Difference between revisions of "2003 USAMO Problems/Problem 4"
(→Solution) |
m (→Solution 2) |
||
(10 intermediate revisions by 7 users not shown) | |||
Line 3: | Line 3: | ||
Let <math>ABC</math> be a triangle. A circle passing through <math>A</math> and <math>B</math> intersects segments <math>AC</math> and <math>BC</math> at <math>D</math> and <math>E</math>, respectively. Lines <math>AB</math> and <math>DE</math> intersect at <math>F</math>, while lines <math>BD</math> and <math>CF</math> intersect at <math>M</math>. Prove that <math>MF = MC</math> if and only if <math>MB\cdot MD = MC^2</math>. | Let <math>ABC</math> be a triangle. A circle passing through <math>A</math> and <math>B</math> intersects segments <math>AC</math> and <math>BC</math> at <math>D</math> and <math>E</math>, respectively. Lines <math>AB</math> and <math>DE</math> intersect at <math>F</math>, while lines <math>BD</math> and <math>CF</math> intersect at <math>M</math>. Prove that <math>MF = MC</math> if and only if <math>MB\cdot MD = MC^2</math>. | ||
− | == Solution == | + | == Solutions == |
− | + | === Solution 1 === | |
− | + | Extend segment <math>DM</math> through <math>M</math> to <math>G</math> such that <math>FG\parallel CD</math>. | |
− | + | <asy> | |
− | <math>MF = MC | + | 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>. | ||
− | + | Because quadrilateral <math>ABED</math> is cyclic, <math>\angle FDA = \angle ABE</math>. It follows that <math>MC = MF</math> if and only if | |
+ | <cmath>180^\circ = \angle FDA + \angle CGF = \angle ABE + \angle CGF,</cmath> | ||
+ | that is, quadrilateral <math>CBFG</math> is cyclic, which is equivalent to | ||
+ | <cmath>\angle CBM = \angle CBG = \angle CFG = \angle DCF = \angle DCM.</cmath> | ||
+ | Because <math>\angle DMC = \angle CMB</math>, <math>\angle CBM = \angle DCM</math> if and only if triangles <math>BCM</math> and <math>CDM</math> are similar, that is | ||
+ | <cmath>\frac{CM}{BM} = \frac{DM}{CM},</cmath> | ||
+ | or <math>MB\cdot MD = MC^2</math>. | ||
+ | === 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>. 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> | <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)); | |
− | |||
− | |||
− | |||
− | draw( | ||
− | draw( | ||
− | draw( | ||
− | dot | ||
− | |||
− | dot | ||
− | |||
− | dot | ||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | dot | ||
− | |||
− | |||
</asy> | </asy> | ||
+ | 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> | ||
+ | implying that <math>AE\parallel CF</math>, so <math>\angle AEF = \angle CFE</math>. Because quadrilateral <math>ABED</math> is cyclic, <math>\angle ABD = \angle AED</math>. Hence | ||
+ | <cmath>\angle FBM = \angle ABD = \angle AED = \angle AEF = \angle CFE = \angle MFD.</cmath> | ||
+ | Because <math>\angle FBM = \angle DFM</math> and <math>\angle FMB = \angle DMF</math>, triangles <math>BFM</math> and <math>FDM</math> are similar. Consequently, <math>\frac{FM}{DM} = \frac{BM}{FM}</math>, or <math>FM^2 = BM\cdot DM = CM^2</math>. Therefore <math>MC^2 = MB\cdot MD</math> implies <math>MC = MF</math>. | ||
+ | |||
+ | Now we assume that <math>MC = MF</math>. Applying Ceva's Theorem to triangle <math>BCF</math> and cevians <math>BM, CA, FE</math> gives | ||
+ | <cmath>\frac{BA}{AF}\cdot\frac{FM}{MC}\cdot\frac{CE}{EB} = 1,</cmath> | ||
+ | implying that <math>\frac{BA}{AF} = \frac{BE}{EC}</math>, so <math>AE\parallel CF</math>. | ||
+ | |||
+ | Consequently, <math>\angle DCM = \angle DAE</math>. Because quadrilateral <math>ABED</math> is cyclic, <math>\angle DAE = \angle DBE</math>. Hence | ||
+ | <cmath>\angle DCM = \angle DAE = \angle DBE = \angle CBM.</cmath> | ||
+ | Because <math>\angle CBM = \angle DCM</math> and <math>\angle CMB = \angle DMC</math>, triangles <math>BCM</math> and <math>CDM</math> are similar. Consequently, <math>\frac{CM}{DM} = \frac{BM}{CM}</math>, or <math>CM^2 = BM\cdot DM</math>. | ||
+ | |||
+ | Combining the above, we conclude that <math>MF = MC</math> if and only if <math>MB\cdot MD = MC^2</math>. | ||
+ | |||
+ | |||
− | + | {{alternate solutions}} | |
− | + | == See also == | |
− | + | {{USAMO newbox|year=2003|num-b=3|num-a=5}} | |
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] | ||
+ | {{MAA Notice}} |
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.