Difference between revisions of "1994 USAMO Problems/Problem 3"
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<math>\angle AEC=\angle ADC</math> because <math>A</math>,<math>C</math>,<math>D</math> and <math>E</math> all lie on the circle, and <math>\angle ADC=\angle CDQ</math>. <math>\angle EAD=\angle ECD</math> because <math>A</math>,<math>C</math>,<math>D</math> and <math>E</math> all lie on the circle. <math>\angle DAC=\angle ECF</math> because <math>CD=EF</math>, and <math>A</math>,<math>C</math>,<math>D</math>,<math>E</math> and <math>F</math> all lie on the circle. Then, | <math>\angle AEC=\angle ADC</math> because <math>A</math>,<math>C</math>,<math>D</math> and <math>E</math> all lie on the circle, and <math>\angle ADC=\angle CDQ</math>. <math>\angle EAD=\angle ECD</math> because <math>A</math>,<math>C</math>,<math>D</math> and <math>E</math> all lie on the circle. <math>\angle DAC=\angle ECF</math> because <math>CD=EF</math>, and <math>A</math>,<math>C</math>,<math>D</math>,<math>E</math> and <math>F</math> all lie on the circle. Then, | ||
− | <math>\angle EAC=\angle EAD+\angle DAC=\angle ECD+\angle ECF=\angle | + | <math>\angle EAC=\angle EAD+\angle DAC=\angle ECD+\angle ECF=\angle DCQ</math> |
Therefore, <math>\triangle AEC</math> and <math>\triangle CDQ</math> are similar, so <math>AC/CE=CQ/QD</math>. | Therefore, <math>\triangle AEC</math> and <math>\triangle CDQ</math> are similar, so <math>AC/CE=CQ/QD</math>. | ||
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Because <math>CD=EF</math> and <math>C</math>,<math>D</math>,<math>E</math> and <math>F</math> all lie on the circle, <math>CF</math> is parallel to <math>DE</math>. So, <math>\triangle CPQ</math> and <math>\triangle EPD</math> are similar, and <math>CQ/DE=CP/PE</math>. | Because <math>CD=EF</math> and <math>C</math>,<math>D</math>,<math>E</math> and <math>F</math> all lie on the circle, <math>CF</math> is parallel to <math>DE</math>. So, <math>\triangle CPQ</math> and <math>\triangle EPD</math> are similar, and <math>CQ/DE=CP/PE</math>. | ||
− | Putting it all together, <math>CP | + | Putting it all together, <math>\frac{CP}{PE}=\frac{CQ}{DE}=\frac{AC}{CE}\cdot \frac{QD}{DE}=(\frac{AC}{CE})^2</math>. |
+ | |||
+ | Borrowed from https://mks.mff.cuni.cz/kalva/usa/usoln/usol943.html | ||
+ | |||
+ | ==See Also== | ||
+ | {{USAMO box|year=1994|num-b=2|num-a=4}} | ||
+ | {{MAA Notice}} | ||
+ | [[Category:Olympiad Geometry Problems]] |
Latest revision as of 07:02, 19 July 2016
Problem
A convex hexagon is inscribed in a circle such that and diagonals , and are concurrent. Let be the intersection of and . Prove that .
Solution
Let the diagonals , , meet at .
First, let's show that the triangles and are similar.
because ,, and all lie on the circle, and . because , and ,,, and all lie on the circle. Then,
Therefore, and are similar, so .
Next, let's show that and are similar.
because ,, and all lie on the circle, and . because ,, and all lie on the circle. because , and ,,, and all lie on the circle. Then,
Therefore, and are similar, so .
Lastly, let's show that and are similar.
Because and ,, and all lie on the circle, is parallel to . So, and are similar, and .
Putting it all together, .
Borrowed from https://mks.mff.cuni.cz/kalva/usa/usoln/usol943.html
See Also
1994 USAMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.