Difference between revisions of "2012 USAMO Problems/Problem 5"
(Created page with "== Problem == Let <math>P</math> be a point in the plane of triangle <math>ABC</math>, and <math>\gamma</math> a line passing through <math>P</math>. Let <math>A'</math>, <math...") |
Knowingant (talk | contribs) (deleting my solution: i found the points a', b', and c' i claimed the coordinates of in the solution are not actually collinear. i messed up some algebra somewhere and i'm not going through this thing again.) |
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==Solution== | ==Solution== | ||
+ | |||
+ | By the [[Law_of_Sines|sine law]] on triangle <math>AB'P</math>, | ||
+ | <cmath>\frac{AB'}{\sin \angle APB'} = \frac{AP}{\sin \angle AB'P},</cmath> | ||
+ | so | ||
+ | <cmath>AB' = AP \cdot \frac{\sin \angle APB'}{\sin \angle AB'P}.</cmath> | ||
+ | |||
+ | <asy> | ||
+ | import graph; | ||
+ | import geometry; | ||
+ | |||
+ | unitsize(0.5 cm); | ||
+ | |||
+ | pair[] A, B, C; | ||
+ | pair P, R; | ||
+ | |||
+ | A[0] = (2,12); | ||
+ | B[0] = (0,0); | ||
+ | C[0] = (14,0); | ||
+ | P = (4,5); | ||
+ | R = 5*dir(70); | ||
+ | A[1] = extension(B[0],C[0],P,reflect(P + R,P - R)*(A[0])); | ||
+ | B[1] = extension(C[0],A[0],P,reflect(P + R,P - R)*(B[0])); | ||
+ | C[1] = extension(A[0],B[0],P,reflect(P + R,P - R)*(C[0])); | ||
+ | |||
+ | draw((P - R)--(P + R),red); | ||
+ | draw(A[1]--B[1]--C[1]--cycle,blue); | ||
+ | draw(A[0]--B[0]--C[0]--cycle); | ||
+ | draw(A[0]--P); | ||
+ | draw(B[0]--P); | ||
+ | draw(C[0]--P); | ||
+ | draw(P--A[1]); | ||
+ | draw(P--B[1]); | ||
+ | draw(P--C[1]); | ||
+ | draw(A[1]--B[0]); | ||
+ | draw(A[1]--B[0]); | ||
+ | |||
+ | label("$A$", A[0], N); | ||
+ | label("$B$", B[0], S); | ||
+ | label("$C$", C[0], SE); | ||
+ | dot("$A'$", A[1], SW); | ||
+ | dot("$B'$", B[1], NE); | ||
+ | dot("$C'$", C[1], W); | ||
+ | dot("$P$", P, SE); | ||
+ | label("$\gamma$", P + R, N); | ||
+ | </asy> | ||
+ | |||
+ | Similarly, | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | B'C &= CP \cdot \frac{\sin \angle CPB'}{\sin \angle CB'P}, \ | ||
+ | CA' &= CP \cdot \frac{\sin \angle CPA'}{\sin \angle CA'P}, \ | ||
+ | A'B &= BP \cdot \frac{\sin \angle BPA'}{\sin \angle BA'P}, \ | ||
+ | BC' &= BP \cdot \frac{\sin \angle BPC'}{\sin \angle BC'P}, \ | ||
+ | C'A &= AP \cdot \frac{\sin \angle APC'}{\sin \angle AC'P}. | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | Hence, | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | &\frac{AB'}{B'C} \cdot \frac{CA'}{A'B} \cdot \frac{BC'}{C'A} \ | ||
+ | &= \frac{\sin \angle APB'}{\sin \angle AB'P} \cdot \frac{\sin \angle CB'P}{\sin \angle CPB'} \cdot \frac{\sin \angle CPA'}{\sin \angle CA'P} \cdot \frac{\sin \angle BA'P}{\sin \angle BPA'} \cdot \frac{\sin \angle BPC'}{\sin \angle BC'P} \cdot \frac{\sin \angle AC'P}{\sin \angle APC'}. | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Since angles <math>\angle AB'P</math> and <math>\angle CB'P</math> are supplementary or equal, depending on the position of <math>B'</math> on <math>AC</math>, | ||
+ | <cmath>\sin \angle AB'P = \sin \angle CB'P.</cmath> | ||
+ | Similarly, | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \sin \angle CA'P &= \sin \angle BA'P, \ | ||
+ | \sin \angle BC'P &= \sin \angle AC'P. | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | By the reflective property, <math>\angle APB'</math> and <math>\angle BPA'</math> are supplementary or equal, so | ||
