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2014 IMO Problems/Problem 3

Revision as of 02:12, 8 September 2018 by Mathdummy (talk | contribs) (→‎Solution)

Problem

Convex quadrilateral $ABCD$ has $\angle{ABC}=\angle{CDA}=90^{\circ}$. Point $H$ is the foot of the perpendicular from $A$ to $BD$. Points $S$ and $T$ lie on sides $AB$ and $AD$, respectively, such that $H$ lies inside $\triangle{SCT}$ and \[\angle{CHS}-\angle{CSB}=90^{\circ},\quad \angle{THC}-\angle{DTC} = 90^{\circ}.\]

Prove that line $BD$ is tangent to the circumcircle of $\triangle{TSH}.$

Solution

[asy] import cse5; import graph; import olympiad; dotfactor = 3; unitsize(1.5inch); path circle = Circle(origin, 1); // draw(circle); pair A = (0,1), C=(0,-1); pair Oo = (0,-0.05); pair Bb = rotate(-8,Oo)*(2,-0.05), Dd =rotate(-8, Oo)*(-2,-0.05); pair B = IP(Dd--Bb, circle, 1); pair D = IP(Dd--Bb, circle, 0); pair H = foot(A, Dd, Bb); pair H1 = bisectorpoint(C,H); pair H2 = foot(H1,C,H); pair K = extension(A,B,H1,H2); pair L = extension(A,D,H1,H2); path circle3 = Circle(K,length(K-C)); path circle4 = Circle(L,length(L-C)); // draw(circle3, dashed); // draw(circle4, dashed); pair T = IP(D--A,circle4, 0); pair S = IP(A--B, circle3, 0); pair O = circumcenter(H,S,T); path circle2 = Circle(O, length(O-H)); pair Q = IP(A--D,circle2, 0); pair R = IP(B--A, circle2, 0); pair G = IP(A--H, circle2, 0); pair Q = rotate(180,D)*C; pair P = rotate(180,B)*C; path arc1 = arc(L,length(L-H), 110,0); path arc2 = arc(K, length(K-H), 225, 45); pair M = extension(C,D,L,K); dot("$C$", C, SE); dot("$D$", D, W); dot("$A$", A, dir(40));dot("$B$", B, E);dot("$H$", H, SW); dot("$T$", T, W); dot("$S$", S, E); dot("$K$",K,E); dot("$L$",L,W); // dot("$R$",R,NE); dot("$Q$",Q,NW); dot("$G$", G, SE); dot("$Q$",Q,NW); dot("$P$",P,NE); dot("$M$",M,S); draw(arc1,dashed); draw(arc2, dashed); draw(circle2); draw(A--D--B--cycle); // draw(A--C, dashed); draw(A--H); draw(C--S--H--cycle); draw(C--T--H); draw(D--C--B); draw(D--L--K--B, dashed); draw(T--G, dashed); draw(K--H, dashed); draw(D--Q--P--B,dashed); draw(Q--H--P,dashed); draw(C--L--H, dashed); draw(C--K,dashed); draw(T--S,dashed); // draw(anglemark(H,S,K)); // draw(anglemark(K,H,S)); label("$x$",C+(0,0.1),N); label("$y$",C+(-0.1,0.3),N); label("$w$",H+(-0.03,0.1), N); label("$z$", H+(0.06,0.12), N ); label("$u$",T+(-0.1,-0.2), S); label("$v$", S+(0,-0.2), S); label("$t$",T+(0.1,-0.3), S); label("$s$", S+(-0.08,-0.2), S); [/asy]

Denote $\angle{HSB}=v$, $\angle{HTD}=u$, $\angle{HSC}=s$, $\angle{HTC}=t$, $\angle{HCS}=x$, $\angle{HCT}=y$, $\angle{AHS}=z$, $\angle{AHT}=w$. Since $\angle{CHS}-\angle{CSB} =90$ and $\angle{CHT}-\angle{CTD}=90$, we have $\angle{CSA}=x+90$, $\angle{CTA}=y+90$.


