Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 14"
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− | To find <math>EC</math>, we can proceed by Power of a Point using point <math>D</math> on circle <math>(ABC)</math> to get <math>DE \cdot DC = DB \cdot DA.</math> Since <math>DC=13</math>, <math>DB = 7</math>, and <math>AD = 12</math>, we have <math>DE=\frac{84}{13}.</math> Since <math>CD=13</math>, we have <cmath>EC=CD-DE=\frac{85}{13} \qquad (2).</cmath> ~samrocksnature | + | To find <math>EC</math>, we can proceed by Power of a Point using point <math>D</math> on circle <math>(ABC)</math> to get <math>DE \cdot DC = DB \cdot DA.</math> Since <math>DC=13</math>, <math>DB = 7</math>, and <math>AD = 12</math>, we have <math>DE=\frac{84}{13}.</math> Since <math>CD=13</math>, we have <cmath>EC=CD-DE=\frac{85}{13} \qquad (2).</cmath> |
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+ | To find <math>BE</math>, we use the Pythagorean Theorem in <math>BED</math>. (We already found <math>\angle BEC=90^\circ</math>, which tells us that supplementary <math>\angle BED = 90^\circ</math>.) By the Pythagorean Theorem, <math>BE^2+DE^2=BD^2.</math> We found that <math>DE=\frac{84}{13}</math>, and since we are given <math>BD=7</math>, we have <cmath>BE=\sqrt{7^2-\left (\frac{84}{13}\right )^2}=\frac{35}{13} \qquad (3).</cmath> | ||
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+ | Our answer, by equation <math>(1)</math>, is <math>\frac{EC}{BE}</math>. From equation <math>(2),</math> <math>EC=\frac{85}{13}</math> and from equation <math>(3),</math> <math>BE=\frac{35}{13}</math>. Therefore, our final answer is <math>\frac{\frac{85}{13}}{\frac{35}{13}}=\frac{17}{7} \implies \boxed{24}.</math> | ||
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+ | ~samrocksnature | ||
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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== | ||
+ | #[[2021 JMPSC Invitationals Problems|Other 2021 JMPSC Invitationals Problems]] | ||
+ | #[[2021 JMPSC Invitationals Answer Key|2021 JMPSC Invitationals Answer Key]] | ||
+ | #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]] | ||
+ | {{JMPSC Notice}} |
Latest revision as of 21:18, 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.