geometry
circumcircle
ratio
geometric transformation
reflection
analytic geometry
homothety

$A_0D\cap(I)$ is the tangency point of the circle passing through B,C tangent to the incircle.
Then we consider the Apollonian circles with the ratio s-a/s-b,etc. It's easy to prove they intersect at exactly 2 points.
The concurrency of $A_0D,B_0E,C_0F$ can be proved by radical axis.
The remaining is to show that I,O have the same power wrt the 3 Apollonian circles. For I it's easy; for O it's simple computation.

I know I have over-compressed the solution... :blush:

You can find an alternative solution in [url=http://groups.yahoo.com/group/Hyacinthos/message/9551?expand=1]Hyacinthos message #9551[/url]:

[quote="[url=http://groups.yahoo.com/group/Hyacinthos/message/9551?expand=1]Hyacinthos message #9551[/url] (CORRECTED)"][b]From:[/b] Darij Grinberg
[b]Subject:[/b] Re: midpoints of AH, BK, CL

Dear Orlando,

In Hyacinthos message #9539, you wrote:

[quote="orl"]Given a triangle $ ABC$ inscribed in a circle with center $ O$. Let $ H$, $ K$, $ L$ be the feet of the altitudes of triangle $ ABC$ from the vertices $ A$, $ B$, $ C$, respectively. Denote by $ A_{0}$, $ B_{0}$, $ C_{0}$ the midpoints of these altitudes $ AH$, $ BK$, $ CL$, respectively. The incircle of triangle $ ABC$ has center $ I$ and touches the sides $ BC$, $ CA$, $ AB$ at the points $ D$, $ E$, $ F$, respectively. Prove that the four lines $ A_{0}D$, $ B_{0}E$, $ C_{0}F$ and $ OI$ are concurrent. (When the point $ O$ concides with $ I$, we consider the line $ OI$ as an arbitrary line passing through $ O$.)[/quote]

The first three lines $ A_{0}D$, $ B_{0}E$, $ C_{0}F$ pass through the excenters $ I_{a}$, $ I_{b}$, $ I_{c}$ of triangle ABC (proof: see below). Hence, these three lines coincide with the lines $ DI_{a}$, $ EI_{b}$, $ FI_{c}$, respectively. But triangles DEF and $ I_{a}I_{b}I_{c}$ are homothetic (since their corresponding sidelines are parallel, what can be easily seen); hence, the lines $ DI_{a}$, $ EI_{b}$, $ FI_{c}$ concur at the homothetic center T of the two triangles DEF and $ I_{a}I_{b}I_{c}$.

Now, O is the nine-point center of triangle $ I_{a}I_{b}I_{c}$, and I is the orthocenter of triangle $ I_{a}I_{b}I_{c}$. Hence, the circumcenter of triangle $ I_{a}I_{b}I_{c}$ is the reflection of the orthocenter in the nine-point center, i. e. the reflection of I in O.

Obviously, the circumcenter of triangle DEF is I.

Since the circumcenters of two homothetic triangles lie on one line with the homothetic center, it follows that the reflection of I in O and the point I lie on one line with the homothetic center T of triangles DEF and $ I_{a}I_{b}I_{c}$. In other words, the point T lies on the line OI.

Hence, altogether, the point T lies on the lines $ DI_{a}$, $ EI_{b}$, $ FI_{c}$ (i. e., on the lines $ A_{0}D$, $ B_{0}E$, $ C_{0}F$), and on the line OI. The proof is complete.

Two notes:

(1) In the above, I have left out the proof that the lines $ A_{0}D$, $ B_{0}E$, $ C_{0}F$ pass through the points $ I_{a}$, $ I_{b}$, $ I_{c}$.
Here is this proof:

The "semiprojection" of a point on a line will mean the midpoint between the point and its orthogonal projection on the line. Of course, if three points are collinear, their semiprojections on any line are collinear, too (in fact, point $ \mapsto$ its semiprojection on a given line is an affine mapping).

If D' is the point of tangency of the A-excircle of triangle ABC with the side BC, and D" is the point diametrically opposite to D' on the A-excircle, then the tangent to the A-excircle at D" is parallel to BC; hence, if this tangent meets AB and AC at B" and C", respectively, the triangles ABC and AB"C" are homothetic. The homothetic center of these two triangles is A, of course.

Now, the incircle of triangle ABC touches BC at D; the incircle of triangle AB"C" is the A-excircle of triangle ABC and touches B"C" at D". Hence, the points D and D" are collinear with the homothetic center A of the two triangles ABC and AB"C".

Now, since the points D, D" and A are collinear, the points D, $ I_{a}$, $ A_{0}$ - being their semiprojections on the line BC - are collinear, too, i. e. the line $ A_{0}D$ passes through $ I_{a}$. Similarly, the lines $ B_{0}E$ and $ C_{0}F$ pass through $ I_{b}$ and $ I_{c}$, respectively, qed..

(2) The point T is the triangle center X(57) in Kimberling's ETC; in fact, T is the isogonal conjugate of the Mitten point X(9).

  Sincerely,
  Darij Grinberg[/quote]

[quote="darij grinberg"]But triangles A0B0C0 and IaIbIc are homothetic (since their corresponding sidelines are parallel, what can be easily seen);[/quote]

What????

This is obviously an incorrect statement (a counterexample can be constructed very easyly). May be you meant that triangles $DEF$ and $I_AI_BI_C$ are homothetic...

[quote="Remike"]What????

This is obviously an incorrect statement (a counterexample can be constructed very easyly). May be you meant that triangles $DEF$ and $I_AI_BI_C$ are homothetic...[/quote]

Okay, I'm sorry, I confused some notations (mainly DEF and A0B0C0). Now it should be correct.

  Darij

[quote="Virgil Nicula"][b]Remark. [/b]The exincircles with the centers $ I_a,\ I_b,\ I_c$ of the $ \triangle ABC$ touch the sides $ BC,\ CA,\ AB$ in the points $ M,\ N,\ P$ respectively. Then $ I\in MA_0\cap NB_0\cap PC_0.$[/quote]
dear Virgil Nicula,do you have a proof for this nice problem?

[quote="Virgil Nicula"][b]Remark. [/b]The exincircles with the centers $I_a,\ I_b,\ I_c$ of the $\triangle ABC$ touch the sides $BC,\ CA,\ AB$ in the points $M,\ N,\ P$ respectively. Then $I\in MA_0\cap NB_0\cap PC_0.$[/quote]
Basically, note that $ \triangle AA_0I \sim \triangle I_aMI \Longrightarrow \frac{AA_0}{r_a}\equal{}\frac{IA}{II_a}$

Let me restatement this problem so that it can fits my solution :lol: :
"let $ ABC$ be the acute triangle with incircle $ (I)$. Denote $ D$, $ E$, $ F$ the tagent points of $ (I)$ with $ BC$, $ CA$, $ AB$, respectively. Let $ A_1$, $ B_1$, $ C_1$ be the projections of $ A$, $ B$, $ C$ onto $ BC$, $ CA$, $ AB$, respectively. Suppose $ A_2$, $ B_2$, $ C_2$ are the midpoints of $ AA_1$, $ BB_1$, $ CC_1$, respectively.
Prove that: $ DA_2$, $ EB_2$, $ FC_2$, $ OI$ are concurent, where $ O$ is the circumcenter of triangle $ ABC$."

[i]Proof:[/i]
Let $ X$ be the intersection of $ EF$ with $ BC$. It is well- known that $ (X,D,B,C) \equal{} \minus{} 1$ $ (1)$. Denote $ X'$ the intersection of $ AD$ with $ EF$. From $ (1)$, we also get that $ (X,X',E,F) \equal{} \minus{} 1$. Therefore $ X$ is the pole  of $ AD$ wrt $ (I)$. Hence, $ IX$ is perpendicular to $ AD$. It is easy to notice that triangle $ IXD$ is similar to triangle $ DAA_1$. Now, let me call $ A_3$ is the midpoint of $ XD$, from the notice above, we get, $ IA_3$ is also perpendicular to $ DA_2$. Therefore, $ A_3$ is the pole of $ DA_2$ wrt $ (I)$. Define the same for $ B_3$, $ C_3$.
So now, in order to prove $ DA_2$, $ EB_2$, $ FC_2$ are concurent, we need to prove that $ A_3$, $ B_3$, $ C_3$ are collinear.
Indeed, because, $ (X,D,B,C) \equal{} \minus{} 1$, $ A_3$ is the midpoint of $ XD$, we get, $ A_3D^2 \equal{} \overline {A_3B}.\overline {A_3C}$, which also means that the power of $ A_3$ wrt $ (I)$ and $ (ABC)$ are equal. With the same argument for $ B_3$, $ C_3$. Hence, $ (A_3B_3C_3)$ is the radical axis wrt $ (I)$ and $ (ABC)$, further, $ A_3$, $ B_3$, $ C_3$ are collinear. As the result, $ DA_2$, $ EB_2$, $ FC_2$ are concurent.
Let $ P$ be the intersection of these lines, we get that, $ P$ is the pole of $ (A_3B_3C_3)$ wrt $ (I)$. Therefore, $ IP$ is perpendicular to $ (A_3B_3C_3)$. $ OI$ is also perpendicular to $ (A_3B_3C_3)$ (Because $ (A_3B_3C_3)$ is the radical axis wrt $ (I)$ and $ (ABC)$), where $ O$ is the circumcenter of triangle $ ABC$. Hence, $ O,I,P$ are collinear. Which leads to the result that $ DA_2$, $ EB_2$, $ FC_2$, $ OI$ are concurent.
Our proof is completed.

Here is a solution combining projective geometry and homogeneous/barycentric coordinates.  The calculations fit on two pages when I performed them by hand.

[asy]import graph; size(12cm); real lsf=0.5; pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real xmin=-10.98,xmax=12.96,ymin=-6.14,ymax=7.52;
pen zzttqq=rgb(0.6,0.2,0), qqttff=rgb(0,0.2,1), zzzzzz=rgb(0.6,0.6,0.6), zzzzqq=rgb(0.6,0.6,0), qqzzzz=rgb(0,0.6,0.6), aqgreen=rgb(0,0.4,0.4);
pair A=(-1.88,-3), B=(9.12,-3), C=(0.12,6), L=(0.12,-3), O=(3.62,0.5), I=(1.86,0), L_C=(-8.78,-3), C_0=(0.12,1.5), F=(1.86,-3), P=(0.81,-0.29), P_C=(1.11,-3), K_C=(2.74,-3);
D(A--B--C--cycle,zzttqq);
D(A--B,zzttqq); D(B--C,zzttqq); D(C--A,zzttqq); D(CR(I,3),linetype("4 4")+qqttff); D(C--L,zzzzzz); D(C_0--F,zzzzqq); D(O--L_C,linewidth(1.2)+green); D(C--P_C,qqzzzz); D(L_C--A); D(C--K_C,qqzzzz); D(F--I,linetype("0 3 4 3")+zzzzzz); D(CR(O,6.52),linewidth(1.3)+aqgreen);
D(A); MP("A",(-1.81,-2.89),SW*lsf); D(B); MP("B",(9.2,-2.89),SE*lsf); D(C); MP("C",(0.19,6.12),NW*lsf); D(L); MP("L",(0.07,-3.52),NE*lsf); D(O); MP("O",(3.69,0.62),NE*lsf); D(I); MP("I",(1.95,0.13),NE*lsf); D(L_C); MP("L_C",(-8.98,-2.79),NE*lsf); D(C_0); MP("C_0",(0.19,1.62),NE*lsf); D(F); MP("F",(1.91,-3.48),NE*lsf); D(P); MP("P",(0.89,-0.17),NE*lsf); D(P_C); MP("P_C",(0.99,-3.42),NE*lsf); D(K_C); MP("K_C",(2.73,-3.44),NE*lsf);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);  [/asy]

Let $P = C_0F \cap IO$, $L_C = IO \cap AB$ and $K_C = AI \cap AB$.  Define $P_C = AP \cap AB$.  Let $C_\infty$ be a point at infinity on line $AL$.  Our goal is to compute the barycentric coordinates of $P_C$ with respect to $\triangle ABC$.

We know that $K_C = (a:b:0)$ and $F = (s-b: s-a: 0)$.  The coordinates of $L_C$ can be computed and simplified using the identities $aS_A - cS_C = s(c-a)(s-b)$, $I=(a:b:c)$ and $O=(a^2S_A: b^2S_B: c^2S_C)$ as \begin{align*}
	L_C &= \left( \det \left[ \begin{array}{cc} a^2S_A & c^2S_C \\ a & c \end{array}\right] :
		\det \left[ \begin{array}{cc} b^2S_B & c^2S_C \\ b & c \end{array} \right] :
			0\right) \\
	&= \left( a(aS_A - cS_C) : b(bS_B - cS_C) : 0 \right) \\
	&= \left( a(a-c)(s-b) : b(b-c)(s-a) : 0 \right)
\end{align*}
Now, note that $(C,L; C_{\infty}, C_0) = -1$ is a harmonic bundle.  Taking a perspective at $F$ with onto line $IO$, we see that $(CF \cap IO, L_C; I, P) = -1$ is a harmonic bundle.  Taking perspective at $C$ onto line $AB$, we see that $(F, L_C; K_C, P_C) = -1$.  There's a [url=http://en.wikipedia.org/wiki/Cross-ratio#Definition_in_homogeneous_coordinates]lemma[/url] that states that if we have $K_C = F+L_C$ when adding componentwise, then $P_C = -F + L_C$.  So, we want to find a real $r$ such that \[ \frac{r(s-b) + a(a-c)(s-b)}{r(s-a) + b(b-c)(s-a)} = \frac{a}{b} \] because this will give us the coordinates of $P_C$ for free.  Solving for $r$ (this is the hardest part), \begin{align*}
	r( b(s-b) - a(s-a) ) &= ab( (b-c)(s-a) - (a-c)(s-b) ) \\
	\implies r(a-b)(s-c) &= \frac{1}{2} ab( (b-c)(b+c-a) - (a-c)(a+c-b) ) \\
	&= \frac{1}{2} ab( (b^2-c^2)-(a^2-c^2) + b(a-c) - a(b-c) ) \\
	&= ab(b-a)(b+c-a) \\
	\implies r &= -ab
\end{align*}
So what of $P_C$?  We get \[
	\frac{-r(s-b) + a(a-c)(s-b)}{-r(s-a) + b(b-c)(s-a) } = \frac{ab(s-b) + a(a-c)(s-b)}{ab(s-a) + b(b-c)(s-a) } = \frac{a(s-b)}{b(s-a)}. \] Hence, $P_C = (a(s-b) : b(s-a) : 0)$.

