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...") |
m |
||
(6 intermediate revisions by 6 users not shown) | |||
Line 1: | Line 1: | ||
==Problem== | ==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, 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). | + | 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== | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | <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 <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. | ||
+ | |||
+ | ==Solution 2== | ||
+ | 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>. | ||
+ | |||
+ | 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== | ||
Line 8: | Line 54: | ||
{{USAJMO newbox|year=2012|beforetext=|before=First Problem|num-a=2}} | {{USAJMO newbox|year=2012|beforetext=|before=First Problem|num-a=2}} | ||
+ | |||
+ | [[Category:Olympiad Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 19:50, 19 December 2023
Contents
[hide]Problem
Given a triangle , let and be points on segments and , respectively, such that . Let and be distinct points on segment such that lies between and , , and . Prove that , , , are concyclic (in other words, these four points lie on a circle).
Solution
Since , the circumcircle of triangle is tangent to at . Similarly, since , the circumcircle of triangle is tangent to at .
For the sake of contradiction, suppose that the circumcircles of triangles and are not the same circle. Since , , so the power of A with respect to both circumcircles is the same. Thus, lies on the radical axis of both circles. However, both circles pass through and , so the radical axis of both circles is . Hence, lies on , which is a contradiction.
Therefore, the two circumcircles are the same circle. In other words, , , , and all lie on the same circle.
Solution 2
Note that (as in the first solution) the circumcircle of triangle is tangent to at . Similarly, since , the circumcircle of triangle is tangent to at .
Now, suppose these circumcircles are not the same circle. They already intersect at and , so they cannot intersect anymore. Thus, AS must touch the two circumcircles at points and , with on the circumcircle of triangle . By Power of a Point, and . Hence, because , , 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.