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

Revision as of 14:43, 6 June 2022

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)

Solution 3 (Complex Bash)

$[asy] unitsize(0.8cm); draw((0,0)--(14,0)--cycle); draw((0,0)--(5,12)--cycle); draw((5,12)--(14,0)--cycle); draw((2.071,4.97)--(14,0)--cycle); draw((8.96,6.72)--(0,0)--cycle); draw((2.071,4.97)--(9,0)--cycle); draw((8.96,6.72)--(9,0)--cycle); draw((0,0)--(0,2.083)--cycle); draw((14,0)--(14,6.75)--cycle); draw((0,2.083)--(14,6.75)--cycle); draw((0,2.083)--(2.071,4.97)--cycle); draw((0,2.083)--(9,0)--cycle); draw((8.96,6.72)--(14,6.75)--cycle); draw((14,6.75)--(9,0)--cycle); draw(circle((4.5,1.0415),4.61895)); draw(circle((11.5,3.375),4.20007)); label((0,0),"$C$",SW); label((14,0),"$B$",SE); label((5,12),"$A$",N); label((8.96,7),"$N$",N); label((2.071,4.97),"$M$",NW); label((8.25,0),"$W$",S); label((5,4),"$H$",N); label((0,2.083),"$Y$",NW); label((14,6.75),"$X$",NE); [/asy]$

