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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...")
 
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==Problem==
 
==Problem==
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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, such that <math>AP = AQ</math>.  Let <math>S</math> and <math>R</math> be distinct points on segment <math>\overline{BC}</math> such that <math>S</math> lies between <math>B</math> and <math>R</math>, <math>\angle BPS = \angle PRS</math>, and <math>\angle CQR = \angle QSR</math>.  Prove that <math>P</math>, <math>Q</math>, <math>R</math>, <math>S</math> are concyclic (in other words, these four points lie on a circle).
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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, such that <math>AP = AQ</math>.  Let <math>S</math> and <math>R</math> be distinct points on segment <math>\overline{BC}</math> such that <math>S</math> lies between <math>B</math> and <math>R</math>, <math>\angle{BPS} = \angle{PRS}</math>, and <math>\angle{CQR} = \angle{QSR}</math>.  Prove that <math>P</math>, <math>Q</math>, <math>R</math>, <math>S</math> are concyclic (in other words, these four points lie on a circle).
  
 
==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>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.
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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.
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==Solution 2==
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Note that (as in the first solution) 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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Now, suppose these circumcircles are not the same circle. They already intersect at <math>R</math> and <math>S</math>, so they cannot intersect anymore. Thus, AS must touch the two circumcircles at points <math>M</math> and <math>N</math>, with <math>M</math> on the circumcircle of triangle <math>PRS</math>. By Power of a Point, <math>AQ^2 = AM \cdot AS</math> and <math>AP^2 = AN \cdot AS</math>. Hence, because <math>AP = AQ</math>, <math>AM = AN</math>, 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==
 
==See also==
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{{USAJMO newbox|year=2012|beforetext=|before=First Problem|num-a=2}}
 
{{USAJMO newbox|year=2012|beforetext=|before=First Problem|num-a=2}}
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[[Category:Olympiad Geometry Problems]]
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{{MAA Notice}}

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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