Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 14"
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+ | ==Solution 2== | ||
+ | Note <math>ABEC</math> is cyclic, so <math>\angle BEC = \angle BED = 90^o</math>. By Power Of A Point we have <math>7(7+5)=DE \cdot DC \implies DE=\frac{84}{13}</math>, <math>EC=\frac{85}{13}</math>. Now, note <cmath>\angle ABC = \angle BCA</cmath><cmath>\qquad =\angle BEA</cmath><cmath>= \angle AEC</cmath>Therefore, by Angle Bisector Theorem, <cmath>\frac{FC}{BF}=\frac{EC}{BE}=\frac{85}{35} \implies \frac{17}{7} \implies 17+7=\boxed{24}</cmath> | ||
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+ | ~Geometry285 | ||
==See also== | ==See also== |
Latest revision as of 20:18, 11 July 2021
Contents
Problem
Let there be a such that , , and , and let be a point on such that Let the circumcircle of intersect hypotenuse at and . Let intersect at . If the ratio can be expressed as where and are relatively prime, find
Solution
We claim that is the angle bisector of .
Observe that , which tells us that is a triangle. In cyclic quadrilateral , we have and Since , we have . This means that , or equivalently , is an angle bisector of in .
By the angle bisector theorem and our We seek the lengths and .
To find , we can proceed by Power of a Point using point on circle to get Since , , and , we have Since , we have
To find , we use the Pythagorean Theorem in . (We already found , which tells us that supplementary .) By the Pythagorean Theorem, We found that , and since we are given , we have
Our answer, by equation , is . From equation and from equation . Therefore, our final answer is
~samrocksnature
Solution 2
Note is cyclic, so . By Power Of A Point we have , . Now, note Therefore, by Angle Bisector Theorem,
~Geometry285
See also
- Other 2021 JMPSC Invitationals Problems
- 2021 JMPSC Invitationals Answer Key
- All JMPSC Problems and Solutions
The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition.