Difference between revisions of "2005 USAMO Problems/Problem 3"
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== Solution == | == Solution == | ||
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+ | === Solution 1 === | ||
Let <math>B_1'</math> be the second intersection of the line <math>C_1A</math> with the circumcircle of <math>APC</math>, and let <math>Q'</math> be the second intersection of the circumcircle of <math>B_1' C_1P</math> and line <math>BC</math>. It is enough to show that <math>B_1'=B_1</math> and <math>Q' =Q</math>. All our angles will be directed, and measured mod <math>\pi</math>. | Let <math>B_1'</math> be the second intersection of the line <math>C_1A</math> with the circumcircle of <math>APC</math>, and let <math>Q'</math> be the second intersection of the circumcircle of <math>B_1' C_1P</math> and line <math>BC</math>. It is enough to show that <math>B_1'=B_1</math> and <math>Q' =Q</math>. All our angles will be directed, and measured mod <math>\pi</math>. | ||
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It follows that <math>Q'</math> is the intersection of <math>BC</math> and the line parallel to <math>AC</math> and passing through <math>C_1</math>. Hence <math>Q' = Q</math>. Then <math>B_1'</math> is the second intersection of the circumcircle of <math>APC</math> and the line parallel to <math>AB</math> passing through <math>Q</math>. Hence <math>B_1' = B_1</math>, as desired. <math>\blacksquare</math> | It follows that <math>Q'</math> is the intersection of <math>BC</math> and the line parallel to <math>AC</math> and passing through <math>C_1</math>. Hence <math>Q' = Q</math>. Then <math>B_1'</math> is the second intersection of the circumcircle of <math>APC</math> and the line parallel to <math>AB</math> passing through <math>Q</math>. Hence <math>B_1' = B_1</math>, as desired. <math>\blacksquare</math> | ||
+ | |||
+ | Motivation: One can notice that if you take such a <math>B_1'</math> then <math>PQB_1'C_1</math> is cyclic, and that similarly <math>PQB_1C_1'</math> is also cyclic. One gets the intuition that only one such circle should exist where the other chord passes through A, and so sets up a ghost point for <math>Q</math>, which works. ~cocohearts | ||
+ | |||
+ | ===Solution 2=== | ||
+ | ''Lemma''. <math>B_1, A, C_1</math> are collinear. | ||
+ | |||
+ | Suppose they are not collinear. Let line <math>B_1 A</math> intersect circle <math>ABP</math> (i.e. the circumcircle of <math>ABP</math>) again at <math>C_2</math> distinct from <math>C_1</math>. Because <math>\angle C_2 B_1 Q = \angle C_2 AB = \angle C_2 PB = 180^\circ - \angle C_2 PQ</math>, we have that <math>B_1 C_2 PQ</math> is cyclic. Hence <math>\angle C_2 QP = \angle C_2 B_1 P = \angle P B_1 A = \angle C</math>, so <math>C_2 Q // AC</math>. Then <math>C_2</math> must be the other intersection of the parallel to <math>AC</math> through <math>Q</math> with circle <math>ABP</math>. Then <math>C_2</math> is on segment <math>C_1 Q</math>, so <math>C_2</math> is contained in triangle <math>ABQ</math>. However, line <math>AB_1</math> intersects this triangle only at point <math>A</math> because <math>B_1</math> lies on arc <math>AC</math> not containing <math>P</math> of circle <math>APC</math>, a contradiction. Hence, <math>B_1, A, C_1</math> are collinear, as desired. | ||
+ | |||
+ | As a result, we have <math>\angle C_1 B_1 Q = \angle C_1 AB = \angle C_1 PB = 180^\circ - \angle C_1 PQ</math>, so <math>B_1 C_1 PQ</math> is cyclic, as desired. | ||
+ | |||
+ | === Solution 3 === | ||
+ | |||
+ | Due to the parallel lines, <math>m\angle C_1QB_1=m\angle A.</math> Therefore, it suffices to prove that <cmath>m\angle C_1PB_1=m\angle C_1BA+m\angle ACB_1=m\angle A.</cmath> Note that <math>m\angle BC_1A+m\angle AB_1C=180^{\circ}</math> by the cyclic quadrilaterals. Now, the condition simplifies to proving <math>C_1,A,B_1</math> collinear. | ||
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+ | Use barycentric coordinates. Let <cmath>A=(1,0,0), B=(0,1,0), C=(0,0,1), P=(0,p,1-p), Q=(0,q,1-q).</cmath> By the parallel lines, <cmath>C_1=(1-q-z',q,z'), B_1=(q-y',y',1-q)</cmath> for some <math>y',z'.</math> As <math>A</math> is a vertex of the reference triangle, we must prove that <math>q(1-q)=y'z'.</math> | ||