If we define $P_A$ and $P_B$ analogously, we see that they concur at the point \[ P = (a(s-b)(s-c) : b(s-c)(s-a) : c(s-a)(s-b) ). \] It remains to show that this point actually lies on $IO$, which is surprisingly easy, as we have \[
\det \left[ \begin{array}{ccc} a^2S_A & b^2S_B & c^2S_C \\ a & b & c \\ a(s-b)(s-c) & b(s-c)(s-a) & c(s-a)(s-b) \end{array} \right] \] which simplifies as \[ \sum_{\text{cyc}} a(s-b)(s-c) \left( bc(bS_B - cS_C) \right)
	= abc(s-a)(s-b)(s-c) \sum_{\text{cyc}} (b-c)
	= 0.
\] This implies the points are collinear, so we're done.

[quote][b]Remark. [/b]The exincircles with the centers $ I_a,\ I_b,\ I_c$ of the $ \triangle ABC$ touch the sides $ BC,\ CA,\ AB$ in the points $ M,\ N,\ P$ respectively. Then $ I\in MA_0\cap NB_0\cap PC_0.$
[/quote]
$I\in MA_0\iff$ $\frac {ID}{A_0H}=\frac{MD}{MH}\iff$ $\frac{r}{\frac {h_a}{2}}=\frac {|b-c|}{\frac {s|b-c|}{a}}\iff$ $\frac {2r}{h_a}=\frac as\iff$ $2sr=ah_a(=2S)$  [b]O.K.[/b]

[hide="Solution"]Let $I_aI_bI_c$ be the excentral triangle of triangle $ABC$. By [url=http://www.artofproblemsolving.com/Forum/blog.php?u=53544&b=74638]this[/url], it suffices to show that $I_aD$, $I_bE$, $I_cF$, and $OI$ concur, or equivalently, since angle chasing yields triangles $DEF$ and $I_aI_bI_c$ are homothetic, that their homothetic center $H$ lies on $OI$. Since the homothety sending the triangles to one another also sends their circumcenters to one another, it follows that $H$ lies on $OV$, where $V$ is the Bevan point of triangle $ABC$. But $OV$ is the Euler line of triangle $I_aI_bI_c$ and $I$ is its orthocenter, so we're done. $\blacksquare$[/hide]

Dear Mathlinkers,
also a link

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=497757

Sincerely
Jean-Louis

I use Three line Coordinates maybe this way is similar to Evan chen's solution,but still....
I use $A$ $B$ $C$ to represent the $cosA$ $cosB$ $cosC$
because $A_0$ is the midpoint of the altitudes $AH$,then we can get $A_0$ $(1,C,B)$ ,and we can get $D$$(0,1+C,1+B)$
then line $A_0D$ is $(C-B,-1-B,1+C)$ similarily we can get line $B_0E$ is $(1+A,A-C,-1-C)$ and line $C_0F$ is $(-1-C,1+B,B-A)$
notice that incenter $I$ is $(1,1,1)$ and circumcentre $O$ is $(A,B,C)$ then line $OI$ is $(C-B,A-C,B-A)$ the rest part is to calculate the Det which is easy only takes 5minutes

It is well known (and proveable using cross ratio) that $A_0D$ passes through the A-excenter, call it $I_a$. Now we want to prove $DI_a, EI_b, F_c$ concur at $OI$, where $I_a, I_b, I_c$ are the excenters. But just note that $I_bI_c\parallel EF$ and similarly to other two sides, so $\triangle I_aI_bI_c$ and $\triangle DEF$ are homothetic. So the three lines concur at the center of homothety, say $P$. $P$ also lie on the line through circumcenter of the two triangles, which is line $IX_{40}$, which coincide with $OI$, because $O$ is midpoint of $IX_{40}$. So $P, O, I$ colinear and we are done.

Midpoints are nasty to work with, so we delete them from the problem by instead considering the lines $AI_A$, $BI_B$, and $CI_C$, as the midpoint of the altitude from A, D and $I_A$ are collinear. Observing that the sides of $\triangle{DEF}$ are parallel to those of $\triangle{I_AI_BI_C}$, we see that there is a homothety mapping to $I_A$ to $D$, and so on for the other two vertices. This already implies that $A_0D$, $B_0E$ and $C_0F$ are concurrent. Let this point of concurrency be $T$.

Denote the circumcenters of  $\triangle{DEF}$ and $\triangle{I_AI_BI_C}$ as $I$ and $O_{I}$ respectively. Let the incenter of $\triangle{I_AI_BI_C}$, which is also the circumcenter of$\triangle{ABC}$, be $O$. $O_I$, otherwise known as the [i][url=http://mathworld.wolfram.com/BevanPoint.html]Bevan Point[/url][/i] of $\triangle{ABC}$, is on line connecting containing the incenter and circumcenter of $\triangle{ABC}$, implying that $O_{I}$, $O$ and $I$ are collinear. Our homothety however, maps the circumcenter of $\triangle{I_AI_BI_C}$ to the circumcenter of $\triangle{DEF}$, implying that $T$, $O_I$, and $I$ are collinear. It follows that $T$, $I$ and $O$ are
collinear, implying that $IO$ also passes through $T$ as desired. $\square$

Let $I_A$ be the $A$-excenter, $I_B$ be the $B$-excenter, and $I_C$ be the $C$-excenter of $\triangle{ABC}$. Let $\omega$ be the incircle of $\triangle{ABC}$, $\Gamma$ be the circumcircle of $\triangle{ABC}$, and $\Omega$ be the circumcircle of $\triangle{I_AI_BI_C}$ (centered at $P$). Let $Q$ be the exsimilicenter of $\Gamma$ and $\Omega$. Note that $Q$ lies on $OP$.

If $M_A$ is the midpoint of arc $BC$ of $\Gamma$ which does not contain $A$, then the incenter-excenter lemma gives that $M_A$ is the midpoint of $II_A$. Similar statements hold for $B$ and $C$, so $\Gamma$ maps to $\Omega$ under a homothety centered at $I$ with ratio $2$. In particular $P$ lies on $OI$, so $Q$ lies on $OI$. Furthermore, the tangent to $\Omega$ at $I_A$ is parallel to the tangent to $\Gamma$ at $M_A$, which is parallel to $BC$.

Now, let $A_1$ be the foot of $A$ onto $BC$, $D_1$ be the tangency point of the $A$-excircle of $\triangle{ABC}$ onto $BC$, and $D_2$ be the antipode of $D_1$ with respect to the $A$-excircle of $\triangle{ABC}$. Then the homothety centered at $A$ sending $\omega$ to the $A$-excircle of $\triangle{ABC}$ sends $D$ to $D_2$, so $A$, $D$ and $D_2$ are collinear. Now, consider the spiral similarity sending $AA_1$ to $D_2D_1$. Because $D\in AD_2$ and $D\in A_1D_1$, $D$ is the center of this. But this maps $A_0$ to $I_A$, so $A_0$, $D$, and $I_A$ are collinear. Thus, we can instead consider $DI_A$.

Now, I claim that $Q$ lies on $DI_A$. Consider the homothety centered at $P$ sending $\omega$ to $\Omega$. It sends $D$ to $I_A$ because these are the ``bottom-most'' points (their tangents are parallel to $BC$). Thus, $Q$ lies on $DI_A$.

But then by symmetry, $Q$ lies on $EI_B$ and $FI_C$ also, so $A_0D$, $B_0E$, $C_0F$ concur at $Q$ on $OI$.

I didn't see this solution while skimming so here's a cevian nest solution:

[hide=Solution]
[img]https://lh3.googleusercontent.com/-wl09OCn5UkU/XLZeaVKSWoI/AAAAAAAAgYU/WY_SjMjXQ7AL_CipEG-I3PjRMhcEgTD7gCK8BGAs/s0/2019-04-16.png[/img]

Let $I_A,I_B,I_C$ denote the $A,B,C$-excenters.
-----
First we will show that $A_0D, B_0E,$ and $C_0F$ are concurrent, which is equivalent to showing that $I_AD,I_BE,$ and $I_CF$ are concurrent since it is well known that $I_A,D,$ and $A_0$ are collinear. However, we know that $I_AA,I_BB,$ and $I_CC$ are concurrent at the incenter w.r.t. $\triangle ABC$ and that $AD,BE,$ and $CF$ are concurrent at the Gergonne point w.r.t. $\triangle ABC,$ so by Cevian Nest, we know that the three lines meet at some point $Z.$
-----
Now we will show that $IO$ passes through $Z$ as well. To do so, we consider the homothety centered at $Z$ mapping $\triangle DEF$ to $\triangle I_AI_BI_C;$ since $I$ is the circumcenter w.r.t. $\triangle DEF,$ it is mapped to the circumcenter of $I_AI_BI_C$ (let this point be $I'$). But $I$ is the orthocenter w.r.t. $\triangle I_AI_BI_C$ and $O$ is its nine-point center, so $I,O,I',$ and $Z$ are collinear just as we wanted (in particular, they lie upon the Euler line of $\triangle I_AI_BI_C$). $\blacksquare$[/hide]

Here is my solution for this problem
[b]Solution[/b]
Let $X$, $Y$, $Z$ be second intersections of $DA_0$, $EB_0$, $FC_0$ with $(I)$; $D'$ is reflection of $D$ through $I$; $S$ $\equiv$ $AD$ $\cap$ $(I)$ $(S \not \equiv D)$
Since: $D(DSD'X) = D(AHD'A_0) = - 1$ then: $DXSD'$ is harmonic quadrilateral
But: $DFD'E$ is harmonic quadrilateral so: $EF$, $XD'$, $BC$ and tangent at $S$ of $(I)$ concurrent
Hence: $X(BCDD') = - 1$ or $XD$ is bisector of $\angle{BXC}$
Then: $(BCX)$ tangents $(I)$ at $X$
Similarly: $(CAY)$ tangents $(I)$ at $Y$, $(ABZ)$ tangents $(I)$ at $Z$
Let $A_1$, $A'$ be midpoint of $\stackrel\frown{BC}$ which not containing $A$ of $(O)$, midpoint of $\stackrel\frown{BC}$ which not containing $X$ of $(BCX)$; similarly, we have: $B_1$, $B'$, $C_1$, $C'$
We have: $IA'^2 - IC'^2 = \overline{A'D} . \overline{A'X} - \overline{C'F} . \overline{C'Z} = BA'^2 - BC'^2$
Then: $C'A'$ $\perp$ $IB$
But: $C_1A_1$ $\perp$ $IB$ so: $C_1A_1$ $\parallel$ $C'A'$
Similarly: $B_1C_1$ $\parallel$ $B'C'$, $A_1B_1$ $\parallel$ $A'B'$
Hence: $O$ is circumcenter of $\triangle A'B'C'$
Let $J$ be external homothetic center of $(I)$ and $(A'B'C')$ then: $J$; $D$; $A'$ are collinear
But: $D$, $X$, $A'$ are collinear so: $D$, $J$, $X$, $A'$ are collinear or $D$, $J$, $X$ are collinear
Similarly: $E$, $J$, $Y$ are are collinear; $F$, $J$, $Z$ are collinear
Hence: $DX$, $EY$, $FZ$ concurrent at a point $J$ lies on $IO$ or $DA_0$, $EB_0$, $FC_0$ concurrent at a point $J$ lies on $IO$

[url=https://artofproblemsolving.com/community/c776104h1811576_x25_is_the_pole_of_the_orthic_axis_wrt_abc]Lemma #2, Result #3, Result #4 (here)[/url]

See that ${ A_{0} –D –I _ {A}}$
We know that isotomic conjugate of Gergonne point is Nagel point, Due to which$ \Delta ABC$ forms a cevian nest WRT its excentral triangle.
$ I_{A}D,I_{B}E, I_{C}F$ concur
Now homothety centred at $X$ taking
$\Delta DEF \leftrightarrow \Delta I_{A} I_B I {C}$
finishes the question $\blacksquare $

[quote=orl]Given a triangle $ABC$. Let $O$ be the circumcenter of this triangle $ABC$. Let $H$, $K$, $L$ be the feet of the altitudes of triangle $ABC$ from the vertices $A$, $B$, $C$, respectively. Denote by $A_{0}$, $B_{0}$, $C_{0}$ the midpoints of these altitudes $AH$, $BK$, $CL$, respectively. The incircle of triangle $ABC$ has center $I$ and touches the sides $BC$, $CA$, $AB$ at the points $D$, $E$, $F$, respectively. Prove that the four lines $A_{0}D$, $B_{0}E$, $C_{0}F$ and $OI$ are concurrent. (When the point $O$ concides with $I$, we consider the line $OI$ as an arbitrary line passing through $O$.)[/quote]

To prove that the four lines $A_{0}D$, $B_{0}E$, $C_{0}F$ and $OI$ are concurrent. First we will prove that the lines $A_{0}D$, $B_{0}E$, $C_{0}F$ are concurrent.

Recall that by [b][color=#f00]Midpoint of Altitudes Lemma[/color][/b], the set of points $\{A_0, D, I_A \}$, $\{B_0, E, I_B \}$ and $\{C_0, F, I_C \}$ are collinear $-(\bigstar)$. Hence it suffices to show that the lines $\overline{I_AD}, \overline{I_BE}$ and $\overline{I_CF}$ are concurrent. Note that in $\triangle I_AI_BI_C$, the lines $AI_A, BI_B$ and $CI_C$ are concurrent at $I$ since the points $\{A, I, I_A \}$, $\{B, I, I_B \}$ and $\{C, I, I_C \}$ are collinear (by [b][color=#f00]Fact5[/color][/b]). Next note that the lines $AD, BE$ and $CF$ are concurrent at the "[color=#9a00ff][b]Gergonne Point[/b][/color]". So, by the [b][color=#9a00ff]Cevian Nest (Theorem)[/color][/b] we have the lines $\overline{I_AD}, \overline{I_BE}$ and $\overline{I_CF}$ concurrent, at some point. Let us denote that concurrency point by $X$.

So, $(X=\overline{I_AD} \cap \overline{I_BE} \cap \overline{I_CF})$. Next we will prove that the line $OI$ passes through $X$ as well.