For any point $P,$ let $p$ denote the complex number for point $P.$ First off, let $C$ be the origin. Now, since $WY$ is the diameter of the circumcircle of triangle $CMW$ , we must have $\angle{YMW}=\angle{YCH}=90^{\circ}.$ Since angles inscribed in the same arc are congruent, $\angle{MYW}=\angle{MCW}=\angle{C}.$ This means that $\frac{|w-m|}{|y-m|}=\tan{\angle{C}}.$ Combining this with the fact that $\angle{YMW}$ is right, we find that $i(y-m)\tan{\angle{C}}=(w-m).$ Solving, we find that $y=\frac{w+(i\tan{\angle{C}}-1)m}{i\tan{\angle{C}}}.$ We wish to simplify $(i\tan{\angle{C}}-1)m$ first. Note that $m=\frac{|CM|}{|CA|}\cdot(a)=\frac{(|BC|)\cos{\angle{C}}}{|CA|}\cdot(|CA|)(\cos{\angle{C}}+i\sin{\angle{C}})=(|BC|)\cos{\angle{C}}(\cos{\angle{C}}+i\sin{\angle{C}}).$ This means that $\begin{align*} (i\tan{\angle{C}}-1)m&=(i\tan{\angle{C}}-1)((|BC|)\cos{\angle{C}}(\cos{\angle{C}}+i\sin{\angle{C}}))\\ &=\left(\frac{i\sin{\angle{C}}}{\cos{\angle{C}}}-1\right)((|BC|)\cos{\angle{C}}(\cos{\angle{C}}+i\sin{\angle{C}}))\\ &=(i\sin{\angle{C}}-\cos{\angle{C}})(|BC|)(\cos{\angle{C}}+i\sin{\angle{C}})\\ &=-|BC| \end{align*}$ This means that $y=\frac{w-|BC|}{i\tan{\angle{C}}}=\frac{|BC|-w}{\tan{\angle{C}}}i.$ Now, since $WX$ is a diameter of the circumcircle of triangle $NBW,$ we must have $\angle{WNX}=\angle{WBX}=90^{\circ}.$ Since angles inscribed in the same arc are congruent, $\angle{NXW}=\angle{NBW}=\angle{B}.$ This means that $\frac{|x-n|}{|w-n|}=\cot{\angle{B}}.$ Combining this with the fact that $\angle{WNX}$ is right, we find that $(x-n)=i\cot{\angle{B}}(w-n).$ Solving, we find that $x=wi\cot{\angle{B}}+n(1-i\cot{\angle{B}}).$ We wish to simplify $n(1-i\cot{\angle{B}})$ first. Note that $n=e^{(90-\angle{B})i}|CN|=(\cos{(90-\angle{B})}+i\sin{(90-\angle{B})})(|BC|)\sin{\angle{B}}=(|BC|)\sin{\angle{B}}(\sin{\angle{B}}+i\cos{\angle{B}})).$ This means that $\begin{align*} n(1-i\cot{\angle{B}})&=(|BC|)\sin{\angle{B}}(\sin{\angle{B}}+i\cos{\angle{B}}))(1-i\cot{\angle{B}})\\ &=(|BC|)\sin{\angle{B}}(\sin{\angle{B}}+i\cos{\angle{B}}))\left(1-\frac{\cos{\angle{B}}}{\sin{\angle{B}}}i\right)\\ &=(|BC|)(\sin{\angle{B}}+i\cos{\angle{B}})(\sin{\angle{B}}-i\cos{\angle{B}})\\ &=(|BC|)(1) \end{align*}$ This means that $x=|BC|+wi\cot{\angle{B}}=|BC|+\frac{w}{\tan{\angle{B}}}i.$ This means that the line through complex numbers $x$ and $y$ satisfy the equation $\Im{(z)}=\left(\frac{\frac{w}{\tan{\angle{B}}}-\frac{|BC|-w}{\tan{\angle{C}}}}{|BC|}\right)\Re{(z)}+\frac{|BC|-w}{\tan{\angle{C}}}.$ If there is a fixed point to the line, then the real and imaginary values of the point must not contin $w$ . If the fixed point is $c,$ then we have $\left(\frac{\frac{1}{\tan{\angle{B}}}+\frac{1}{\tan{\angle{C}}}}{|BC|}\right)\Re{(c)}-\frac{1}{\tan{\angle{C}}}=0,$ after comparing the coefficient of $w.$ This means that $\begin{align*} \Re{(c)}&=\frac{\frac{1}{\tan{\angle{C}}}}{\frac{\frac{1}{\tan{\angle{B}}}+\frac{1}{\tan{\angle{C}}}}{|BC|}}=\frac{\frac{\cos{\angle{C}}}{\sin{\angle{C}}}}{\frac{\frac{\cos{\angle{B}}}{\sin{\angle{B}}}+\frac{\cos{\angle{C}}}{\sin{\angle{C}}}}{|BC|}}\\ &=\frac{\frac{\cos{\angle{C}}}{\sin{\angle{C}}}}{\frac{\frac{\cos{\angle{B}}\sin{\angle{C}}+\sin{\angle{B}}\cos{\angle{C}}}{\sin{\angle{B}}\sin{\angle{C}}}}{|BC|}}=\frac{\frac{\cos{\angle{C}}}{\sin{\angle{C}}}}{\frac{\sin{(\angle{B}+\angle{C})}}{|BC|\sin{\angle{B}}\sin{\angle{C}}}}\\ &=\frac{\frac{\cos{\angle{C}}}{\sin{\angle{C}}}}{\frac{\sin{(180-\angle{A})}}{|BC|\sin{\angle{B}\sin{\angle{C}}}}}=\frac{|BC|\cos{\angle{C}}\sin{\angle{B}}\sin{\angle{C}}}{\sin{\angle{A}\sin{\angle{C}}}}\\ &=\frac{|BC|}{\sin{\angle{A}}}\cdot(\cos{\angle{C}}\sin{\angle{B}}). \end{align*}$ From the Law of Sines, we find that $\frac{|BC|}{\sin{\angle{A}}}=\frac{|AC|}{\sin{\angle{B}}}.$ Substituting this, we find that $\begin{align*} \Re{(c)}&=\frac{|BC|}{\sin{\angle{A}}}\cdot(\cos{\angle{C}}\sin{\angle{B}})\\ &=\frac{|AC|}{\sin{\angle{B}}}\cdot(\cos{\angle{C}}\sin{\angle{B}})\\ &=|AC|\cos{\angle{C}}. \end{align*}$ This means that $\begin{align*} \Im{(c)}&=\left(\frac{\frac{w}{\tan{\angle{B}}}-\frac{|BC|-w}{\tan{\angle{C}}}}{|BC|}\right)|AC|\cos{\angle{C}}+\frac{|BC|-w}{\tan{\angle{C}}}. \end{align*}$ Since the $w$ 's cancel out, we can just discard everything with $w$ in it. Thus, $\begin{align*} \Im{(c)}&=\left(\frac{-\frac{|BC|}{\tan{\angle{C}}}}{|BC|}\right)|AC|\cos{\angle{C}}+\frac{|BC|}{\tan{\angle{C}}}\\ &=\left(-\frac{1}{\tan{\angle{C}}}\right)|AC|\cos{\angle{C}}+\frac{|BC|}{\tan{\angle{C}}}\\ &=\frac{1}{\tan{\angle{C}}}\left(|BC|-|AC|\cos{\angle{C}}\right). \end{align*}$ Since $|BC|=|AC|\cos{\angle{C}}+|AB|\cos{\angle{B}},$ we have $|BC|-|AC|\cos{\angle{C}}=|AB|\cos{\angle{B}}.$ Thus, $\Im{(c)}=\frac{1}{\tan{\angle{C}}}\cdot|AB|\cos{\angle{B}}.$ In conclusion, $c=|AC|\cos{\angle{C}}+\left(\frac{1}{\tan{\angle{C}}}\cdot|AB|\cos{\angle{B}}\right)i,$ which is the fixed point $XY$ always passes through. However, by inspection, $c=H=\text{the orthocenter}.$ Therefore, we conclude that $X,Y,H$ are collinear for all acute triangles $ABC.$

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