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+ | Now we find the circumcircle of <math>\triangle APC.</math> Let its equation be <cmath>-a^2yz-b^2xz-c^2xy+(x+y+z)(ux+vy+wz)=0.</cmath> Substituting in <math>A,C</math> gives <math>u=w=0.</math> Substituting in <math>P</math> yields <cmath>-a^2p(1-p)+vp=0\implies v = a^2(1-p).</cmath> Substituting in <math>B_1</math> yields | ||
+ | <cmath>0=-a^2y'(1-q)-b^2(q-y')(1-q)-c^2y'(q-y')+a^2(1-p)y'=0.</cmath> | ||
+ | <cmath>c^2y'^2+(a^2q-a^2p+b^2(1-q)-c^2q)y'-b^2q(1-q)=0</cmath> | ||
+ | |||
+ | Similarly, the circumcircle of <math>\triangle APB</math> is <cmath>-a^2yz-b^2xz-c^2xy+(x+y+z)a^2pz.</cmath> Now, we substitute <math>C_1</math> into the equation | ||
+ | |||
+ | <cmath>0=-a^2qz'-b^2z'(1-q-z')-c^2q(1-q-z')+a^2pz'=0</cmath> | ||
+ | <cmath>b^2 z'^2+(a^2p-a^2q-b^2(1-q)+c^2q)z'-c^2q(1-q)=0</cmath> | ||
+ | |||
+ | Let <math>k=a^2p-a^2q-b^2(1-q)+c^2q,</math> and note that both <math>y',z'</math> are negative. By the quadratic formula, we have | ||
+ | <cmath>y'=\frac{k-\sqrt{k^2+4b^2c^2q(1-q)}}{2c^2}, z'=\frac{-k-\sqrt{k^2+4b^2c^2q(1-q)}}{2b^2}</cmath> | ||
+ | Multiplying these two, we have <cmath>y'z'=\frac{4b^2c^2q(1-q)}{4b^2c^2}=q(1-q),</cmath> as desired. <math>\blacksquare</math> | ||
{{alternate solutions}} | {{alternate solutions}} | ||
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[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 00:17, 11 October 2021
Problem
(Zuming Feng) Let be an acute-angled triangle, and let and be two points on side . Construct point in such a way that convex quadrilateral is cyclic, , and and lie on opposite sides of line . Construct point in such a way that convex quadrilateral is cyclic, , and and lie on opposite sides of line . Prove that points , and lie on a circle.
Solution
Solution 1
Let be the second intersection of the line with the circumcircle of , and let be the second intersection of the circumcircle of and line . It is enough to show that and . All our angles will be directed, and measured mod .
Since points are concyclic and points are collinear, it follows that But since points are concyclic, It follows that lines and are parallel. If we exchange with and with in this argument, we see that lines and are likewise parallel.
It follows that is the intersection of and the line parallel to and passing through . Hence . Then is the second intersection of the circumcircle of and the line parallel to passing through . Hence , as desired.
Motivation: One can notice that if you take such a then is cyclic, and that similarly is also cyclic. One gets the intuition that only one such circle should exist where the other chord passes through A, and so sets up a ghost point for , which works. ~cocohearts
Solution 2
Lemma. are collinear.
Suppose they are not collinear. Let line intersect circle (i.e. the circumcircle of ) again at distinct from . Because , we have that is cyclic. Hence , so . Then must be the other intersection of the parallel to through with circle . Then is on segment , so is contained in triangle . However, line intersects this triangle only at point because lies on arc not containing of circle , a contradiction. Hence, are collinear, as desired.
As a result, we have , so is cyclic, as desired.
Solution 3
Due to the parallel lines, Therefore, it suffices to prove that Note that by the cyclic quadrilaterals. Now, the condition simplifies to proving collinear.
Use barycentric coordinates. Let By the parallel lines, for some As is a vertex of the reference triangle, we must prove that
Now we find the circumcircle of Let its equation be Substituting in gives Substituting in yields Substituting in yields
Similarly, the circumcircle of is Now, we substitute into the equation
Let and note that both are negative. By the quadratic formula, we have Multiplying these two, we have as desired.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See also
- <url>Forum/viewtopic.php?p=213011#213011 Discussion on AoPS/MathLinks</url>
2005 USAMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
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