Next recall that by [color=#f00][b]Duality of Orthocenters and Excenters Lemma[/b][/color], [b]$\triangle ABC$ is the orthic triangle of $\triangle I_AI_BI_C$[/b] and[b] $I$ is the orthcenter of $\triangle I_AI_BI_C$[/b]. Also from $(\bigstar)$ we can conclude that there exists a homothety $\psi$ that maps $\triangle DEF$ to $\triangle I_AI_BI_C$. And hence the point $I$, which is the circumcenter of $\triangle DEF$ gets mapped to the circumcenter of $\triangle I_AI_BI_C$ (let us denote it by $O'$), and hence $X, I, O'$ are collinear $ - (\star)$. Next note that in $\triangle I_AI_BI_C$, the points: $I$ ([b]the orthocenter[/b]), $O'$ ([b]the circumcenter[/b]) and [b]the Nine-Point center[/b] must be collinear, as all of these lie on the [color=#9a00ff][b]Euler Line[/b][/color] $ - (\star\star)$. Hence from $(\star)$ and $(\star\star)$ we can conclude that $X, I, O'$ and [b]the Nine-Point center of $\triangle I_AI_BI_C$[/b] are collinear. $-(\bigstar\bigstar)$

Now, note that in $\triangle I_AI_BI_C$, $AI_A, BI_B$ and $CI_C$ are the altitudes, that is $A, B$ and $C$ are the feet of the altitudes, and the circumcircle of $\triangle ABC$ passes through these $3$ points only, hence $\odot(ABC)$ is the Nine-Point Circle of $\triangle I_AI_BI_C$ and thus the center of $\odot(ABC)$, $O$ is the Nine-Point center of $\triangle I_AI_BI_C$. Hence from $(\bigstar\bigstar)$ we can say that $X, I, O'$ and $O$ are collinear, and thus $OI$ passes through $X$ as well. $\quad \blacksquare$

[quote=a1267ab]Anyone want to try this generalization?

Let $ABC$ be a triangle with circumcenter $O$. Let $D, E, F$ be the feet of the altitudes from $A, B, C$ respectively. Denote by $A_{0}$, $B_{0}$, $C_{0}$ the midpoints of $AD, BE, CF$. Let $P$ be a point on the Darboux cubic of $ABC$, and $X, Y, Z$ be the feet from $P$ to $BC, CA, AB$. Prove that $A_0X, B_0Y, C_0Z, OP$ concur.

($P$ being on the Darboux cubic just means that $AX, BY, CZ$ concur.)[/quote]

Let $P'$ be the reflection of $P$ over $O$.

Claim: There exist points $A',B',C'$ on $PX,PY,PZ$ such that $A\in B'C', B\in C'A'$, and $C\in A'B'$.

Proof: Let $A'B'C'$ be the antipedal triangle of $P'$. Since $P$ and $P'$ are reflections over $O$, $P$ is the isogonal conjugate of $P'$ with respect to $A'B'C'$. Thus, $A'P\perp BC$, so $A'$ lies on $PX$. Similarly, $B',C'$ lie on $PY,PZ$ respectively. $\square$

By cevian nest, $AA',BB',CC'$ concur at some point $Q$. Projecting $(AD;A_0P_{\infty})$ through $X$ onto $AA'$, we see that $Q$ lies on $A_0X$. Similarly, $Q$ lies on $B_0Y$ and $C_0Z$. But by Sondat's theorem, $P, Q$, and $P'$ are collinear, so $Q$ lies on $OP$. Thus, $A_0Z, B_0Y, C_0Z, OP$ concur ar $Q$. $\square$

Let $I_a$, $I_b$, $I_c$ be the $A$-excenter, $B$-excenter, and $C$-excenter of $ABC$ respectively, and define $O_1$ as the circumcenter of $I_aI_bI_c$. In addition, we label the feet of the altitudes from $A, B, C$ as $X, Y, Z$ respectively. The Midpoint of Altitudes Lemma implies $I_a \in A_0D$, $I_b \in B_0E$, and $I_c \in C_0F$.

[b][color=#f00]Claim:[/color][/b] $DEF$ and $I_aI_bI_c$ are homothetic.

[i]Proof.[/i] Properties of internal and external bisectors yield $CI \perp \overline{I_aCI_b}$. Now, since $CD = CE$ and $\angle ECI = \angle ICD$, we can conclude $IC \perp DE$, so $\overline{I_aCI_b} \parallel DE$. Analogous processes give $I_bI_c \parallel EF$ and $I_cI_a \parallel FD$, as required. $\square$

Thus, $\overline{A_0DI_a}$, $\overline{B_0EI_b}$, and $\overline{C_0FI_c}$ concur at a homothetic center of $DEF$ and $I_aI_bI_c$, which we denote with $J$.

[b][color=#f00]Claim:[/color][/b] $OI$ is the Euler Line of $I_aI_bI_c$.

[i]Proof.[/i] It's well-known that $ABC$ is the orthic triangle of $I_aI_bI_c$. Hence, $I$ is the orthocenter of $I_aI_bI_c$.

Now, we also know $(ABC)$ is the Nine-Point Circle of $I_aI_bI_c$, so $O$ is the Nine-Point Center of $I_aI_bI_c$. The desired result follows readily. $\square$

Hence, we have $O_1 \in OI$.

[b][color=#f00]Claim:[/color][/b] $J \in IO_1$.

[i]Proof.[/i] Because $I$ is the circumcenter of $DEF$ and $O_1$ is the circumcenter of $I_aI_bI_c$, we know $IO_1$ passes through any of the homothetic centers between the two triangles. $\square$

Now, we can conclude $J \in \overline{OIO_1}$, which finishes. $\blacksquare$

[b]An Alternate Start:[/b] You can also use the Cevian Nest Lemma on $I_aI_bI_c$ and $ABC$ to show that $I_aD, I_bE, I_cF$ concur. In fact, this was how I initially proved the concurrency.

Denote $I_A,I_B,I_C$ as the excenters of $A,B,C$ respectively.

By Lemma 4.14 (EGMO), we have that $A_0,B_0,C_0$ lie on $I_AD,I_BE,I_CF$ respectively. Hence if suffices to show that $I_AD,I_BE,I_CF$$,$$OI$ are concurrent.

[b]Claim:[/b] There exists a homotethy sending $\triangle I_AI_BI_C$ to $\triangle DEF$.

[b]Proof:[/b] It suffices to show $I_AI_B \parallel DE$, and the rest of the corresponding sides follow with the same logic. We will show $\angle I_BCB = \angle EDB$. Since $CD=CE$, we have $\angle EDC = 90 - \frac{\angle C}{2}$, so $\angle EDB = 90 + \frac{\angle C}{2}$. It is easy to see that $\angle I_BCB = 90 + \frac{\angle C}{2}$, so we are done.

All is left now is to show that $OI$ passes through $X$. Observe that $I$ is the circumcenter of $\triangle DEF$. Label $O'$ as the circumcenter of $\triangle I_AI_BI_C$. If the center of homotethy is $X$, then $O'I$ passes through $X$.

It is well known that $O'$ lies on $OI$, the Euler line, so $OI$ also passes through $X$.$\blacksquare$

Let $I_A , I_B , I_C$ be the excenters corresponding to vertices $A,B,C$ respectively.
It is well known that $A_0 , D , I_A$ , $B_0 , E , I_B$ , $C_0 , F , I_C$ are collinear.
Also note that $\triangle DEF$ and $\triangle I_AI_BI_C$ are homothetic, thus $DI_A , EI_B , FI_C$ are concurrent at the homothety center say $P$.
Let $K$ be the center of $(I_AI_BI_C)$. Since $O$ is the 9-point center of $(I_AI_BI_C)$ and $I$ is the orthocenter of $\triangle I_AI_BI_C$, we have $I,O,K$ are collinear. But, since $P$ is the center of homothety, we know $P,I,K$ are collinear. Thus, $P$ lies on $OI$, as desired.

Let $I_A,I_B,I_C$ denote the $A,B,C$-excenters respectively, and rename $H,K,L$ as $X,Y,Z$ respectively.. It is well-known that
[list]
[*] $A_0,D,I_A$ collinear,
[*] $B_0,E,I_B$ collinear,
[*] $C_0,F,I_C$ collinear,
[/list]
so it suffices to show that $\overline{I_AD},\overline{I_BE},\overline{I_CF},\overline{OI}$ concur.
Since $\overline{AI_A},\overline{BI_B},\overline{CI_C}$ are concurrent at the orthocenter of $\triangle I_AI_BI_C$ and $\overline{AD},\overline{BE},\overline{CF}$ are concurrent at the Gergonne point of $\triangle ABC$, by Cevian nest $\overline{I_AD},\overline{I_BE},\overline{I_CF}$ concur, say at point $P$.
I now claim that $\overline{I_BI_C}$ and $\overline{EF}$ are parallel. It is well-known that $\overline{AI_A} \perp \overline{I_BI_C}$ and $A,I_A,I$ are collinear, so it suffices to show that $\overline{AI} \perp \overline{EF}$. But this follows as $IE=IF$ and $AE=AF$, so $AEIF$ is a kite.
Since $\overline{I_CI_A} \parallel \overline{FD}$ and $\overline{I_AI_B} \parallel \overline{DE}$ similarly, with the previous concurrency we find that $\triangle I_AI_BI_C$ and $\triangle DEF$ are homothetic. Now consider the homothety sending $\triangle DEF$ to $\triangle I_AI_BI_C$. This sends the circumcenter of $\triangle DEF$ (which is $I$) to the circumcenter of $\triangle I_AI_BI_C$. Since $\triangle ABC$ is the orthic triangle of $\triangle I_AI_BI_C$, we find that $I$ (the orthocenter of $\triangle I_AI_BI_C$) is collinear with the circumcenter of $\triangle I_AI_BI_C$ and the circumcenter $O$ of $\triangle DEF$, which is the 9-point center of $\triangle I_AI_BI_C$, hence $P,O,I$ are collinear as desired. $\blacksquare$

[b]Claim: [/b] $\triangle I_A I_BI_C\sim \triangle DEF$.
[b]Proof: [/b] Note that the sides are parallel, for example $EF\parallel I_BI_C$ since both are perpendicular to $AI$. Additionally,
\[\angle EDF = 180 - \angle FDB - \angle EDC = 180 - (90 - \frac12 B) - (90-\frac12 C) = 180 - (90 + \frac12 A) = 180-\angle BIC = \angle BI_A C\]
$\square$.

Now, consider the homothety centered at $X$ that sends $\triangle DEF$ to $\triangle I_A I_B I_C$. It clearly sends $D\to I_A, E\to I_B, F\to I_C$, and $I$ goes to $O'$, the circumcenter of $\triangle A'B'C'$. Note that $I, O$ are the orthocenter and nine point center of $\triangle I_A I_B I_C$. Thus, the circumcenter of $\triangle I_AI_BI_C$ lies on $IO$, so $X,I,O_{(ABC)}$ are also collinear and we're done. $\blacksquare$.

[hide=Partial Solution]Note that $A_0D$ passes through $I_A,$ the $A$-excenter of triangle $ABC.$ Now,
\begin{align*}
\frac{\sin{\angle{CI_AD}}}{\sin{\angle{BI_AD}}}&=\frac{CD\cdot\frac{\sin{\angle{DCI_A}}}{DI_A}}{BD\cdot\frac{\sin{\angle{DBI_A}}}{DI_A}}\\
&=\frac{(s-c)\cdot\sin{\left(\frac{180-\angle{C}}{2}\right)}}{(s-b)\cdot\sin{\left(\frac{180-\angle{B}}{2}\right)}}\\
&=\frac{(s-c)\cdot\cos{\left(\frac{\angle{C}}{2}\right)}}{(s-b)\cdot\cos{\left(\frac{\angle{B}}{2}\right)}}.\\
\end{align*}
Similarly, we find that
\begin{align*}
\frac{\sin{\angle{BI_CE}}}{\sin{\angle{AI_BE}}}&=\frac{(s-b)\cdot\cos{\left(\frac{\angle{B}}{2}\right)}}{(s-a)\cdot\cos{\left(\frac{\angle{A}}{2}\right)}}\text{ and }\\
\frac{\sin{\angle{AI_BF}}}{\sin{\angle{CI_BF}}}&=\frac{(s-a)\cdot\cos{\left(\frac{\angle{A}}{2}\right)}}{(s-c)\cdot\cos{\left(\frac{\angle{C}}{2}\right)}}.
\end{align*}
Thus,
\begin{align*}
\frac{\sin{\angle{CI_AD}}}{\sin{\angle{BI_AD}}}\cdot\frac{\sin{\angle{BI_CE}}}{\sin{\angle{AI_BE}}}\cdot\frac{\sin{\angle{AI_BF}}}{\sin{\angle{CI_BF}}}&=\frac{(s-c)\cdot\cos{\left(\frac{\angle{C}}{2}\right)}}{(s-b)\cdot\cos{\left(\frac{\angle{B}}{2}\right)}}\cdot\frac{(s-b)\cdot\cos{\left(\frac{\angle{B}}{2}\right)}}{(s-a)\cdot\cos{\left(\frac{\angle{A}}{2}\right)}}\cdot\frac{(s-a)\cdot\cos{\left(\frac{\angle{A}}{2}\right)}}{(s-c)\cdot\cos{\left(\frac{\angle{C}}{2}\right)}}\\
&=1.
\end{align*}
Thus, by the converse of Ceva's Theorem on triangle $I_AI_BI_C,$ $I_AD, I_BE,$ and $I_CF$ are concurrent. Since $A_0$ lies on $I_AD,$ $B_0$ lies on $I_BE,$ and $C_0$ lies on $I_CF,$ this means that $A_0D,B_0E,C_0F$ are concurrent.[/hide]

Can I have a hint on how to prove that the point of concurrency lies on $OI?$ Thanks in advance!

The results of this entire problem are quite well-known nowadays. Let $\triangle I_aI_bI_c$ be the excentral triangle of $\triangle ABC$. Note that $A_0,D,I_a$ (and $B_0,E,I_b$ and $C_0,F,I_c$) are colinear by the well-known Midpoints of Altitudes Lemma (Lemma 4.14 in EGMO). Hence, it suffices to show that $DI_a,EI_b,FI_c$ concur on $\overline{OI}$. However, these three lines meet at $X_{57}$, the Isogonal Mittenpunkt point, which is known to lie on $\overline{OI}$. Done. $\blacksquare$.

But suppose that we didn't know what the Isogonal Mittenpunkt point was...

Set $\triangle ABC$ as the reference triangle with $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. We also have $D=(0:s-c:s-b)$ and $I_a=(-a:b:c)$. Foot of $A$-altitude has coordinates $H=\left(0:a^2+b^2-c^2:a^2-b^2+c^2\right)$. We can normalize by multiplying each component by $\frac{1}{2a^2}$. Applying the midpoint formula on $\overline{AH}$ gives $A_0=\left(2a^2:a^2+b^2-c^2:a^2-b^2+c^2\right)$.

