Difference between revisions of "2014 IMO Problems/Problem 3"
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==Solution== | ==Solution== | ||
+ | </asy></center> | ||
+ | |||
+ | ==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+x = -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{ALK}=w+x</math>; hence, | ||
+ | <cmath>\frac{\sin{(w+x)}}{\sin{(y+z)}}=\frac{AL}{AK} \qquad \qquad (3) </cmath> | ||
+ | Coming <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}} | {{alternate solutions}} |
Revision as of 23:18, 7 September 2018
Contents
[hide]Problem
Convex quadrilateral has . Point is the foot of the perpendicular from to . Points and lie on sides and , respectively, such that lies inside and
Prove that line is tangent to the circumcircle of
Solution
</asy>
Solution
Denote , , , , , , , . Since and , we have , .
Since , the tangent of the circumcircle of at point is perpendicular to ; therefore, the circumcenter of (point ) is on . Similarly, the circumcenter of (point ) is on . In addition, is the perpendicular bisector of .
Extend to meet circumcircle of at , and extend to meet circumcircle of at . Then, since , and are the perpendicular bisector of and , respectively; hence is the circumcenter of . Since and are midpoints on and , ; also, , so . Since is the circumcenter, is also the perpendicular bisector of . Hence,
We have Hence, , or Since quadrilaterals and are cyclic, we have , ; so, Hence, Similarly,
Now we apply law of Sines repeatedly on pairs of triangles. For and , , , , ; hence, For , , ; hence, For , , and similarly, ; hence, Coming , we have Therefore, , and . Let the circumcircle of meets at . We have, And, This proves is the diameter of the circle and the center of the circle is on AH.
Solution by .
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 |