Since $\angle{CHS} = \angle{CSB}+90$, the tangent of the circumcircle of $\triangle{CSH}$ at point $S$ is perpendicular to $SB$; therefore, the circumcenter of $\triangle{CSH}$ (point $K$) is on $AB$. Similarly, the circumcenter of $\triangle{CTH}$ (point $L$) is on $AD$. In addition, $KL$ is the perpendicular bisector of $CH$.

Extend $CB$ to meet circumcircle of $\triangle{CSH}$ at $P$, and extend $CD$ to meet circumcircle of $\triangle{CTH}$ at $Q$. Then, since $\angle{ADC}=\angle{ABC}=90$, $AD$ and $AB$ are the perpendicular bisector of $CQ$ and $CP$, respectively; hence $A$ is the circumcenter of $\triangle{PCQ}$. Since $B$ and $D$ are midpoints on $CP$ and $CQ$, $PQ \parallel BD$; also, $AH \perp BD$, so $AH \perp PQ$. Since $A$ is the circumcenter, $AH$ is also the perpendicular bisector of $PQ$. Hence, \[HP=HQ\]

We have \[u=\angle{CHT}-90=(90-w+\angle{DHC}) - 90\] \[v=\angle{CHS}-90 = (90-z+\angle{BHC}) - 90\] Hence, $u+x = -w-z+\angle{DHC}+\angle{BHC} = 180 -(w+z)$, or \[w+z=180-u-v\] Since quadrilaterals $QTHC$ and $PSHC$ are cyclic, we have $\angle{THQ}=\angle{TCQ}=90-u$, $\angle{SHP}=\angle{SCP}=90-v$; so, \[\angle{PHQ}=\angle{THQ}+\angle{SHP}+w+z=90-u+90-v+w+z = 2(w+z)\] \[w+z=\frac{1}{2}\angle{PHQ}=\angle{AHQ}=w+\angle{THQ}=w+\angle{TCQ}\] Hence, \[\angle{TCQ}=z \qquad \qquad\] Similarly, \[\angle{SCP}=w\]

Now we apply law of Sines repeatedly on pairs of triangles. For $\triangle{QCH}$ and $\triangle{PCH}$, $\angle{HQC}=\angle{HTC}=t$, $\angle{HCQ}=y+z$, $\angle{HPC}=s$, $\angle{HCP}=x+w$; hence, \[\frac{\sin{t}}{\sin{(y+z)}}=\frac{HC}{HQ}=\frac{HC}{HP}=\frac{\sin{s}}{\sin{(x+w)}} \qquad \qquad (1)\] For $\triangle{LHK}$, $\angle{HLK}=\frac{1}{2}\angle{HLC}=t$, $\angle{HKL}=\frac{1}{2}\angle{HKC}=s$; hence, \[\frac{\sin{s}}{\sin{t}}=\frac{LH}{KH}=\frac{LT}{KS} \qquad \qquad (2)\] For $\triangle{LAK}$, $\angle{ALK}=90-\angle{DML}=90-\angle{CMK}=\angle{MCH}=y+z$, and similarly, $\angle{ALK}=w+x$; hence, \[\frac{\sin{(w+x)}}{\sin{(y+z)}}=\frac{AL}{AK} \qquad \qquad (3)\] Coming $(1), (2), (3)$, we have \[\frac{AL}{AK}=\frac{LT}{KS}=\frac{AL-AT}{AK-AS}=\frac{AT}{AS}\] Therefore, $TS \parallel KL$, and $\angle{ATS} = \angle{ALK}=y+z$. Let the circumcircle of $\triangle{THS}$ meets $AH$ at $G$. We have, \[\angle{ATG}=\angle{ATS}-\angle{GTS}=(y+z)-\angle{AHS}= y\] And, \[\angle{GTH}=\angle{ATH}-\angle{ATG}=(90+y) - y = 90\] This proves $HG$ is the diameter of the circle and the center of the circle is on AH. $\square$

Solution by $Mathdummy$.

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

See Also

2014 IMO (Problems) • Resources
Preceded by
Problem 2 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 4
All IMO Problems and Solutions
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