We claim that $A_0,D,I_a$ are colinear. To prove this, we just need to verify $$\begin{vmatrix}-a&b&c\\ 0&s-c&s-b\\ 2a^2&a^2+b^2-c^2&a^2-b^2+c^2\end{vmatrix}=0$$ which is true upon expansion.

We can do similar things to show that $B_0,E,I_b$ and $C_0,F,I_c$ are colinear (just permute variables).  Thus, it suffices to show that $DI_a,EI_b,FI_c$ concur on the line $OI$.

The equation of line $DI_a$ is given by $$\begin{vmatrix}x&y&z\\ 0&s-c&s-b\\ -a&b&c\end{vmatrix}=0 \iff (c-b)(-a+b+c)x+a(a-b+c)y-a(a+b-c)z=0. $$ We can find the equtaions of lines $EI_b,FI_c$ by permuting variables.

To see that $DI_a,EI_b,FI_c$ concur, we need to check that $$\begin{vmatrix}(c-b)(-a+b+c)&a(a-b+c)&-a(a+b-c)\\ -b(-a+b+c)&(a-c)(a-b+c)&b(a+b-c)\\ c(-a+b+c)&-c(a-b+c)&(b-a)(a+b-c)\end{vmatrix}=0$$ however this is quite clear since $$\begin{vmatrix}(c-b)(-a+b+c)&a(a-b+c)&-a(a+b-c)\\ -b(-a+b+c)&(a-c)(a-b+c)&b(a+b-c)\\ c(-a+b+c)&-c(a-b+c)&(b-a)(a+b-c)\end{vmatrix}=(a+b-c)(a-b+c)(-a+b+c)\begin{vmatrix}(c-b)&a&-a\\ -b&(a-c)&b\\ c&-c&(b-a)\end{vmatrix}=0.$$

So the lines concur. Their concurrency point is the non-trivial solution to $$\begin{bmatrix}(c-b)(-a+b+c)&a(a-b+c)&-a(a+b-c)\\ -b(-a+b+c)&(a-c)(a-b+c)&b(a+b-c)\\ c(-a+b+c)&-c(a-b+c)&(b-a)(a+b-c)\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$$ which we can solve with Cramer's Rule to be $$\left(\frac{a}{-a+b+c} ,\frac{b}{a-b+c},\frac{c}{a+b-c}\right).$$

Finally, it suffices to check that this point lies on $OI$. Since $I=(a:b:c)$ and $O=(a^2(b^2+c^2-a^2):b^2(c^2+a^2-b^2):c^2(a^2+b^2-c^2))$, we just need to verify that $$\begin{vmatrix}a&b&c\\ a^2(b^2+c^2-a^2)&b^2(c^2+a^2-b^2)&c^2(a^2+b^2-c^2)\\ \frac{a}{-a+b+c}&\frac{b}{a-b+c}&\frac{c}{a+b-c}\end{vmatrix}=0 $$ which (after factoring out $abc$) is true upon expansion.

[hide = Solution] We first use a simple lemma which pretty much kills this problem.

$\bold{Lemma}$: Let $ABC$ be a triangle and let $AD$ be the altitude from $A$ to $BC$. Let $M$ be the midpoint of $AD$ and let $E$ denote the tangent of the incircle of $ABC$ to side $\overline{BC}$. Let $I_A$ be the $A$-excenter of $ABC$. Then we have that $I_A, E, D$ are collinear.
[hide = Proof] Consider the homothety centered at $I_A$ that sends incenter $I$ to $A$. Then this homothety trivially sends $E$ to $D$. $\square$ [/hide]

We then have that (using the notation of the original problem) $A_0, D, I_A$ are collinear. Similarly $B_0, E, I_B$ and $C_0, F, I_C$ are collinear. We then have that the homothety sending $D \mapsto I_A$ sends $\triangle DEF \mapsto \triangle I_AI_BI_C$. We obviously then have that these points concur at some point $T$.

Denote by $O'$ as the circumcenter of $I_AI_BI_C$. Then we have that the same homothety sends $I$ to $O'$. Therefore $IO'$ is the Euler line of $I_AI_BI_C$. Now as $(ABC)$ is the nine-point circle ($\astrosun_9$) of $I_AI_BI_C$, we have that the circumcenter of $ABC$ is the $N_9$ point of $I_AI_BI_C$. Hence $I, O, O'$ are collinear. (Further we actually have $IO:OO' = 1:1$).

Now the original homothety that sent $(DEF)$ to $(I_AI_BI_C)$ is so that $T, I, O'$ are collinear. (We have the map sends $I$ to $O'$ and $T$ to $T$). We are done. $\blacksquare$ [/hide]

Note that $A_0$, $D$, $I_A$ are collinear and so on. As such, it remains to show that the lines $I_AD$, $I_BE$, $I_CF$ are collinear. Since $\triangle I_AI_BI_C \sim \triangle DEF$, that follows and the lines intersect at the homothety center.
Then, $I$ and $O$ are the orthocenter and nine-point center of $I_AI_BI_C$ respectively. As such, the circumcenter of $I_AI_BI_C$ lies on $IO$. Since $I$ is the circumcenter of $DEF$, it follows that $IO$ also is concurrent.

Let $\triangle I_AI_BI_C$ be the excentral triangle. By the midpoints of altitudes lemma, we have $A_0, D$, and $I_A$ collinear, and analogous results for the others. Therefore it now suffices to show that $DI_A, EI_B, FI_C, OI$ are collinear. But $\overline{DE} \parallel \overline{I_AI_B}$ as they're both perpendicular to $\overline{IC}$, and similar for the other sides, so $DI_A, EI_B$, and $FI_C$ concur at the center of homothety taking $\triangle DEF$ to $\triangle I_AI_BI_C$. But it's well-known that the Euler line of both triangles is $OI$, and as such the result follows.

Solved with [b]kingu[/b]. Very fun problem.

It is well known (Midpoint of the Altitudes) that $A_0D$ passes through $I_A$ and similarly for the other sides. Then, we can eliminate the points $A_0$ , $B_0$ and $C_0$ from the picture. It remains to show that $I_AD$ , $I_BE$ and $I_CF$ are concurrent on $IO$. For this, note that since $\triangle DEF$ and $\triangle I_AI_BI_C$ are homothetic, it follows that $I_AD$ , $I_BE$ and $I_CF$ are concurrent, say at point $R$. Then, note that since $\triangle M_AM_BM_C \sim \triangle DEF$ are homothetic, and the circumcenters of these two triangles are $O$ and $I$ respectively, their orthocenters must also be concurrent at this concurrency point. Clearly, $I$ is the orthocenter of $\triangle M_AM_BM_C$ ($AI \perp M_BM_C$ etc. as mentioned before) and thus, the orthocenter of $\triangle DEF$ must also lie on $IO$. Then, since $\triangle DEF$ and $\triangle I_AI_BI_C$ are homothetic, the line joining their orthocenters must be concurrent at $X$ as well. It is well known that the orthocenter of $\triangle I_AI_BI_C$ is $I$. We previously showed that the orthocenter of $\triangle DEF$ lies on $IO$ from which it follows that $IO$ passes through $X$, as the problem requires.


Finitely many polygons are placed in the plane. If for any two polygons of them, there exists a line through origin $O$ that cuts them both, then these polygons are called "properly placed". Find the least $m \in \mathbb{N}$, such that for any group of properly placed polygons, $m$ lines can drawn through $O$ and every polygon is cut by at least one of these $m$ lines.

This looks very familiar. I think something similar appeared on the forum recently.

First of all, notice that $m=1$ doesn't work: We may assume WLOG that none of the polygons contain $O$. Consider a line $AB$ through $O$ with $A,B$ on opposite sides of $O$, and rays $(OC,(OD$ on the same side of $AB$, with $(OC$ between $(OB$ and $(OD$. We can now take our "polygons" to be $BC,CD,DA$ (they can, of course, be "thickened" into proper polygons and not just segments).

Next, we can show that $m=2$ works: For each polygon in the collection consider the minimal region contained between two lines through $O$ which contains the polygon. Now pick a polygon $\mathcal P$ in the collection whose region is minimal wrt inclusion, in the sense that there's no other polygon whose region is properly included in the $\mathcal P$'s region. The two lines through $O$ which bound the region of $\mathcal P$ have the required properties, as can easily be seen.

I hope it's correct.


Prove the first case of FLT for Sophie Germain primes

You mean $ a^{n}\plus{}b^{n}\equal{}c^{n}$, but for $ n\in \mathbb{P}$ or $ a,b,c\in \mathbb{P}$ or both?

To write it down completely:

Let $ p>2$ be a prime such that $ 2p\plus{}1$ is prime, too. Prove that if $ x^p\plus{}y^p\equal{}z^p$ for integers $ x,y,z$, then $ p|xyz$.

I have proved that $ 2p\plus{}1$ divides $ xyz$. This is because mod $ 2p\plus{}1$ $ x^p$, $ y^p$, $ z^p$ are congruent to $ \minus{}1$ $ 1$ or $ 0$. Then the result follows

Looking modulo $ 2p\plus{}1$, we see that $ (2p\plus{}1) \mid xyz$. Does this help?


Find all pairs of positive integers $ (m,n)$ such that the sum of all the solutions to the equation $ \lfloor x\rfloor\equal{}m\left\{nx\right\}$ amounts to $ 22$.

[hide="My solution"]Let's write $ nx = nk + r + \alpha$, where $ k,r\in\mathbb{Z},0\leqslant r\leqslant n - 1, \alpha\in\mathbb{R},0\leqslant\alpha < 1$.

Then $ \{nx\} = \alpha$, and $ x = k + {r + \alpha\over n}$, but as $ 0\leqslant r + \alpha < n$, we get $ \lfloor x\rfloor = k$

Hence by the initial equation, $ k = m\alpha\iff\alpha = {k\over m}$. From the definition of the fractional part, this yields $ 0\leqslant k\leqslant m - 1$.

We now get $ x = k + {r + \alpha\over n} = k + {r + {k\over m}\over n} = k + {mr + k\over mn} = {(mn + 1)k + mr\over mn}$ for $ 0\leqslant k\leqslant m - 1,0\leqslant r\leqslant n - 1$

Now let's check if any two solutions are equal:

$ (mn + 1)k_1 + mr_1 = (mn + 1)k_2 + mr_2\implies (mn + 1)|k_1 - k_2| = m|r_1 - r_2|$

But $ 0\leqslant r\leqslant n - 1$ yields $ |r_1 - r_2|\leqslant n - 1$, hence $ |k_1 - k_2|\leqslant {m(n - 1)\over mn + 1} < 1$, therefore $ k_1 = k_2$, which in turn yields $ r_1 = r_2$

Therefore all the solutions are pairwise distinct, so we can simply drive the sums through the expression for $ x$:

\begin{eqnarray*}S & = & \sum_{k = 0}^{m - 1}\sum_{r = 0}^{n - 1}{(mn + 1)k + mr\over mn} \\
& = & \sum_{k = 0}^{m - 1}\left(n\cdot{(mn + 1)k\over mn} + {1\over n}\cdot{n(n - 1)\over 2}\right) \\
& = & \sum_{k = 0}^{m - 1}\left({(mn + 1)k\over m} + {n - 1\over 2}\right) \\
& = & {mn + 1\over m}\cdot {m(m - 1)\over 2} + {m(n - 1)\over 2} \\
& = & {(mn + 1)(m - 1) + m(n - 1)\over 2} \\
& = & {m^2n - mn + m - 1 + mn - m\over 2} \\
& = & {m^2n - 1\over 2}\end{eqnarray*}

So we must solve $ {m^2n - 1\over 2} = 22\iff m^2n = 45 = 3^2\cdot 5$ in positive integers. The only solutions are obviously $ (m,n) = (1,45)$ and $ (m,n) = (3,5)$, with the corresponding equations $ \lfloor x\rfloor = \{45x\}$ and $ \lfloor x\rfloor = 3\{5x\}$[/hide]


Let $a,b>0$ be integers with $a$ odd. Define the sequence $U_n$ by :
$U_0 = b$ and for each $n \geq 0$ :
- If $U_n$ is even then $U_{n+1} = \frac 1 2 U_n$
- If $U_n$ is odd then $U_{n+1} = a + U_n$.

1) Prove that there exists $n$ such that $U_n \leq a$
2) Prove that the sequence becomes periodic.

Pierre.

Well, the sequence is $\le$ than the one obtained by adding $a$ and alternatively dividing by $2$. That sequence (define it, for example, as $x_0=b,\ x_{2n+1}=\frac{x_{2n}}2,\ x_{2n+2}=x_{2n+1}+a$ if $b$ is even and the other way around if it's odd) has a subsequence which tends to $a$, and because $u_n$ are integers, it means that $(u_n)$ has a subsequence of terms $\le a$. Part b) follows easily, since $u_{n+1}$ only depends on $u_n$, and since there are infinitely many $u_n\le a$, we must have $u_m=u_n$ for some $m>n\in\mathbb N$

if we assume that [tex]U_n>=a[/tex] for all n
just we can consider the n that minimise the quantity [tex]U_n-a[/tex]
and discuss it's parity.
about the second question
we can easyly proove that there exist an infinity of n st [tex]U_n<=a[/tex]
and so there exist to distinct n and m st : [tex]U_n=U_m[/tex]
and the proof is complete
bye


Show that we cannot find compact sets $A_1, A_2, A_3, ... $ such that (1) all elements of $A_n$ are rational and (2) given any compact set $K$ whose members are all rationals, $K$ contained in some $A_n$.

Choose any rational point $c$. The set $I_n=[c-\frac 1n,c+\frac 1n]\cap \mathbb Q$ cannot be contained in $A_n$, because then $A_n$ wouldn't be closed (it'c closure would include $[c-\frac 1n,c+\frac 1n]$). Now, from each $I_n$, choose a rational $x_n\not\in A_n$. The set $\{x_n\}_n\cup\{c\}$ is closed, bounded, and it contains only rational points, but it's not contained in any of the $A_n$'s.


1) If A, B, & C are sets, then which of the following is true? (A' denotes the complement of A)

A)  A U B = intersection of B and A

B) (A U B)'= C

C)  Intersection of (A U B) and C = A U (Intersection of B and C)

D) (A U B')'=A' U B

E) (Intersection of A' and B) U B'= (Intersection of A and B)'

2) A sector with interior angle 135 degrees is cut from a circle of radius 8 and discarded. The remainder of the circle is formed into a cone by taping together the straight edges. What is the radius of the base of the cone?

3) What are the 2 divisor of 4^12-1 between 60 and 80?

4) Angie can paddle her canoe at 6 miles an hour in still water. When she tries to paddle directly across an 80 foot wide river, she ends up on the opposite bank 60 feet downstream from the starting point. How fast is the river flowing?

