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

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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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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>AP^2 = AQ^2</math>, so the power of A with respect to both circumcircles is the same. Thus, <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.
  
 
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.
 
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.

Latest revision as of 19:50, 19 December 2023

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 $AP = AQ$, $AP^2 = AQ^2$, so the power of A with respect to both circumcircles is the same. Thus, $A$ 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.

Solution 2

Note that (as in the first solution) 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 $Q$.

Now, suppose these circumcircles are not the same circle. They already intersect at $R$ and $S$, so they cannot intersect anymore. Thus, AS must touch the two circumcircles at points $M$ and $N$, with $M$ on the circumcircle of triangle $PRS$. By Power of a Point, $AQ^2 = AM \cdot AS$ and $AP^2 = AN \cdot AS$. Hence, because $AP = AQ$, $AM = AN$, a contradiction because then, as they lie on the same line segment, M and N must be the same point! (Note line segment, not line.) Hence, the two circumcircles are the same circle.

See also

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

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