z_{i}$ according to their $z$-coordinates in common.
Let $a_{i}$ be the number of unique projected points from each $Z_{i}$ to the $yz$-plane
Let $b_{i}$ be the number of unique projected points from each $Z_{i}$ to the $xz$-plane
This provides the following: $|Z_{i}| \le a_{i}b_{i}\;$ [Equation 1]
We also know that $|S|=\sum_{i=1}^{n}|Z_{i}|\;$ [Equation 2]
Since $a_{i}$ be the number of unique projected points from each $Z_{i}$ to the $yz$-plane,
if we add them together it will give us the total points projected onto the $yz$-plane.
Therefore, $|S_{x}|=\sum_{i=1}^{n}a_{i}\;$ [Equation 3]
likewise, $|S_{y}|=\sum_{i=1}^{n}b_{i}\;$ [Equation 4]
We also know that the total number of elements of each $Z_{i}$ is less or equal to the total number of elements in $S_{z}$
That is, $|Z_{i}| \le |S_{z}|\;$ [Equation 5]
Multiplying [Equation 1] by [Equation 5] we get:
$|Z_{i}|^{2} \le a_{i}b_{i}|S_{z}|$
Therefore, $|Z_{i}| \le \sqrt{a_{i}b_{i}}\sqrt{|S_{z}|}$
Adding all $|Z_{i}|$ we get:
$\sum_{i=1}^{n}|Z_{i}| \le \sqrt{|S_{z}|}\sum_{i=1}^{n}\sqrt{a_{i}b_{i}}\;$[Equation 6]
Substituting [Equation 2] into [Equation 6] we get:
$|S| \le \sqrt{|S_{z}|}\sum_{i=1}^{n}\sqrt{a_{i}b_{i}}$
$|S|^{2} \le |S_{z}| \left( \sum_{i=1}^{n}\sqrt{a_{i}b_{i}} \right)^{2}$
Since, $\sum_{i=1}^{n}\sqrt{a_{i}b_{i}} \le \sqrt{\sum_{i=1}^{n}a_{i}}\sqrt{\sum_{i=1}^{n}b_{i}}$,
Then, $|S|^{2} \le |S_{z}| \left( \sqrt{\sum_{i=1}^{n}a_{i}}\sqrt{\sum_{i=1}^{n}b_{i}} \right)^{2}$
$|S|^{2} \le |S_{z}| \left( \sum_{i=1}^{n}a_{i}\right)\left( \sum_{i=1}^{n}b_{i}\right)\;$ [Equation 7]
Substituting [Equation 3] and [Equation 4] into [Equation 7] we get:
$|S|^{2} \le |S_{x}| \cdot |S_{y}| \cdot |S_{z}|$
~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 $\mathbb{N} = \{1,2,3, \ldots\}$. Determine if there exists a strictly increasing function $f: \mathbb{N} \mapsto \mathbb{N}$ with the following properties:
(i) $f(1) = 2$;
(ii) $f(f(n)) = f(n) + n, (n \in \mathbb{N})$.

Solution:
Here is my Solution https://artofproblemsolving.com/community/q2h62193p16226748
Find as ≈ Ftheftics


Let $S$ be the set of real numbers strictly greater than $-1$.  Find all functions $f:S \to S$ satisfying the two conditions:
1. $f(x+f(y)+xf(y)) = y+f(x)+yf(x)$ for all $x$ and $y$ in $S$;
2. $\frac{f(x)}{x}$ is strictly increasing on each of the intervals $-1<x<0$ and $0<x$.

Solution:
The only solution is $f(x) = \frac{-x}{x+1}.$
Setting $x=y,$ we get

Therefore, $f(s) = s$ for $s = x+f(x)+xf(x).$
Note: If we can show that $s$ is always $0,$ we will get that $x+f(x)+xf(x) = 0$ for all $x$ in $S$ and therefore, $f(x) = \frac{-x}{x+1}.$
Let $f(s)=s.$ Setting $x=y=s,$ we get

If $t=2s+s^2,$ we have $f(t)=t$ as well.
Consider $s > 0.$ We get $t = 2s + s^2 > s.$ Since $\frac{f(x)}{x}$ is strictly increasing for $x>0$ and $t > s$ in this domain, we must have $\frac{f(t)}{t} > \frac{f(s)}{s}$ but since $f(s) = s$ and $f(t) = t,$ we also have that $\frac{f(s)}{s} = 1 = \frac{f(t)}{t}$ which is a contradiction. Therefore $s \leq 0$
Consider $-1 < s < 0.$ Using a similar argument, we will get that $t = 2s + s^2 < s$ but also $\frac{f(s)}{s} = 1 = \frac{f(t)}{t},$ which is a contradiction.
Hence, $s$ must be $0.$ Since $s=0,$ we can conclude that $x+f(x)+xf(x) = 0$ and therefore, $f(x) = \frac{-x}{x+1}$ for all $x$.


Let $ABCDEF$ be a convex hexagon with $AB=BC=CD$ and $DE=EF=FA$, such that $\angle BCD=\angle EFA=\frac{\pi}{3}$. Suppose $G$ and $H$ are points in the interior of the hexagon such that $\angle AGB=\angle DHE=\frac{2\pi}{3}$. Prove that $AG+GB+GH+DH+HE\ge CF$.

Solution: Draw $AE$ and $BD$ to make equilateral $\triangle EFA$ and $\triangle BCD$, and draw points $I$ and $J$ such that $IA=IB$, $JD=JE$, directed angle $\measuredangle IAB=-\measuredangle CDB$, and directed angle $\measuredangle JDE=-\measuredangle FAE$ to make equilateral $\triangle AIB$ and $\triangle DJE$. Notice that $G$ is on the circumcircle of $\triangle AIB$ and $H$ is on the circumcircle of $\triangle DJE$. By Ptolemy, $GA+GB=GI$ and $HD+HE=HJ$. So,  Notice that octagon $AIBCDJEF$ is symmetric about $\overline{BE}$. So, $IG+GH+HJ\ge IJ=CF$. --tigerzhang