5) A person walks one mile, veers 24 degrees to the right and walks a mile, etc. How far will the person walk before returning to the original position?

Tare and theone each get credit for 5. (I see a tie coming on....)

We'll start at 2:00cst, that's best for me actually, and go from there...

[quote="Leeyuhang"]30 min intervals? do you meam each round or problem?[/quote]
round. and no, this is the 6th.

but i've only seen one thread that actually says 5th edition.

Somehow the first round for the 5th edition got deleted and so id the second, oh well. What're you gonna do, call Murder and Scully? :twisted: </jk>

so... rite now is it

theone & tare --2
leeyuhang --1

??????????

just checkin...

No we both have 3 and leehuyang's got 1.

[quote="notfunny"]noon will work for me

n this time im gonna win[/quote]
cough...
ackkkkk.... :P
cough
cough

It would be interesting if tare and I tied for all of the next contest questions and we'd have like 23 each.

Mmm, then we'd have to have a tiebreaker round which I'm hoping we won't because my computer's slow, can we do it over the phone? I got 3-way calling! :)

yeah but what about long distance dailing? theone lives in california.

We can afford it. I mean how long will it take? The whole round takes about 5-10 minutes so unless we veer off to non-math topics it should take only like 2 minutes (literally! :lol:)


function
modular arithmetic
number theory theorems
number theory

[quote="dieuhuynh"]$f: N\to N$ and $f(m+n)=f(n) mod m$. Has function $f$  propertys?[/quote]

$f: \mathbf{N}\to \mathbf{N}$ and $f(m+n)\equiv f(n) \pmod{m}$

 :)

And what is your questions? To determine all such functions? ..

[quote="dieuhuynh"]have you theorem about it?[/quote]

Well, I don't understand you. Do you want us to find all such functions, or ?? I don't understand your problem.

first ,sorry because my English.I want some one post a few problem form that.

By setting m=1, you get:
$f(1+n)=f(n) mod 1=0$. So for every $x\geq 2$, you get $f(x)=0$
Now, we set n=0 and m=1:
$f(1)=f(0) mod 1=0$
If we wanted to find out about f(0), we had to set m=n=0. But then, we had f(0) mod 0, but a division by zero is not allowed. Hence I do this for $f: N^+\to N$ and not for $f: N\to N$ (I know taht this isn't the question).
So now, for all positive integers x we have f(x)=0.

[quote="dieuhuynh"]$f: N\to N$ and $f(m+n)=f(n) mod m$. Has function $f$  propertys?[/quote]

The first thing is to show that there are functions with $(f(x)-f(y))/(x-y) \in \Bbb{Z}$ that are [i]not[/i] polynomials.   Second, I think there is some difference between the case $N \to N$ and $Z \to Z$.  Third, for the case $N \to N$ there is some Mahler expansion similar to p-adic functions,  $f = \Sigma a_{n}{x \choose n}$.

[quote="Blender"]By setting m=1, you get:
$f(1+n)=f(n) mod 1=0$. So for every $x\geq 2$, you get $f(x)=0$[/quote]

$f(1+n) \equiv f(n) (mod\ 1) \equiv0 (mod\ 1)$ <=> $1|f(n+1)$
you are do : [quote]$f(1+n) \equiv 0 (mod\ 1)$ <=> $f(n+1)=0$ :!:  :?: [/quote]

A problem :
$a,b$ is positive integer and $f(0)=2, f(1)=a$ and $f(n+2)=af(n+1)+bf(n)$ Prove that
 $f(p^{k}) = f(p^{k-1}) mod p^{k}$ :)


geometry
3D geometry
sphere
algebra solved
algebra

Let's define a band as being the region between 2 parallel and distinct lines in the plane. Does there exist a set of points in the plane s.t.

a) the distance between every 2 is >= 1.
b) every band in the plane contains at least a point from the set.

I have an answer, but I haven't even bothered to prove it (the answer was given to me by my teacher and the problem was spoilt for me  :D ).

I know the following truly WONDERFUL Iranian problem about bands:

Define the width of a bind as the distance between these two parallel lines.
Suppose a circle of radius 1 is covered completely by finitely many bands.
Show that the sum of the widths of these bands is at least 2.

Unfortunately ( :D ) I happen to know the idea (indeed a WONDERFUL one) for the problem Arne posted. I say unfortunately because this spoils the problem:

Look at the circle as the projection of a sphere and the pands as projections of sphere zones (or segments, I knever knew exactly how those are called  :) ). We do this because in space, the volume of such a sphere zone only depends on the distance between the 2 parallel planes = the distance between the parallel lines of the band which is the projection of the zone. These zones cover the sphere completely, so see what you get from here.

[quote="grobber"]We do this because in space, the volume of such a sphere zone only depends on the distance between the 2 parallel planes = the distance between the parallel lines of the band which is the projection of the zone.[/quote]

I think you mean the area of the sphere zone, not its volume :)

Hmmm Do I? Now you got me all confused  :D Could you post the actual solution? I don't know the formulas for the volumes and areas and stuff by heart, and I would have to deduce them.

Suppose two parallel planes with distance d define a sphere zone.
The area of this sphere zone is 2 \pi R d, where R is the radius of the sphere.

If the circle is covered, the sphere is covered also.
Let di, 1 \leq i \leq k be the widths of the bands.
Then obviousy \sumi2 \pi R di \geq 4 \pi R2 since the sphere is covered.
We are given R = 1 so only \sumidi \geq 2 remains :).

Yup, so I stand corrected. Do you have any more information about this? Where was it proposed? It seems to be part of maths folklore.

Oh, don't you people forget about the problem I posted! :D

It was problem 6 of the second round of the Iranian olympiad 2002.
It was also used afterwards in a Belgian selection test for IMO 2003 :)

Iranian problems seem really nice. Where would one find the last Iranian olympiads? (I mean above 2000).

I've got all 2002 problems and some 2001 problems.

How about posting them here somewhere? If you have enough time, and if it's not illegal! :D  :D

P.S. What a successful topic! :D

I'll type them in (I have them on copies) - maybe they can be put online or so.

The Iran 2002 problems (18 in total) will be soon in the downloads section :)


inequalities
real analysis
real analysis unsolved

Let us define $ x_k=(\sum_{i_1+..+i_k=n}{a_1^{i_1}...a_n^{i_n}})/C(n+k-1,n)$ where $ C(n,k)=n!/(k!(n-k)!)$. Prove that the sequence $\sqrt[k]{x_k}$ is increasing. Application: deduce great Myth's inequality!  :). Here $a_1,...,a_n$ are positive real numbers.

We were discussing something like that and Fiachra proved some result.
BTW, there is a typo: $i_1+...+i_n=k$ and you must divide to $C(n+k-1,n-1)$.

Yes, sorry. Indeed, but the result he proved I think is weaker than this one.  ;) . I cannot find the topic, though.

[quote="harazi"]Let us define $ x_k=(\sum_{i_1+..+i_k=n}{a_1^{i_1}...a_n^{i_n}})/C(n+k-1,n)$.[/quote]
The summation should be over $i_1+\ldots+i_n=k$. That was quite clear, though, so sorry for spamming  :)


calculus
integration
LaTeX
geometry
special factorizations

For how many integral values of M is M^2 + 100M + 1000 a perfect square?

choice are:
1) 16
2) 8
3) 4
4) 3
5) Less than 3

PS: I'm a newbie here, hey how do you guys write those mathematical symbols like M^2 and square roots symbol etc.
Thanks in advance[/hide]

[hide="Hint"]
$ q^2 \equal{} m^2 \plus{} 100m \plus{} 1000 \equal{} m^2 \plus{} 100m \plus{} 2500 \minus{} 1500 \equal{} (m\plus{}50)^2 \minus{} 1500$

Let $ p \equal{} m\plus{}50$. We then have: $ q^2 \equal{} p^2 \minus{} 1500 \iff p^2 \minus{} q^2 \equal{} 1500 \iff (p\plus{}q)(p\minus{}q) \equal{} 1500$ for $ (p,q) \in \mathbb{N}^2$
[/hide]

If you go on the left-hand side menu, look in the LaTeX area and it will teach you how to use LaTeX code. Enclose mathematical formula code within dollar signs to typeset it.

[hide="solution"]We write $ m^{2} \plus{} 100m \plus{} 1000 \equal{} k^{2}$, for some nonnegative integer $ k$. Completing the square and rearranging, we get $ (m \minus{} 50)^{2} \minus{} k^{2} \equal{} 1500$, or $ (m \minus{} 50 \minus{} k)(m \minus{} 50 \plus{} k) \equal{} 1500$. If we let $ a \equal{} m \minus{} 50 \minus{} k,b \equal{} m \minus{} 50 \plus{} k$, we get that $ m \equal{} \dfrac{a \plus{} b}{2} \plus{} 50,k \equal{} \dfrac{b \minus{} a}{2}$, so any $ a,b$ such that $ b > a$ and $ 2|(a \plus{} b)$ will lead to a solution. $ 2|(a \plus{} b)$ implies that either $ a,b$ are both odd or both even. However, they cannot both be odd, since they multiply to an even number, 1500. Therefore, $ a$ and $ b$ are two even numbers, with $ a < b$, that multiply to 1500. We find that 1500 has 16 even factors, and since $ a < b$, there are 8 such pairs.[/hide]

(edit) Looks like The Zuton Force beat me to it.


probability
function
logarithms
combinatorics unsolved
combinatorics

Given $n$ points in the plane we add $k \geq n-1$ edges randomly; what is  the probability that the graph formed is planar.

Do you mean, (a) "is the abstract graph formed planar" , or (b) "do two edges cross each other"?

For (a), there is phase transition; for some smoothly increasing function $f(n) \to \infty$, the probability goes to $0$ if  $k >> f(n)$, and the probability goes to 1 if $k << f(n)$.  I don't remember how large $f$ has to be, something like $n \log (n)$, and I think that nonplanarity develops earlier (i.e. after adding fewer edges) in a random graph than the existence of any fixed nonplanar graph.  I am certain that $f(n)$ exists, because (non)planarity is a monotone graph property.  I believe it is also known or conjectured that the phase transition is sharp; if $k/{f(n)}$ is bounded away from $1$ then we also get the 0-1 behavior.   Bollobas' book on random graphs should have the answer to this.

For (b), I guess it might depend on the position of the $n$ points.

i meant (a) thanks i will look it on bollobas's Random Graphs.


LaTeX
search
function
trigonometry
set theory
\/closed

hi,

looks nice (haven't checked out the new features yet - sorry, no time...).

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=frac&t=38958

this is what happens if one highlights a string that appears in the latex code. [edit: fixed now, thanks valentin.]

also, i think searching for latex code yields pretty crazy results, but i am not sure... would experiment a bit more but i am heavily busy right now.

by the way, this here, IMHO, really sucks:

[quote]You cannot make another search so soon after your last; please try again in a short while.[/quote]

is it really necessary or just some standard phpBB setting? in order to prevent bot attacks, one could use image authentification (it would be less aggravating)...

  dg

  darij

Yeah, that is a normal setting for all forums. There is a flood limit for search (searches impose high loads on the server). 10 users searching the db every 5 seconds would result in a major slowdown.

That limit is really annoying if you want to search for something properly.

But the real big problem is: the search doesn't work. Don't know why or what or who or..., but it simply impossible to use the search:
When I look e.g. for "search function bug", I get exactly 1000 matches, all of them useless (e.g. what has http://www.mathlinks.ro/Forum/viewtopic.php?highlight=bug+function+search&t=91594 to do with that search¿¿¿).

My Browser is Firefox 1.5.0.4, operating system Windoof XP.

Search works just fine. You guys were not paying attention when searching and it is partially my fault. Before I had selected the default to bee search for [b]all[/b] entered terms, now it's set to [b]any[/b].

1000 is the limit of search results.

For example a [url=http://www.mathlinks.ro/Forum/search.php?ppage=0&search_id=733053861&highlight=search+function+bug&sort_dir=desc&start=50]proper search for topics that contain the words search + function + bugs[/url] return 216 results as it should.

I will take a look at the latex part later on.

the flood limit is still annoying, because a search for a math problem usually takes much more than 1 attempt.

as for the automatic "or" instead of "and" search, may be it is the reason why searching for latex code got hard (are signs like - considered to separate words?)...

  darij

Yeah I think so. I tried searching for frac and I didn't get any errors. I might just take that mode out.

Sorry for the above, I was just to much used to the "and"  :blush:

[quote="darij grinberg"]
as for the automatic "or" instead of "and" search, may be it is the reason why searching for latex code got hard (are signs like - considered to separate words?)...
[/quote]
No. I tried to search for e^ 1e \frac with the "and" option and I got 20 pages of posts most of which should not be there. Right before going to the new server I used exactly the same search and it gave me just a few results among which I easily found the post I wanted. Now, when I reduced the search to e^ or for e^* in College Playground>>Unsolved and proposed problems I got the "no posts or topics met your search criteria" message while I know for sure that there are tons of formulae with the exponential function there. So, the $\LaTeX$ part of the search is clearly malfunctioning.

Not quite, you can't search for less than 4 characters now. Again this is an anti-flood system. Rules will be set, and users with special auths (like moderators, old users, etc.) will be able to search as before (basically).

Hmmm... But there are plenty of English words that are less than 4 characters long. For instance, with new restrictions the search for "set theory" became impossible: the engine just ignores the word "set" and gives you the results for "theory", while it is clear that "set theory" is a more refined and meaningful phrase than just "theory". Shouldn't the restriction be based on the total length of the words in the search rather than on the length of each word separately? Seems like in an attempt to solve the flood problem a lot of minor inconveniences have been created. Is the situation with the flooding really so bad that we should make all these sacrifices?

I repeat myself: this is not the final version. I haven't looked over the search part yet, because there are still a couple of other things higher in the priority list, but I will deal with this.

Most probably guests will have the same search as it is now, whereas registred and logged on users will have search as it was before the new server.

is it just because of the string length?

a problem was posted yesterday: http://www.mathlinks.ro/Forum/viewtopic.php?t=97755 . i know exactly that it was discussed before, so i search for *\frac{1}{a}+\frac{1}{c}* (what is surely more than 4 letters long). 1000 matches...

on the other hand, with all the search engine bots registered, maybe it will be possible in the future to search on ML by using google for instance?

  darij

The problem is that currently the search
[code]*\frac{1}{a}+\frac{1}{c}*[/code]
is interpreted as
[code]*frac*[/code]

AH... so indeed { and } as delimiters...

what about a special environment allowing to search for strings of arbitrary symbols? for instance, by searching for [string]\sin a + \sin b[/string], i get exactly those topics that contain the string "\sin a + \sin b"? this would be very useful for searching latex code...

  darij

Darij, I will come back to this in a couple of days. For now I have really more pressuring bugs to fix.

