Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 11"
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==Solution== | ==Solution== | ||
− | {{ | + | Remark that since <math>ABDE</math> is cyclic we have <math>\angle CED=\angle DBF</math>, and similarly <math>\angle BFD=\angle DCE</math>. Therefore by AA similarity <math>\triangle DBF\sim\triangle DEC</math>. Thus there exists a spiral similarity sending <math>B</math> to <math>E</math> and <math>F</math> to <math>D</math>, so by a fundamental theorem of spiral similarity <math>\triangle BDE\sim\triangle FDC</math>. The angle equality condition gives <math>\angle CFD=\angle EBD=\angle DCF</math>, so <math>\triangle CDF</math> is isosceles and <math>DC=7</math>. Similarly, <math>BD=8</math>. Finally, note that the congruent side lengths actually imply <math>\triangle DBF=\triangle DEC</math>, so <math>EC=BF</math>. |
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+ | Let <math>x=CE=BF</math> and <math>b=AC</math>. Remark that from the perimeter condition <math>AB=45-b</math>. Now from Power of a Point we have the system of two equations <cmath>\begin{cases}7\cdot 15&=xb,\8\cdot 15&=x(45-b).\end{cases}</cmath> Expanding the second equation and rearranging variables gives <math>45x=8\cdot 15+bx=15^2\implies x=5</math>. Back-substitution yields <math>AC=21</math> and consequently <math>AB=24</math>. Thus <math>AE=16</math> and <math>AF=19</math>, so the desired ratio is <math>\tfrac{16}{19}\implies\boxed{035}</math>. | ||
==See Also== | ==See Also== | ||
{{Mock AIME box|year=Pre 2005|n=3|num-b=10|num-a=12}} | {{Mock AIME box|year=Pre 2005|n=3|num-b=10|num-a=12}} |
Latest revision as of 14:32, 22 March 2015
Problem
is an acute triangle with perimeter . is a point on . The circumcircles of triangles and intersect and at and respectively such that and . If , then the value of can be expressed as , where and are relatively prime positive integers. Compute .
Solution
Remark that since is cyclic we have , and similarly . Therefore by AA similarity . Thus there exists a spiral similarity sending to and to , so by a fundamental theorem of spiral similarity . The angle equality condition gives , so is isosceles and . Similarly, . Finally, note that the congruent side lengths actually imply , so .
Let and . Remark that from the perimeter condition . Now from Power of a Point we have the system of two equations Expanding the second equation and rearranging variables gives . Back-substitution yields and consequently . Thus and , so the desired ratio is .
See Also
Mock AIME 3 Pre 2005 (Problems, Source) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |