Difference between revisions of "2012 USAMO Problems/Problem 5"
(→Solution) |
|||
Line 4: | Line 4: | ||
==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. | ||
==See also== | ==See also== |
Revision as of 09:13, 26 April 2012
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.
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 |