Any suggestions on how we best search latex for the time being?

Like I look for a topic on integer sols of w^4=x^4+y^4+z^4... but x^4+y^4 gives no matches at all, just like any other latex fragment.


Here is the problem of English from Japanese university entrance exam.

Strange to say, what do you think how the problem is difficult for non- native English speaker?

Please answer the question. Any comments would be appreciated.

Thanks in advance.

kunny

Wow. That would be hard for native English speakers. I don't think it's appropriate at all to test fluency in English - it seems more like an IQ test or something - but who am I to make that decision?

I would definitely think a lot of the vocabulary is difficult for non-native speakers, and the fill-in-the-blanks with nearly identical phrases, and IIIA.

My biggest problem with this test is trying to get around giving you choices (IIIA and VA). It seems to me that the only difference between that and actually giving choices is luck and logic.

Thank you for your commemt, jb05. :) This year's Waseda University entrance exam seems to be difficult, anyway this English test is difficult,  I think.

Hmm, isn't it something like CPE level? :)

It has been passed one years since the last year's problem of English.

Here is this year's problem of Waseda University.

Could you teach answer to the problem?

Thanks in advance.

kunny

After a quick look through the tests, they both seemed easy enough and I could reasonably expect a well-read (native English speaker) middle-schooler to do well on the tests.

It does look difficult for a second language test , though.

You said you wanted help with the test; what sections are you talking about, specifically?

Without specific knowledge,  I'd say the best thing you could do is read a lot of books that are in English and perhaps watch some television series.  The vocabulary seems to be the most important thing for tests like these, so I'd pay particular attention to that.


You have triangle ABC, where angle ABC=angle BCA=40. The angle bisector of angle ABC intersects side AC at D, and is extended to E such that DE=DA. Find the measure of angle DEC.[/code]

[hide="Solution"]
By elementary means, we can find that $ m\angle EDC \equal{} 120^{\circ},m\angle B \equal{} 20^{\circ},m\angle BCD \equal{} 100^{\circ}$. We then extent $ \overline{BC}$ to a point $ F$ such that $ \triangle BCF$ is right.

Let $ m\angle DEC \equal{} y, m\angle DCE \equal{} x, m\angle ECF \equal{} x_1,m\angle CEF \equal{} y_1$.

We have that $ 20^{\circ} \plus{} y \plus{} y_1 \equal{} 90^{\circ},100^{\circ} \plus{} x \plus{} x_1 \equal{} 180^{\circ},x_1 \plus{} y_1 \equal{} 90^{\circ}$, and $ x \plus{} y \equal{} 60^{\circ}$.

Solving, we have $ y \equal{} m\angle DEC \equal{} 20^{\circ}$.
[/hide]

Angle EDC is 60 by elementary methods. My teacher said the answer was 80. Let me see if your method can still apply.

If BC is extended to F, then B,C,and F are collinear and not a triangle.

:oops:

I meant $ \triangle BFE$. I don't see where my proof is wrong otherwise, though.

How is angle BCD 100 degrees? C is on AC so angle BCD is 40 degrees.
IF only CA was a bisector of angle BCE...

Let $ F$ be on $ BC$ and $ BF\equal{}BD.$ Then $ \angle CFD\equal{}100^\circ, CF\equal{}FD$ and $ \Delta CFD \sim \Delta BAC$.
So $ \frac{CD}{CF}\equal{}\frac{BC}{BA}$ and as $ BD$ is angle bisector $ \frac{BC}{BA}\equal{}\frac{CD}{DA}$ or $ CF\equal{}DA$.
We have $ BC\equal{}BF\plus{}CF\equal{}BD\plus{}CF\equal{}BD\plus{}DA\equal{}BD\plus{}DE$ or $ BC\equal{}BE$ so $ \angle BEC\equal{}80^\circ$.

OH WHOOPS

I read the problem as [quote]
You have triangle ABC, where angle ABC=angle [b]BAC[/b]=40. The angle bisector of angle ABC intersects side AC at D, and is extended to E such that DE=DA. Find the measure of angle DEC.[/quote]

In which case my answer would be correct... here's a scanned diagram of my messy work with such conditions.

[hide="outline of a different solution"] (stolen from WOOT) We show that AD+DB=BC. Consider the point F on BC such that BD=BF. It suffices to show that AD=FC. By angle chase, DF=FC. Also by angle chase, ABFD is cyclic, and BD bisects <ABF so AD=DF, and thus AD+DB=BC.

DA=DE, so BE=BC and the result follows.[/hide]

[hide="quick, I think"]Because $ BD$ is an angle bisector, we can flip $ \triangle DAB$ over the angle bisector and $ A'$ will fall on $ BC$ (yay).  Now $ DA'\equal{}DE$ so $ EDA'$ is isosceles! And a quick angle chase gives us $ EDA'$ is bisected.  Thus $ ECA\equal{}ACB$ and we are done. [/hide]


inequalities
number theory unsolved
number theory

Let $ S$ be the set of odd integers greater than $ 1$. For each $ x \in S$, denote by $ \delta (x)$ the unique integer satisfying the inequality $ 2^{\delta (x)}<x<2^{\delta (x) \plus{}1}$. For $ a,b \in S$, define:

$ a \ast b\equal{}2^{\delta (a)\minus{}1} (b\minus{}3)\plus{}a.$

Prove that if $ a,b,c \in S$, then:
$ (a)$ $ a \ast b \in S$ and
$ (b)$ $ (a \ast b)\ast c\equal{}a \ast (b \ast c)$.

(a)$ \delta(x)\ge1\forall x \in S, 2\mid b \minus{} 3 \implies 2 \nmid 2^{\delta(a) \minus{} 1)}(b \minus{} 3) \plus{} a \ge a > 1 \implies a \ast b \in S$.
(b)just extend that relation using $ \delta(a \ast b) \equal{} \delta(a) \plus{} \delta(b) \minus{} 1$ (infact $ 2^\delta(a) \plus{} 2^{\delta(a) \minus{} 1}(b \minus{} 3) < a\ast b < 2^{\delta(a) \plus{} 1} \plus{} 2^{\delta(a) \minus{} 1}(2^{\delta(b) \plus{} 1} \minus{} 4) \equal{} 2^{\delta(a) \plus{} \delta(b)}$).


Hi all how you doin? :)
Guys i had a question a bout the song "You Were My Everything" by Aviation, i searched a lot in google but couldnt find anythin useful :( , its a beautiful song and seems its related to a movie or something, does anyone know which movie?
would be really thankful if someone help me with this one... :)

If you put it into wikipedia, it will most likely have a list of references in pop culture.  If it doesn't show up on that list, then you're imagining things.

finally i figured it out, but not by wikipedia, it was really hard :D , the singer is [b]Chuckie Akenz[/b], Wow he's really cool, specially his other song feat. Kangel called :"Love Hurts" is amazing...he is a Vietnamese-Canadian Rapper and i think he's one of the Bests! Wow :lol:

[img]http://i17.tinypic.com/6xbp4dx.jpg[/img]


If we are currently in November, what month will it be 200 months from now?

200%12=8
so in 8 months it will be july.

remainder upon division of 200 days by 12 is 8... july

[hide]its July. (years dont matter  :rotfl:) [/hide]


geometry
rectangle
perpendicular bisector
geometry unsolved

let $ABC$ be triangle with$\angle{BAC} =135^{o}$and AM;BN be two of its altitudes (M on Bc ; N on CA). the line MN cuts the perpendicular bisector of AC at P.Let D and E be the midpoints respectively of NP and BC .
 Prove that$ADE$ is a right isosceles triangle .

[quote="hucht"]If yopu are reffering to triangle DEP, the problem is wrong  :|[/quote]

Probably triangle $AED$

Daniel

[quote="dhernandez"][quote="hucht"]If you are reffering to triangle DEP, the problem is wrong  :|[/quote]

Probably triangle $AED$

Daniel[/quote]

No, $\triangle AED$ is isosceles but isn't rectangle.

[quote="hucht"]
No, $\triangle AED$ is isosceles but isn't rectangle.[/quote]

Can't see what is wrong in my solution then

1) $\angle BAC = 135, \quad \angle BNC = 90, \quad \angle AMB = 90 \Rightarrow ANBM$ cyclic hence $\angle ANP = \angle ABC$
A litlle bit of angle chasing now
2) Also $\angle NAB = \angle NBA = 45 \Rightarrow \angle NMB = \angle CMP = 45$
3) $AF = FC = FM$ ($FM$ median of $AMC$ rectangle) $\Rightarrow \angle FMC = \angle ACB$ hence using 2) $\angle FMP = 45 + \angle ACB = 90 - \angle ABC$ cause $\angle ABC + \angle ACB = 45$
4) Also by 1) $\angle FPD = 90 - \angle ABC$ hence $\triangle FDP$ isosceles
5) Because of 3) and 4) the circumference with center $F$ and radius $FA$ passes through the four points $A$, $M$, $P$ and $C$. Therefore $\angle APM = \angle ACB$
6) Using 1) and 5) you have that triangles $APN$ and $ABC$ are similar with $AD$ and $AE$ respective medians hence triangles $ABE$ and $ADN$ are similar
7) $\angle ADN = \angle AEB$ hence $ADME$ cyclic and finally $\angle ADE = \angle AME = 90$, $\angle AED = \angle AMN = 45 \Rightarrow \triangle ADE$ is isosceles and rectangle

Daniel


For $x \ge 0$, we let $\lfloor{x}\rfloor$ denote the integer part of $x$, and we let $\langle x \rangle$ denote the fractional part of $x$. Thus $\lfloor{\pi}\rfloor = 3$ while$\langle \pi \rangle = .14159265...$ Find the area of the region described by $x \ge 0$, $y \ge 0$, and $100\langle x \rangle \ge \lfloor{x}\rfloor + \lfloor{y}\rfloor$.

Hmm... I did a lot of stuff that I'm not very confident about and don't want to spend a lot of time writing up since it's probably wrong... but, is it this?
[hide]
1717
[/hide]

I got[hide]$1691.25$.  I found the area was made up of 100 trapezoids, of which could be found by evaluating the sum $\sum_{k=1}^{100}\frac{(102-k)(\frac{100-k}{100})}{2}$  Can someone confirm if this is correct?[/hide]


Trong số các bạn VN có bạn chien than, tinh thần hăng hái rất cao, nên đã post bài rất nhiều. Tuy nhiên những bài của bạn phần nhiều được lấy từ các sách báo, tạp chí.

Điều này không nên làm chút nào. Nếu em muốn biết thêm về những bài đó thì hãy tìm đọc ngay trong diễn đàn ML này bằng cách search.
-----------------------------------
[color=blue]Tuân: Tôi edited tên topic :)[/color]

Em nó là học sinh cấp 2 mà Thuận? Nó chưa thể tham gia thảo luận ngay được, cần có thời gian. Mà chưa chắc hắn đã biết sử dụng search đâu. Hay ông dạy cho em nó luôn trong này đi :P

search bằng cách nào hả 2 thày  :D Chẳng lẽ lại mở từng topic để tìm ạ  :roll:  :P

[quote="N.T.TUAN"]Đây này, nhưng phải chọn từ khoá khéo vào :P[/quote]
Thế mình gõ công thức toán vào thì nó có search không thầy :)
Search tên topic biết bao giờ xong :P

Được thì phải, nhưng bỏ hai cái dấu [code]'$'[/code] đi. Mà đừng gọi tớ là ''thầy'', gọi là anh đi :D

gọi thày thì lớp đàn em như tụi em mới trẻ ra được  :rotfl: Với lại thày là thày trò là trò cứ gọi là thày đi mọi người cho thày Tuân đã già lại còn già hơn  :P

Ôi!  :(  Cái thằng ranh này, khéo tôi chết già với nó thôi :D Gọi là anh đeeeeeeeê...! :P Mà này, spam vừa thôi, không là anh thịt chú đấy! Bài này của anh xem như là bài nhắc nhở, nghe chưa?

chien than,
Trên kia anh đã thanh minh giúp chú, nhưng chú làm cái gì thế này? http://www.mathlinks.ro/Forum/viewtopic.php?t=147507
Đấy là box dành cho các sinh viên bậc Đại học. Chú post linh tinh vào đấy làm gì? Tên topic đặt cũng lung tung nữa. Chú phải cẩn thận với các post của mình, anh nghĩ chú chỉ nên tham gia các box từ Pre-Olympiad trở xuống mà thôi, lúc nào đủ mạnh hãy tham gia các box cao hơn.

Em Chien Than này sao khó nhắc nhở quá. Có một điều đơn giản thế mà em không chịu tiếp thu. Em vẫn làm theo cách trước đây (không chấp nhận được) là thế nào????

