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Difference between revisions of "2013 IMO Problems/Problem 4"

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Probably a simpler solution than above.
 
Probably a simpler solution than above.
  
−
As above, let <math>T = \omega_1 \cap \omega_2 \neq W.</math> By Miquel <math>AMTHN</math> is cyclic. Then since <math>\angle WCY = \angle WBX = 90^{\circ}</math> we know, because <math>W,B,X,T \in \omega_1</math> and <math>W,C,Y,T \in \omega_2,</math> that <math>\angle WTY = \angle WTX = 90^{\circ},</math> thus <math>X,T,Y</math> are collinear.
+
As above, let <math>T = \omega_1 \cap \omega_2 \neq W.</math> By Miquel <math>MTHN</math> is cyclic. Then since <math>\angle WCY = \angle WBX = 90^{\circ}</math> we know, because <math>W,B,X,T \in \omega_1</math> and <math>W,C,Y,T \in \omega_2,</math> that <math>\angle WTY = \angle WTX = 90^{\circ},</math> thus <math>X,T,Y</math> are collinear.
 
There are a few ways to finish.
 
There are a few ways to finish.
  

Revision as of 01:00, 25 February 2018

Problem

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

Hint

Draw a good diagram, or use the one below. What do you notice? (In particular, what do you want to be true? How do you prove it true?)


Solution 1

[asy] //Original diagram by suli, August 2014. Feel free to make edits, but please leave this comment in place. import olympiad; import math; unitsize(10); pair A = (15,25), B = (0,0), C = (20,0), W = (12,0); pair H = orthocenter(A, B, C); pair N = extension(C,H, A,B); pair M = extension(B,H, A,C); pair L = extension(A,H, B,C); path p1 = circumcircle(N, B, W); path p2 = circumcircle(C, M, W); pair X = intersectionpoints(B--(0,10), p1)[1]; pair Y = intersectionpoints(C--(20,10), p2)[1]; pair T = intersectionpoints(p1,p2)[0]; //Time to start drawing dot(A); dot(B); dot(C); dot(W); dot(H); dot(M); dot(L); dot(X); dot(N); dot(Y); dot(T); dot(circumcenter(N,B,W)); dot(circumcenter(C,M,W)); draw(p1); draw(p2); draw(A--B--C--Y--W--X--B); draw(C--A); draw(A--L); draw(B--M); draw(C--N); label("A", A, E); label("B", B, S); label("C", C, E); label("L", L, S); label("M", M, S); label("N", N, S); label("H", H, E); label("T", T, E); label("W", W, S); label("X", X, E); label("Y", Y, E); [/asy]

Let $T$ be the intersection of $\omega_1$ and $\omega_2$ other than $W$.

Lemma 1: $T$ is on $XY$.

Proof: We have $\angle{XTW} = \angle{YTW} = 90^\circ$ because they intercept semicircles. Hence, $\angle{XTY} = \angle{XTW} + \angle{YTW} = 180^\circ$, so $XTY$ is a straight line.

Lemma 2: $T$ is on $AW$.

Proof: Let the circumcircles of $NBW$ and $MWC$ be $\omega_1$ and $\omega_2$, respectively, and, as $BNMC$ is cyclic (from congruent $\angle{BNC} = \angle{BMC} = 90^\circ$), let its circumcircle be $\omega_3$. Then each pair of circles' radical axises, $BN, TW,$ and $MC$, must concur at the intersection of $BN$ and $MC$, which is $A$.

Lemma 3: $YT$ is perpendicular to $AW$.

Proof: This is immediate from $\angle{YTW} = 90^\circ$.

Let $AH$ meet $BC$ at $L$, which is also the foot of the altitude to that side. Hence, $\angle{ALB} = 90^\circ.$

Lemma 4: Quadrilateral $THLW$ is cyclic.

Proof: We know that $NHLB$ is cyclic because $\angle{BNH}$ and $\angle{BLH}$, opposite and right angles, sum to $180^\circ$. Furthermore, we are given that $NTWB$ is cyclic. Hence, by Power of a Point,

\[AT * AW = AN * AB = AH * AL.\]

The converse of Power of a Point then proves $THLW$ cyclic.

Hence, $\angle{WTH} = 180^\circ - \angle{WLH} = 90^\circ$, and so $HT$ is perpendicular to $AW$ as well. Combining this with Lemma 3's statement, we deduce that $T, H, Y$ are collinear. But, as $X$ is on $YT$ (from Lemma 1), $X, Y, H$ are collinear. This completes the proof.

$\blacksquare$

--Suli 13:51, 25 August 2014 (EDT)

Solution 2

Probably a simpler solution than above.

As above, let $T = \omega_1 \cap \omega_2 \neq W.$ By Miquel $MTHN$ is cyclic. Then since $\angle WCY = \angle WBX = 90^{\circ}$ we know, because $W,B,X,T \in \omega_1$ and $W,C,Y,T \in \omega_2,$ that $\angle WTY = \angle WTX = 90^{\circ},$ thus $X,T,Y$ are collinear. There are a few ways to finish.

(a) \[BX \perp BC \perp AH \iff \angle NBX = \angle NAH\] \[\iff \angle NTX = \angle NTH \iff H \in TX\]so $H,X,Y$ are collinear, as desired $\square$

(b) Since $BNMC$ is cyclic we know $AN \cdot AB = AM \cdot AC$ which means $p(A,\omega_1) = p(A, \omega_2)$ so $A$ is on the radical axis, $TW,$ hence \[\angle ATX = \angle XBW = 90^{\circ} = \angle AMH = \angle ATH\] so $H$ lies on this line as well and we may conclude $\square$

--mathguy623 03:10, 12 August 2016 (EDT)

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