Difference between revisions of "2012 USAJMO Problems/Problem 1"

(Created page with "==Problem== Given a triangle <math>ABC</math>, let <math>P</math> and <math>Q</math> be points on segments <math>\overline{AB}</math> and <math>\overline{AC}</math>, respectively...")
 
(Solution)
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==Solution==
 
==Solution==
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Since <math>\angle BPS = \angle PRS</math>, the circumcircle of triangle <math>PRS</math> is tangent to <math>AB</math> at <math>P</math>.  Similarly, since <math>\angle CQR = \angle QSR</math>, the circumcircle of triangle <math>QRS</math> is tangent to <math>AC</math> at <math>Q</math>.
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<asy>
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import markers;
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unitsize(0.5 cm);
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pair A, B, C, O, P, Q, R, S;
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A = (2,12);
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B = (0,0);
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C = (14,0);
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P = intersectionpoint(A--B,Circle(A,8));
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Q = intersectionpoint(A--C,Circle(A,8));
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O = extension(P, P + rotate(90)*(A - P), Q, Q + rotate(90)*(A - Q));
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S = intersectionpoint(B--C,arc(O, abs(O - P), 180, 270));
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R = intersectionpoint(B--C,arc(O, abs(O - P), 270, 360));
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draw(A--B--C--cycle);
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draw(Circle(O, abs(O - P)));
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draw(S--P--R);
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draw(S--Q--R);
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label("$A$", A, N);
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label("$B$", B, SW);
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label("$C$", C, SE);
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label("$P$", P, W);
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label("$Q$", Q, NE);
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label("$R$", R, SE);
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label("$S$", S, SW);
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markangle(1, B, P, S, radius=4mm, marker(markinterval(stickframe(n=1,2mm),true)));
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markangle(1, P, R, S, radius=4mm, marker(markinterval(stickframe(n=1,2mm),true)));
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markangle(1, R, Q, C, radius=4mm, marker(markinterval(stickframe(n=2,2mm),true)));
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markangle(1, R, S, Q, radius=4mm, marker(markinterval(stickframe(n=2,2mm),true)));
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</asy>
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For the sake of contradiction, suppose that the circumcircles of triangles <math>PRS</math> and <math>QRS</math> are not the same circle.  Since <math>AP = AQ</math>, <math>A</math> lies on the [[Radical_axis|radical axis]] of both circles.  However, both circles pass through <math>R</math> and <math>S</math>, so the radical axis of both circles is <math>RS</math>.  Hence, <math>A</math> lies on <math>RS</math>, which is a contradiction.
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Therefore, the two circumcircles are the same circle.  In other words, <math>P</math>, <math>Q</math>, <math>R</math>, and <math>S</math> all lie on the same circle.
  
 
==See also==
 
==See also==

Revision as of 18:26, 25 April 2012

Problem

Given a triangle $ABC$, let $P$ and $Q$ be points on segments $\overline{AB}$ and $\overline{AC}$, respectively, such that $AP = AQ$. Let $S$ and $R$ be distinct points on segment $\overline{BC}$ such that $S$ lies between $B$ and $R$, $\angle BPS = \angle PRS$, and $\angle CQR = \angle QSR$. Prove that $P$, $Q$, $R$, $S$ are concyclic (in other words, these four points lie on a circle).

Solution

Since $\angle BPS = \angle PRS$, the circumcircle of triangle $PRS$ is tangent to $AB$ at $P$. Similarly, since $\angle CQR = \angle QSR$, the circumcircle of triangle $QRS$ is tangent to $AC$ at $Q$.

[asy] import markers;  unitsize(0.5 cm);  pair A, B, C, O, P, Q, R, S;  A = (2,12); B = (0,0); C = (14,0); P = intersectionpoint(A--B,Circle(A,8)); Q = intersectionpoint(A--C,Circle(A,8)); O = extension(P, P + rotate(90)*(A - P), Q, Q + rotate(90)*(A - Q)); S = intersectionpoint(B--C,arc(O, abs(O - P), 180, 270)); R = intersectionpoint(B--C,arc(O, abs(O - P), 270, 360));  draw(A--B--C--cycle); draw(Circle(O, abs(O - P))); draw(S--P--R); draw(S--Q--R);  label("$A$", A, N); label("$B$", B, SW); label("$C$", C, SE); label("$P$", P, W); label("$Q$", Q, NE); label("$R$", R, SE); label("$S$", S, SW);  markangle(1, B, P, S, radius=4mm, marker(markinterval(stickframe(n=1,2mm),true))); markangle(1, P, R, S, radius=4mm, marker(markinterval(stickframe(n=1,2mm),true))); markangle(1, R, Q, C, radius=4mm, marker(markinterval(stickframe(n=2,2mm),true))); markangle(1, R, S, Q, radius=4mm, marker(markinterval(stickframe(n=2,2mm),true))); [/asy]

For the sake of contradiction, suppose that the circumcircles of triangles $PRS$ and $QRS$ are not the same circle. Since $AP = AQ$, $A$ lies on the radical axis of both circles. However, both circles pass through $R$ and $S$, so the radical axis of both circles is $RS$. Hence, $A$ lies on $RS$, which is a contradiction.

Therefore, the two circumcircles are the same circle. In other words, $P$, $Q$, $R$, and $S$ all lie on the same circle.

See also

2012 USAJMO (ProblemsResources)
First Problem Followed by
Problem 2
1 2 3 4 5 6
All USAJMO Problems and Solutions
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