Vừa gửi cho em nó cái link của bài viết này rồi đấy, hay ông thử kết hợp PM với nó xem. Chịu thua rồi :D

[quote="vietkhoa"]Ở bên box Pre-Olympiad cũng có bạn VN post bài trên báo THTT khi chưa hết hạn. Như vậy có đúng không ạ???Có lẽ các bác mod box VN nên nhờ mod box Pre-Olympiad can thiệp. :D[/quote]
Chú cứ đưa cái link lên đây, anh em sẽ nhờ họ lock ngay thôi :D

[quote="vietkhoa"]Có link đây ạ. [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=151097]Link[/url][/quote]
cái này thì muộn rồi, vì đã có thành viên post lời giải cho bài toán đó :D

Quả này lại đến tên Chỉnh,xấu hổ dân sư phạn quá :blush:

http://www.mathlinks.ro/Forum/viewtopic.php?p=966726#966726

Chào các bạn.
Tôi cứ thắc mắc tại sao nhiều thành viên số bài post tăng 1 cách khủng khiếp vậy, hóa ra là có những topic để spam thoải mái như thế này.
Tôi có lời này không biết có nên nói ra ko: [hide]có lẽ chúng ta nên dừng mấy cái trò này lại, nó đang tiêu tốn thời gian một cách khá vô ích để làm những việc có ý nghĩa hơn cho cuốc sống.[/hide]
Nếu bạn nào bất bình thì cho tôi xin lỗi nhá.
[i]Những lời nói thật vì mong muốn tốt.[/i]

[quote="N.T.TUAN"]Trong đây không đếm bài.[/quote]
Em ko nghĩ vậy, rõ ràng em đã kiểm tra, số bài post có tăng lên, chứng tỏ vẫn đếm bài.

em là thành viên mới mong các anh chỉ cho em cách post  bài nha  :D

[quote="onlylove_math"][quote="L_Euler"][quote="N.T.TUAN"]Trong đây không đếm bài.[/quote]

Em thấy vẫn đếm anh à, anh điều chỉnh đi, trong này nhiều spam quá!!!  :(  :([/quote]
Đúng đấy , trong này vẫn tính bài mà . Nguy hiểm thật .
Mà anh L_Euler cũng spam nhiều trong này nhỉ :P[/quote]

Lâu lawms mới vào ML, hôm nay Spam cho "tăng bài viết"!!! :rotfl:  :rotfl:

Em không ở VN nên tò mò muốn hỏi là. Các học sinh đạt giải trong IMO thì có được chế độ khen thưởng gì ở trong nước không
Edit: em muốn post ở phần giao lưu cơ, em post nhầm, xin lỗi mọi người

[quote="L_Euler"][quote="onlylove_math"][quote="L_Euler"]
Em thấy vẫn đếm anh à, anh điều chỉnh đi, trong này nhiều spam quá!!![/quote]
Đúng đấy , trong này vẫn tính bài mà . Nguy hiểm thật .
Mà anh L_Euler cũng spam nhiều trong này nhỉ :P[/quote]

Lâu lawms mới vào ML, hôm nay Spam cho "tăng bài viết"!!! :rotfl:[/quote]
Chất lượng hơn số lượng... Nhưng mình thích số lượng ;))

[quote="novae"]nick pinks liên tục post các bài trên THTT lên đây  :mad:[/quote]

ko ai ngăn được tên pinks sao? Thày Tuấn can thiệp đi chứ ạ! Chẳng công bằng tí nào... bức xúc quá, lâu không lên hết ông taylorcute lại đến ông pink..... :mad:

Tiếp tục đến mem này [url]http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=120429[/url].
Mọi người có giải pháp gì để ngăn chặn những tên mặt dày này ?

[quote="Love_Math1994"]Hs vn toan post de tren Thtt :D  :D  :D de roi lai la hs vietnam len ns la moi ng khoa topic,bao moi  ng k dc giai bai :rotfl: bon nc ngoai biet no khinh chet[/quote]
Anh nói chuẩn.


limit
floor function
function
number theory unsolved
number theory

[quote="shencaili"]No,because $ \frac{k}{\pi}$ is uniformly distributed modulo 1.
Then for example,the dense in $ (a,b)$ would be $ ((arcsin(b)-arcsin(a))+(arcsin(b-1)-arcsin(a-1)))/\pi$.[/quote]

Yes, I know $ k/\pi$ is uniformly distributed.  I am sorry shencaili, but I don't understand the rest of your post.  Could you please elaborate or explain what the next sentence is saying.  Thank you.

[quote="baz"]Prove or disprove that the sequence $ a_{k}=\cos(k)$ is uniformly distributed modulo 1.
Thank you.[/quote]
i have a question please,
[u]when we say that $ (u_{n})_{n\in\mathbb{N}}$ is uniformly distributed modulo $ 1$[/u].

[quote="baz"]Let $ (u_{n})$ be a sequence of real numbers. Then $ (u_{n})$ is uniformly distributed modulo 1 if for all $ a,b$ such that $ 0\le a<b\le 1$,
\[ \lim_{n\rightarrow \infty}\frac{\#\{r \le n : \{u_{r}\}\in [a,b] \}}{n}=b-a \]
where $ \{u\}=u-\lfloor u \rfloor$ denotes the fractional part of $ u$.[/quote]
there aren't another definition?

Why do you want another¿ Descriptively, an interval of length $ \epsilon$ is supposed to contain asymptotically $ \epsilon$ of the sequence' elements.
The definition is equivalent to some others, but all of them not shorter if that's what you don't like.

To settle the problem as mentioned above, here a more general version:
Let $ f: [0,1] \to [0,1]$ be a function and $ x_{n}$ be an equally distributed sequence. Then for $ f(x_{n})$ to be equally distributed, it is necessary that $ V(f[a,b]) = b-a$ for $ 0 \leq a \leq b \leq 1$ if $ V$ exists (here $ V$ means the volume of that set).

If anyone could prove the original statement, the sequence $ a_{k}=\cos(k)$ is uniformly distributed modulo 1, or explain shencaili's post, then I would really appreciate it.

Thank you everyone for your help and interest.

Sorry, what's not clear about mine...  :maybe:

shencaili wrote $ (k/\pi)_{k}$ being uniformly distributed implies ??the dense (the what?) in $ (a,b)$ would be $ ((arcsin(b)-arcsin(a))+(arcsin(b-1)-arcsin(a-1)))/\pi.$

I am not certain what you mean by volume, V?
where shencaili got $ ( (arcsin(b)-arcsin(a))+(arcsin(b-1)-arcsin(a-1)) )/\pi$?
and how this all fits together?

alright, I see what you mean by volume (you are just measuring the interval lengths)
and shencaili meant $ \arccos(\cdot)$ NOT $ \arcsin(\cdot)$.
Also, the volume of $ \{\cos(n)\}_{n}$ on $ [a,b]$ would be
\[ \frac{arccos(b)-arccos(a)+arccos(b-1)-arccos(a-1)}{\pi}\]
which is not $ =b-a$ for the right choices of $ a,b$; hence the sequence $ \{\cos(n)\}_{n}$ is not uniformly distributed.


geometry
cyclic quadrilateral
geometry proposed

Let $A, B, C, D$ - four various points of a plane.  Prove, that the straight lines which are passing through the middle of pieces $BC, AC$ and $AB$, accordingly, in parallel to $AD, BD$ and $CD$, accordingly, are crossed in one point or are parallel.
Sorry for my English. :)

Denote by $P,Q,R$ the midpoints of segments $BC$,$AC$,$AB$ , respectively.
$X$ is common point of lines passing through $P$ and $Q$.

Triangle $PQX$ is similar to $ABD$ in scale $\frac{1}{2}$, so $2QX$ = $BD$. But $2QR$ = $BC$, $QR$ is parallel to $BC$ and $QX$ is parallel to$DB$. It means that $RX$ is parallel to $DC$. Q.E.D.

My note: if $ABCD$ is a cyclic quadrilateral, point $X$ lay on circle corcumscribed about triangle $PQR$.


[quote="MCrawford"]On three tests, Molly scored 84, 83 and 91.  On the same three tests, Milly scored 86, 87 and 88.  What is the positive difference between the average scores?[/quote]

[hide]
(84+83+91)/3=258/3=86
(86+87+88)/3=261/3=87
87-86=1
The positive difference between the average scores is 1. :)
[/hide]

[quote="MCrawford"]On three tests, Molly scored 84, 83 and 91.  On the same three tests, Milly scored 86, 87 and 88.  What is the positive difference between the average scores?[/quote]

[hide="solution"]
|86 - 87| = 1
[/hide]

Another method
[hide]
Subract the test scores from each other

84-86=-2
83-87=-4
91-88=3

And find the absolute value of that average, [b]1[/b].
[/hide]

[quote="tarquin"]Here in Texas we have a lot of TMSCA Number Sense tests and competitions. There is a "trick" for getting averages quickly. It's so simple to understand though, I don't even know if it should be considered a trick. Just look at one number, take its difference from the other numbers, add the differences together, divide by however many numbers are there and add it to the number that you were comparing everything to. It is actually quite quick. [hide]So in this problem you can look at one of Molly's scores, 84, take its difference from the other 2, -1 and 7, add those together, get 6, divide by 3, get 2, add it to 84, 86. Then Molly's is found easily as 87. 1.[/hide][/quote]

That is the same trick that I explained...It is extremely helpful, and saves time. I'd reccomend that many of you become familiar with it.

Does anyone want to try and prove it?

[quote="deroundup378"]
answer is 1
[/quote]

instead of just saying the answer, explain how you do it. the explanation part becomes important comes proof oriented test in the future.

[quote="tarquin"]Take any number x, and the differences it has with other members of the set a, -b, and c. The mean of the set is (x + a + x - b + x + c)/3 or (3x + a - b + c)/3. Distributing the 1/3, we get  x + (1/3)a + (1/3)*-b + (1/3)c or x + 1/3(a - b + c).[/quote]
Good job... I hope that everyone sees how this works.

Although I would think it is more correct to use this instead:

Take any number x, and the differences it has with other members of the set a, [b]+[/b]b, and c. The mean of the set is (x + a + x [b]+[/b] b + x + c)/3 or (3x + a [b]+[/b]b + c)/3. Distributing the 1/3, we get  x + (1/3)a + (1/3)*[b]+[/b]b + (1/3)c or x + 1/3(a [b]+[/b] b + c)

Considering the fact that b=negative number or positive number...You really dont need to have the negative b there.  The difference also totally depends on the case we are given.  You won't want to use -b in a case where the difference is positive.   :D

You find the averages and subtract to get 1.


function
geometry
3D geometry
sphere
vector
conics
ellipse

SPAM ON MY FRIENDS! ITS TIME FOR VERSION 2 of the infamous word association game... same rules, so here goes:
NOML

911(o wow My post count hit exactly 911 now!!!!)

You post the first word that pops into your head when you see the last person's word.

Murther mother.

Simpsons. This message is no longer too small.

[quote="mustafa"]D'oh

Why isn't this in G&FF? Is this another plot to enlarge the number of posts people have?[/quote]
Cookie dough.

[quote="Peter VDD"]No, this is a dead topic and you just revived it...[/quote]
Yep. :rotfl:

Oh my.  I guess I forgot to check the post dates. Why isn't this entire forum locked or something? Nobody has posted in in for over a year, except for my careless mistake.  :oops_sign:  :whistling:

The forum is not locked because NOML still exists. The format is changing, so there's a period of dormancy during the reorganization.

To be honest, I wouldn't be distressed if this topic was locked or moved to GFF though. The game was for NOML players, and since NOML isn't going right now and anyone can post here, now it's pretty much spam.  :|


search
parameterization
geometry
geometric transformation
quadratics
ratio
Euler

Is it possible that a,b,c,d are positive integer and a-b, a-c, a-d, b-c, b-d, c-d are perfect square.

This is really brutual numbers! thank you.

You are welcome.

I managed to systematically search through all the smaller values of the parameters I used, and the smallest example I found is:
\[ a\equal{}548.777.477\]
\[ b\equal{}108.784.901\]
\[ c\equal{}9.144.577\]
\[ d\equal{}1\]
with
$ a\minus{}b\equal{}439.992.576\equal{}20.976^2$
$ a\minus{}c\equal{}539.632.900\equal{}23.230^2$
$ a\minus{}d\equal{}548.777.476\equal{}23.426^2$
$ b\minus{}c\equal{}99.640.324\equal{}9.982^2$
$ b\minus{}d\equal{}108.784.900\equal{}10.430^2$
$ c\minus{}d\equal{}9.144.576\equal{}3.024^2$

Of course, it is not guaranteed that this is the generally smallest example; it is the smallest one only among those ones which fit into my parametrization.

Here is one more interesting observation: the quadruple $ (a,b,c,d)\equal{}(548.777.477,539.632.901,439.992.577,1)$ is also a solution, although it has two common values with the previous example. Therefore, the quadruple in question doesn't need to be uniquely determined even if two members (furthermore, two [u]fixed[/u] members!) are known!

For the end, notice that the problem is actually equivalent to asking whether there exist three perfect squares whose pairwise differnces are also perfect squares (because we could always fix $ d$ and obtain the rest by translation, and the latter problem corresponds to $ d\equal{}0$).

A technique I would try:  $ a \minus{} b, b \minus{} c, c \minus{} d, a \minus{} d$ are a [url=http://en.wikipedia.org/wiki/Pythagorean_quadruple]Pythagorean quadruple[/url] and can be parameterized; then there are only two extra conditions $ a \minus{} c, b \minus{} d$.  If the parameterization is properly assigned, I believe we can build up a solution from Pythagorean triples alone.

[quote="Zubr"]If it isn't boring and hard for you. Could you tell me how you find this numbers?[/quote]
No problem.

I fixed $ d \equal{} 1$. Then $ c \minus{} 1 \equal{} k^2\Rightarrow c \equal{} k^2 \plus{} 1$. Since $ b \minus{} k^2 \minus{} 1 \equal{} u^2$ and $ b \minus{} 1 \equal{} v^2(\Rightarrow b \equal{} v^2 \plus{} 1)$, it follows $ (v \minus{} u)(v \plus{} u) \equal{} k^2$. Therefore $ v \minus{} u$ is a divisor of $ k^2$, say $ \delta$, and $ v \plus{} u \equal{} \frac {k^2}{\delta}$. Now one has $ v \equal{} \frac {\delta^2 \plus{} k^2}{2\delta}$ and it follows $ b \equal{} \left(\frac {\delta^2 \plus{} k^2}{2\delta}\right)^2 \plus{} 1$. We now need to fit $ a$. Notice:
$ a \minus{} \left(\frac {\delta^2 \plus{} k^2}{2\delta}\right)^2 \minus{} 1 \equal{} p^2$
$ a \minus{} k^2 \minus{} 1 \equal{} q^2$
$ a \minus{} 1 \equal{} r^2$
This gives
\[ q^2 \plus{} k^2 \equal{} r^2 \equal{} \left(\frac {\delta^2 \plus{} k^2}{2\delta}\right)^2 \plus{} p^2\quad(1)
\]
We now need to search for a number which could be written as a sum of two squares on two ways so that two of the summands (one from each pair) are in a way correlated. In order to do this, I remembered the identity
\[ \begin{array}{rl} (X^2 \plus{} Y^2)(A^2 \plus{} B^2) & \equal{} (XB \plus{} YA)^2 \plus{} (XA \minus{} YB)^2 \equal{} \\
& \equal{} (XB \minus{} YA)^2 \plus{} (XA \plus{} YB)^2\end{array}\quad(2)
\]
However, since our number is a square itself, too, this must also be taken into consideration. The simplest way I saw was to make $ X^2 \plus{} Y^2 \equal{} Z^2$ and $ A^2 \plus{} B^2 \equal{} C^2$, what further means $ X \equal{} \chi^2 \minus{} \omega^2$, $ Y \equal{} 2\chi\omega$, $ A \equal{} \alpha^2 \minus{} \beta^2$ and $ B \equal{} 2\alpha\beta$ (we shall not need $ Z$, $ C$ anymore). Now we want to fit $ |XA \minus{} YB| \equal{} k$ and $ |XB \minus{} YA| \equal{} \frac {\delta^2 \plus{} k^2}{2\delta}$ (no special reason why these two summands are chosen), what implies $ (\delta^2 \plus{} k^2)|XA \minus{} YB| \equal{} 2\delta k|XB \minus{} YA|$. I focused on the last relation: I defined
\[ \begin{array}{rl}Q(\alpha,\beta,\chi,\omega,k,\delta) & \equal{} (\delta^2 \plus{} k^2)\Big((\chi^2 \minus{} \omega^2)(\alpha^2 \minus{} \beta^2) \minus{} 2\chi\omega\cdot 2\alpha\beta\Big) \minus{} \\
& \minus{} 2\delta k\Big((\chi^2 \minus{} \omega^2)\cdot 2\alpha\beta \minus{} 2\chi\omega(\alpha^2 \minus{} \beta^2)\Big)\end{array}
\]
and started to search for its integer zeroes (actually, rational ones would be enough, because we could always scale everything, but that doesn't matter).

