Difference between revisions of "2003 USAMO Problems/Problem 4"
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Extend segment <math>DM</math> through <math>M</math> to <math>G</math> such that <math>FG\parallel CD</math>. | Extend segment <math>DM</math> through <math>M</math> to <math>G</math> such that <math>FG\parallel CD</math>. | ||
− | + | <asy> | |
− | < | + | defaultpen(fontsize(10)+0.6); size(250); |
− | + | var theta=22, r=0.58; | |
+ | pair B=origin, A=dir(theta), C=A+(rotate(78)*0.8*A), O=IP(CR(B,r),CR(A,r)); | ||
+ | path c=CR(O,r); | ||
+ | pair D=IP(c,A--C), E=IP(c,B--C), F=extension(A,B,D,E), M=extension(B,D,C,F), G=extension(D,M,F,F+C-D); | ||
+ | draw(A--B--C--A^^E--F--C^^A--F^^B--M^^E--M); draw(c); draw(F--G--C^^M--G,gray+0.4); | ||
+ | dot("$A$",A,dir(F-E)); dot("$B$",B,2*dir(B-A)); dot("$C$",C,1.5*dir(C-A)); dot("$D$",D,2.5*dir(250)); dot("$E$",E,2.5*dir(C-A)); dot("$F$",F,dir(F-E)); dot("$M$",M,2.5*dir(255)); dot("$G$",G,dir(G-M)); | ||
+ | </asy> | ||
Then <math>MF = MC</math> if and only if quadrilateral <math>CDFG</math> is a parallelogram, or, <math>FD\parallel CG</math>. Hence <math>MC = MF</math> if and only if <math>\angle GCD = \angle FDA</math>, that is, <math>\angle FDA + \angle CGF = 180^\circ</math>. | Then <math>MF = MC</math> if and only if quadrilateral <math>CDFG</math> is a parallelogram, or, <math>FD\parallel CG</math>. Hence <math>MC = MF</math> if and only if <math>\angle GCD = \angle FDA</math>, that is, <math>\angle FDA + \angle CGF = 180^\circ</math>. | ||
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=== Solution 2 === | === Solution 2 === | ||
− | We first assume that <math>MB\cdot MD = MC^2</math>. Because <math>\frac{MC}{MD} = \frac{MB}{MC}</math> and <math>\angle CMD = \angle BMC</math>, triangles <math>CMD</math> and <math>BMC</math> are similar. Consequently, <math>\angle MCD = \angle MBC</math>. | + | We first assume that <math>MB\cdot MD = MC^2</math>. Because <math>\frac{MC}{MD} = \frac{MB}{MC}</math> and <math>\angle CMD = \angle BMC</math>, triangles <math>CMD</math> and <math>BMC</math> are similar. Consequently, <math>\angle MCD = \angle MBC</math>. This exact condition can also be visualized through power of M with respect to the circumcircle of triangle <math>BDC</math> and getting the angle condition from the alternate segment theorem. |
− | + | <asy> | |
− | < | + | defaultpen(fontsize(10)+0.6); size(250); |
− | + | var theta=22, r=0.58; | |
+ | pair B=origin, A=dir(theta), C=A+(rotate(78)*0.8*A), O=IP(CR(B,r),CR(A,r)); | ||
+ | path c=CR(O,r); | ||
+ | pair D=IP(c,A--C), E=IP(c,B--C), F=extension(A,B,D,E), M=extension(B,D,C,F), G=extension(D,M,F,F+C-D); | ||
+ | draw(A--B--C--A^^E--F--C^^A--F^^B--M^^E--M); draw(c); draw(A--E,gray+0.4); | ||
+ | dot("$A$",A,dir(F-E)); dot("$B$",B,2*dir(B-A)); dot("$C$",C,1.5*dir(C-A)); dot("$D$",D,2.5*dir(250)); dot("$E$",E,2.5*dir(C-A)); dot("$F$",F,dir(F-E)); dot("$M$",M,2.5*dir(255)); | ||
+ | </asy> | ||
Because quadrilateral <math>ABED</math> is cyclic, <math>\angle DAE = \angle DBE</math>. Hence | Because quadrilateral <math>ABED</math> is cyclic, <math>\angle DAE = \angle DBE</math>. Hence | ||
<cmath>\angle FCA = \angle MCD = \angle MBC = \angle DBE = \angle DAE = \angle CAE,</cmath> | <cmath>\angle FCA = \angle MCD = \angle MBC = \angle DBE = \angle DAE = \angle CAE,</cmath> | ||
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Combining the above, we conclude that <math>MF = MC</math> if and only if <math>MB\cdot MD = MC^2</math>. | Combining the above, we conclude that <math>MF = MC</math> if and only if <math>MB\cdot MD = MC^2</math>. | ||
+ | |||
+ | |||
{{alternate solutions}} | {{alternate solutions}} |
Latest revision as of 01:39, 10 January 2023
Contents
[hide]Problem
Let be a triangle. A circle passing through and intersects segments and at and , respectively. Lines and intersect at , while lines and intersect at . Prove that if and only if .
Solutions
Solution 1
Extend segment through to such that . Then if and only if quadrilateral is a parallelogram, or, . Hence if and only if , that is, .
Because quadrilateral is cyclic, . It follows that if and only if that is, quadrilateral is cyclic, which is equivalent to Because , if and only if triangles and are similar, that is or .
Solution 2
We first assume that . Because and , triangles and are similar. Consequently, . This exact condition can also be visualized through power of M with respect to the circumcircle of triangle and getting the angle condition from the alternate segment theorem. Because quadrilateral is cyclic, . Hence implying that , so . Because quadrilateral is cyclic, . Hence Because and , triangles and are similar. Consequently, , or . Therefore implies .
Now we assume that . Applying Ceva's Theorem to triangle and cevians gives implying that , so .
Consequently, . Because quadrilateral is cyclic, . Hence Because and , triangles and are similar. Consequently, , or .
Combining the above, we conclude that if and only if .
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See also
2003 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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.