+ | <cmath>\sin \angle APB' = \sin \angle BPA'.</cmath> | ||
+ | Similarly, | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \sin \angle CPA' &= \sin \angle APC', \ | ||
+ | \sin \angle BPC' &= \sin \angle CPB'. | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | Therefore, | ||
+ | <cmath>\frac{AB'}{B'C} \cdot \frac{CA'}{A'B} \cdot \frac{BC'}{C'A} = 1,</cmath> | ||
+ | so by [[Menelaus'_Theorem|Menelaus's theorem]], <math>A'</math>, <math>B'</math>, and <math>C'</math> are collinear. | ||
+ | |||
+ | ==Solution 2, Barycentric (Modified by Evan Chen)== | ||
+ | |||
+ | We will perform barycentric coordinates on the triangle <math>PCC'</math>, with <math>P=(1,0,0)</math>, <math>C'=(0,1,0)</math>, and <math>C=(0,0,1)</math>. Set <math>a = CC'</math>, <math>b = CP</math>, <math>c = C'P</math> as usual. Since <math>A</math>, <math>B</math>, <math>C'</math> are collinear, we will define <math>A = (p : k : q)</math> and <math>B = (p : \ell : q)</math>. | ||
+ | |||
+ | Claim: Line <math>\gamma</math> is the angle bisector of <math>\angle APA' </math>, <math>\angle BPB'</math>, and <math>\angle CPC'</math>. | ||
+ | This is proved by observing that since <math>A'P</math> is the reflection of <math>AP</math> across <math>\gamma</math>, etc. | ||
+ | |||
+ | Thus <math>B'</math> is the intersection of the isogonal of <math>B</math> with respect to <math>\angle P</math> | ||
+ | with the line <math>CA</math>; that is, | ||
+ | <cmath> B' = \left( \frac pk \frac{b^2}{\ell}: \frac{b^2}{\ell} : \frac{c^2}{q} \right). </cmath> | ||
+ | Analogously, <math>A'</math> is the intersection of the isogonal of <math>A</math> with respect to <math>\angle P</math> | ||
+ | with the line <math>CB</math>; that is, | ||
+ | <cmath> A' = \left( \frac{p}{\ell} \frac{b^2}{k} : \frac{b^2}{k} : \frac{c^2}{q} \right). </cmath> | ||
+ | The ratio of the first to third coordinate in these two points | ||
+ | is both <math>b^2pq : c^2k\ell</math>, so it follows <math>A'</math>, <math>B'</math>, and <math>C'</math> are collinear. | ||
+ | |||
+ | ~peppapig_ | ||
+ | |||
+ | ==See also== | ||
+ | *[[USAMO Problems and Solutions]] | ||
+ | |||
+ | {{USAMO newbox|year=2012|num-b=4|num-a=6}} | ||
+ | {{MAA Notice}} | ||
+ | [[Category:Olympiad Geometry Problems]] |
Latest revision as of 21:32, 2 May 2023
Problem
Let be a point in the plane of triangle , and a line passing through . Let , , be the points where the reflections of lines , , with respect to intersect lines , , , respectively. Prove that , , are collinear.
Solution
By the sine law on triangle , so
Similarly, Hence,
Since angles and are supplementary or equal, depending on the position of on , Similarly,
By the reflective property, and are supplementary or equal, so Similarly, Therefore, so by Menelaus's theorem, , , and are collinear.
Solution 2, Barycentric (Modified by Evan Chen)
We will perform barycentric coordinates on the triangle , with , , and . Set , , as usual. Since , , are collinear, we will define and .
Claim: Line is the angle bisector of , , and . This is proved by observing that since is the reflection of across , etc.
Thus is the intersection of the isogonal of with respect to with the line ; that is, Analogously, is the intersection of the isogonal of with respect to with the line ; that is, The ratio of the first to third coordinate in these two points is both , so it follows , , and are collinear.
~peppapig_
See also
2012 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.