Notice that the above expression is quadratic with respect to $ \alpha$. Therefore, the discriminant must be a perfect square. After fixing $ \omega$ and $ \delta$ (no special reason why these two), the discrimant depends only on $ \chi$ and $ k$. For a guessed $ \omega$ and $ \delta$ (by trial and error) I decided to search $ \{1,2,\dots,100\}\times\{1,2,\dots,100\}$ for $ (\chi,k)$ (of course, this is where a computer comes into the play).

Suppose that we have managed to make the discriminant a perfect square. Two problems may occur:
Firstly, we would have to be extremely lucky in order to $ |XA \minus{} YB| \equal{} k$ and $ |XB \minus{} YA| \equal{} \frac {\delta^2 \plus{} k^2}{2\delta}$ be fitted. However, this is not a big issue: indeed, after dividing $ Q$ by $ k^2$ (what does not affect its zeroes) one could notice that the numbers $ k$ and $ \delta$ are not relevant by itself, only their ratio $ kv \equal{} \frac {\delta}k$ is. Therefore, in this step I [b]rechose[/b] $ k$ and $ \delta$ by taking $ k \equal{} |XA \minus{} YB|$ and $ \delta \equal{} kv|XA \minus{} YB|$, what settles the first problem.
Secondly (and more problematic), it could happen that after all the calculations we get $ a \equal{} b$. There is no good way to prevent this (at least I hadn't seen it), but to simply discard such solutions.

Return to the mentioned trial and error. After a little guessing (by no particular order), I found that $ (\omega,\delta) \equal{} (26,16)$ makes the discriminant a perfect square with
\[ (\chi,k)\in\{(4,15),(13,12)\}\quad(3)
\]
Solving $ Q(\alpha,\beta,4,26,15,16) \equal{} 0$ gives $ \alpha \equal{} \frac {65\beta}{27}$ (the other solution, $ \alpha \equal{} \minus{} \frac {27\beta}{65}$, is discarded). Now it is possible to choose $ \beta \equal{} 27$, and $ \alpha \equal{} 65$ follows. Also follows $ (X,Y,A,B) \equal{} ( \minus{} 660,208,3.496,3.510)$, and since $ kv \equal{} \frac {16}{15}$, $ (k,\delta)$ are rechosen as $ \left(| \minus{} 660\cdot 3.496 \minus{} 208\cdot 3.510|,\frac {16}{15}| \minus{} 660\cdot 3.496 \minus{} 208\cdot 3.510|\right)$, that is $ (k,\delta) \equal{} (3.037.440, 3.239.936)$. Comparing $ (1)$ and $ (2)$ further gives $ q \equal{} |XB \plus{} YA| \equal{} 1.589.432$, $ p \equal{} |XA \plus{} YB| \equal{} 1.577.280$ and $ r \equal{} \sqrt {(X^2 \plus{} Y^2)(A^2 \plus{} B^2)} \equal{} 3.428.168$, and now calculating $ a$, $ b$ and $ c$ gives the ones from my first post.

Note that if we choose $ (\chi,k) \equal{} (13,12)$ at $ (3)$, we will get $ a \equal{} b \equal{} 71.402.501$ (what is the mentioned problem).

After my initial post, I had automatized the whole process to do this for all $ (\omega,\delta)\in\{1,2,\dots,26\}\times\{1,2,\dots,16\}$, and that is how I came up with the numbers from my second message.

The Mathematica notebook I used is attached (it works with version 6.0, I haven't tried it with earlier versions but it should work as well).

It is possible, of course, that I made this too complicated, but hey, it works. :) I am, on the other hand, aware that this may be more complicated than you were expecting. However, I would be glad to additionally explain any problematic part, please don't hesitate to ask.

More news: the whole work described above was indeed unnecessary, because it is not hard too find the actually smallest quadruple. That is $ (a,b,c,d) \equal{} (485.810,34.226,23.410,1)$, with $ (a \minus{} b,a \minus{} c,a \minus{} d,b \minus{} c,b \minus{} d,c \minus{} d) \equal{} (672^2,680^2,697^2,104^2,185^2,153^2)$. In order to do this, we could use the mentioned fact:[quote="Bojan Basic"]For the end, notice that the problem is actually equivalent to asking whether there exist three perfect squares whose pairwise differnces are also perfect squares (because we could always fix $ d$ and obtain the rest by translation, and the latter problem corresponds to $ d \equal{} 0$).[/quote]
The following simple code will work:
[code]JeKvadrat[x_] := Floor[Sqrt[x]]^2 == x;
For[a = 1, a <= 1000, a++,
 For[b = 1, b < a, b++,
  For[c = 1, c < b, c++,
   If[JeKvadrat[a^2 - b^2] && JeKvadrat[a^2 - c^2] &&
     JeKvadrat[b^2 - c^2], Print["{", a, ",", b, ",", c, "}"]]
   ]
  ];
 Print[a]
 ][/code]
which after less than hour finds the triple [color=green]{697,185,153}[/color], giving $ a \equal{} 697^2 \plus{} 1$ and so on.

One more interesting thing to notice is that neither this quadruple is determined by two fixed values, because $ (a,b,c,d) \equal{} (485.810,462.401,451.585,1)$ is another solution.

That's it. I'm over with this one. :)

[quote="Bojan Basic"]More news: the whole work described above was indeed unnecessary, because it is not hard too find the actually smallest quadruple. That is $ (a,b,c,d) \equal{} (485.810,34.226,23.410,1)$, with $ (a \minus{} b,a \minus{} c,a \minus{} d,b \minus{} c,b \minus{} d,c \minus{} d) \equal{} (672^2,680^2,697^2,104^2,185^2,153^2)$.[/quote]
This solution was known to Euler in the context of finding [url=http://mathworld.wolfram.com/PerfectCuboid.html]Perfect Cuboid[/url]. Namely, the triple $ (672,104,153)$ form sides of a brick where the main diagonal as well as all face diagonals, except one, are integer.

btw, I think taking $ d \equal{} 0$ is more natural in this problem and leads to more "symmetric" solutions. Why do you take $ d \equal{} 1$?

Because the problems wants _positive_ integers  :D
Nonetheless, this condition is never needed  :wink:


ratio
projective geometry
geometry proposed
geometry

Dear Mathlinkers,
let ABC be a triangle, Y a point, A'B'C' the Y-cevian triangle of ABC, PQR a triangle inscribed in ABC,
L, M, N the points of intersection of QR and B'C', RP and C'A', PQ and A'B'.
Prouve without Ceva or Menelaus's theorem : PL, QM and RN are concurrent in general position.
Sincerely
Jean-Louis

Dear Mathlinkers,
you can see : http://tech.groups.yahoo.com/group/Hyacinthos/message/18066
and the proof of this new result on
http://groups.yahoo.com/group/Hyacinthos/files/A%20new%20result.doc

But I haven't a synthetic one.
Sincerely
Jean-Louis

[b][size=100][color=DarkBlue]LEMMA. - A triangle $ \bigtriangleup ABC$ is given, with an arbitrary point $ Y$ inwardly to it and let $ \bigtriangleup A'B'C',$ its cevian triangle. Let $ \bigtriangleup PQR,$ an arbitrary triangle inscribed in to $ \bigtriangleup ABC$ $ ($ $ P\in BC$ and $ Q\in AC$ and $ R\in AB$ $ )$ and we denote the points $ L\equiv B'C'\cap QR$ and $ M\equiv A'C'\cap PR$ and $ N\equiv A'B'\cap PQ.$ Prove that the point $ K\equiv PQ\cap MN,$ is the harmonic conjugate of $ L,$ with respect to the points $ Q,\ R.$[/color][/size][/b]

[b][size=100]PROOF OF THE LEMMA.[/size][/b] – We denote the point $ T\equiv BC\cap B'C'$ and then, the points $ B,\ A',\ C,\ T,$ are in harmonic conjugation as well $ ($ from the complete quadrilateral $ AB'YC'BC$ $ ).$

So, the pencil $ B'.BA'CT$ is also harmonic and then the points $ B,\ D,\ Q,\ E,$ where $ D\equiv BQ\cap B'C'$ and $ E\equiv BQ\cap B'A',$ are also in harmonic conjugation.

We have now, that the pencil $ C'.BDQE$ is harmonic and so, we conclude that the points $ R,\ L,\ Q,\ K\equiv QR\cap C'E,$ as the point of intersection of this pencil from the line segment $ QR,$ are also in harmonic conjugation.

Because of the collinearity of the points $ B\equiv RC'\cap QE$ and $ P\equiv RM\cap QN$ and $ A'\equiv C'M\cap EN,$ based on the [b][size=100]Desarques theorem[/size][/b], we conclude that the triangles $ \bigtriangleup C'RM,\ \bigtriangleup EQN$ are perspective ans so, the line segments $ C'E,\ RQ,\ MN,$ are concurrent at one point, here the point $ K.$

Hence, the line segment $ MN,$ passes through the point $ K,$ as the harmonic conjugation of $ L,$ with respect to the points $ Q,\ R$ and the proof of the [b][size=100]Lemma[/size][/b] is cpmpleted.

$ \bullet$ Return now in the proposed problem, we denote the point $ K\equiv QR\cap MN$ and based on the above [b][size=100]Lemma[/size][/b], we have that this point is the harmonic conjugate of $ L,$ with respect to the points $ Q,\ R.$

Let be the points $ D\equiv PR\cap LN$ and $ E\equiv PQ\cap LM$ and it is easy to show that the line segment $ DE,$ passes through the point $ K,$ from the complete quadrilateral $ PDLEMN.$

Because of now, the collinearity of the points $ K\equiv DE\cap MN$ and $ Q\equiv PE\cap SM$ and $ R\equiv PD\cap SN,$ where $ S\equiv QM\cap RN,$ based on the [b][size=100]Desarques theorem[/size][/b], we comclude that the triangles $ \bigtriangleup PDE,\ \bigtriangleup SNM$ are perspective and so, the line segments $ PS,\ DN,\ EM,$ are concurrent at one point, here the point $ L.$

Hence the line segment $ PL,$ passes through the point $ S\equiv QM\cap RN$ and the proof is completed.

$ \bullet$ This proof is dedicated to [b][size=100]Sotiris Louridas[/size][/b].

Kostas Vittas.

I doubt that there is a mistake in the following proof, since the condition that $ \triangle ABC$ and $ \triangle A'B'C'$ are persective has never been used. Please point it out for me :blush: . Thank you.

First we state a lemma:

[b]Lemma.[/b] Let $ M,N$ be two points on the plane. The lines $ a,b,c$ pass through $ M$ and the lines $ a',b',c'$ pass through $ N$. Denote by $ A,B,C$ the intersections of $ a$ and $ a'$, $ b$ and $ b'$, $ C$ and $ C'$. Then, $ A,B,C$ are collinear iff $ (a,b,c,MN)=(a',b',c',MN)$.

[b]Proof of the lemma.[/b] Trivially :D

Now let us come back to the problem. Denote by $ X\equiv QM \cap RN$, $ L^*\equiv MN \cap QR$, $ I\equiv BQ\cap B'C'$, $ J\equiv A'N\cup BQ$, and $ \A_0 = B'C'\cap BC$. First we consider the two triangles $ \triangle C'RM$ and $ \triangle NJQ$ which are perspective through line $ BC$, so by Dersagues's theorem, they must also be perspective through a point. But since $ K=QR\cup MN$, $ K,C',E$ are collinear. By that:

$ (LL^*QR)=C'(LL^*QR)=C'(IJQB)=B'(IJQB)=B'(A_0A'CB)=-1$
$ \Rightarrow (MN,ML,MX,MP)=M(L^*LQR)=-1=N(L^*LRQ)=(NM,NL,NX,NP)$

So by the lemma, $ P,X,L$ are collinear and we are done.

But dear [b][size=100]Mashimaru[/size][/b], in the solution I presented above, which is familiar with the yours one, also has never been used the condition of the perspectivity of $ \bigtriangleup ABC,\ \bigtriangleup A'B'C'$ ( I think there is no problem, because it has been used the argument of the harmonic conjugation in the configuration of these triangles ).

The key ( I think ) of the solution of this nice problem, is the fact that for any mobile point $ P$ on $ BC,$ when $ Q,\ R$ have been considered as fixed on $ AC,\ AB$ respectively, the line segment $ MN,$ where $ M\equiv A'C'\cap PR$ and $ N\equiv A'B'\cap PQ,$ always passes through the point $ K$ on the line segment $ QR,$ as the harmonic conjugate of $ L\equiv B'C'\cap QR,$ with respect to the points $ Q,\ R.$

As a difference with respect to your proof, I preferred to finish the solution, applying the [b][size=100]Desarques theorem[/size][/b], instead of the [b][size=100]Double Ratio[/size][/b] argument in special case of harmonic conjugation you based on, in trying to use fewer theory of arguments.

Best regards, Kostas Vittas.

Dear Kostas,
thank you for your nice synthetic proof of this new result. Now the question : is this result fruitfull?
Sincerely
Jean-Louis

Dear Mathlinkers,
you can see tis result on
http://Perso.orange.fr/jl.ayme  vol. 8 Fontené's theorem p. 8.

Sincerely
Jean-Louis


[hide]So the final distance away is 2 blocks east and and 0 blocks north or south. So it is 2 blocks east.[/hide]

[quote="Interval"]What is the total displacement of a student who walks 3 blocks east, 2 blocks north, 1 block west, and then 2 blocks south?

(1) 0                      (3) 2 blocks west
(2) 2 blocks east      (4) 8 blocks[/quote]
[hide]The north and south cancel out and then 1 block west = -1 block east so then 3-1=2 blocks east.
So the answer is (2) 2 blocks east.[/hide]

[hide]two blocks east, draw a picture
anyways 3 east - 1 west =2 east then add the 2 north and 2 south which cancel out so just 2 east[/hide]

I want thank all of you for your great help.