Difference between revisions of "2014 IMO Problems/Problem 3"

Latest revision as of 02:58, 8 September 2018

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, $u+v = -w-z+\angle{DHC}+\angle{BHC} = 180 -(w+z)$ , 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, $\angle{AKL}=w+x$ ; hence, $\frac{\sin{(w+x)}}{\sin{(y+z)}}=\frac{AL}{AK} \qquad \qquad (3)$ Combining $(1), (2), (3)$ , 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.

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Difference between revisions of "2014 IMO Problems/Problem 3"

Latest revision as of 02:58, 8 September 2018

Problem

Solution

See Also

@@ Line 1: / Line 1: @@
 ==Problem==
-Convex quadrilateral <math>ABCD</math> has <math>\angle{ABC}=\angle{CDA}=90^{\circ}. Point </math>H<math> is the foot of the perpendicual from </math>A<math> to </math>BD<math>. Points </math>S<math> and </math>T<math> lie on sides </math>AB<math> and </math>AD<math>, respectively, such that </math>H<math> lies inside </math>\triangle{SCT} and
+Convex quadrilateral <math>ABCD</math> has <math>\angle{ABC}=\angle{CDA}=90^{\circ}</math>. Point <math>H</math> is the foot of the perpendicular from <math>A</math> to <math>BD</math>. Points <math>S</math> and <math>T</math> lie on sides <math>AB</math> and <math>AD</math>, respectively, such that <math>H</math> lies inside <math>\triangle{SCT}</math> and
 <cmath>\angle{CHS}-\angle{CSB}=90^{\circ},\quad \angle{THC}-\angle{DTC} = 90^{\circ}.</cmath>
+Prove that line <math>BD</math> is tangent to the circumcircle of <math>\triangle{TSH}.</math>
 ==Solution==
+<center><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></center>
+Denote <math>\angle{HSB}=v</math>, <math>\angle{HTD}=u</math>, <math>\angle{HSC}=s</math>, <math>\angle{HTC}=t</math>, <math>\angle{HCS}=x</math>, <math>\angle{HCT}=y</math>, <math>\angle{AHS}=z</math>, <math>\angle{AHT}=w</math>. Since <math>\angle{CHS}-\angle{CSB} =90</math> and <math>\angle{CHT}-\angle{CTD}=90</math>, we have <math>\angle{CSA}=x+90</math>, <math>\angle{CTA}=y+90</math>.
+Since <math>\angle{CHS} = \angle{CSB}+90</math>, the tangent of the circumcircle of <math>\triangle{CSH}</math> at point <math>S</math> is perpendicular to <math>SB</math>; therefore, the circumcenter of <math>\triangle{CSH}</math> (point <math>K</math>) is on <math>AB</math>. Similarly, the circumcenter of <math>\triangle{CTH}</math> (point <math>L</math>) is on <math>AD</math>. In addition, <math>KL</math> is the perpendicular bisector of <math>CH</math>.
+Extend <math>CB</math> to meet circumcircle of  <math>\triangle{CSH}</math> at <math>P</math>, and extend <math>CD</math> to meet circumcircle of  <math>\triangle{CTH}</math> at <math>Q</math>. Then, since <math>\angle{ADC}=\angle{ABC}=90</math>, <math>AD</math> and <math>AB</math> are the perpendicular bisector of <math>CQ</math> and <math>CP</math>, respectively; hence <math>A</math> is the circumcenter of <math>\triangle{PCQ}</math>. Since <math>B</math> and <math>D</math> are midpoints on <math>CP</math> and <math>CQ</math>, <math>PQ \parallel BD</math>; also, <math>AH \perp BD</math>, so <math>AH \perp PQ</math>. Since <math>A</math> is the circumcenter, <math>AH</math> is also the perpendicular bisector of <math>PQ</math>. Hence,
+<cmath>HP=HQ</cmath>
+We have
+<cmath>u=\angle{CHT}-90=(90-w+\angle{DHC}) - 90</cmath>
+<cmath>v=\angle{CHS}-90 = (90-z+\angle{BHC}) - 90</cmath>
+Hence, <math>u+v = -w-z+\angle{DHC}+\angle{BHC} = 180 -(w+z)</math>, or <cmath>w+z=180-u-v</cmath>
+Since quadrilaterals <math>QTHC</math> and <math>PSHC</math> are cyclic, we have <math>\angle{THQ}=\angle{TCQ}=90-u</math>, <math>\angle{SHP}=\angle{SCP}=90-v</math>; so,
+<cmath>\angle{PHQ}=\angle{THQ}+\angle{SHP}+w+z=90-u+90-v+w+z = 2(w+z)</cmath>
+<cmath>w+z=\frac{1}{2}\angle{PHQ}=\angle{AHQ}=w+\angle{THQ}=w+\angle{TCQ}</cmath>
+Hence,
+<cmath>\angle{TCQ}=z \qquad \qquad</cmath>
+Similarly,
+<cmath>\angle{SCP}=w</cmath>
+Now we apply law of Sines repeatedly on pairs of triangles.
+For <math>\triangle{QCH}</math> and <math>\triangle{PCH}</math>, <math>\angle{HQC}=\angle{HTC}=t</math>, <math>\angle{HCQ}=y+z</math>, <math>\angle{HPC}=s</math>, <math>\angle{HCP}=x+w</math>; hence,
+<cmath>\frac{\sin{t}}{\sin{(y+z)}}=\frac{HC}{HQ}=\frac{HC}{HP}=\frac{\sin{s}}{\sin{(x+w)}} \qquad \qquad (1)</cmath>
+For <math>\triangle{LHK}</math>, <math>\angle{HLK}=\frac{1}{2}\angle{HLC}=t</math>, <math>\angle{HKL}=\frac{1}{2}\angle{HKC}=s</math>; hence,
+<cmath>\frac{\sin{s}}{\sin{t}}=\frac{LH}{KH}=\frac{LT}{KS} \qquad \qquad (2) </cmath>
+For <math>\triangle{LAK}</math>, <math>\angle{ALK}=90-\angle{DML}=90-\angle{CMK}=\angle{MCH}=y+z</math>, and similarly, <math>\angle{AKL}=w+x</math>; hence,
+<cmath>\frac{\sin{(w+x)}}{\sin{(y+z)}}=\frac{AL}{AK} \qquad \qquad (3) </cmath>
+Combining <math>(1), (2), (3)</math>, we have
+<cmath>\frac{AL}{AK}=\frac{LT}{KS}=\frac{AL-AT}{AK-AS}=\frac{AT}{AS}</cmath>
+Therefore, <math>TS \parallel KL</math>, and <math>\angle{ATS} = \angle{ALK}=y+z</math>.
+Let the circumcircle of <math>\triangle{THS}</math> meets <math>AH</math> at <math>G</math>. We have,
+<cmath>\angle{ATG}=\angle{ATS}-\angle{GTS}=(y+z)-\angle{AHS}= y</cmath>
+And,
+<cmath>\angle{GTH}=\angle{ATH}-\angle{ATG}=(90+y) - y = 90</cmath>
+This proves <math>HG</math> is the diameter of the circle and the center of the circle is on AH. <math>\square</math>
+Solution by <math>Mathdummy</math>.
 {{alternate solutions}}