Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 9"
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==Solution== | ==Solution== | ||
From the fact that <math>AD=DB</math> and <math>\angle ADC = 2\angle ABC,</math> we find that <math>\triangle ABC</math> is a right triangle with a right angle at <math>C;</math> thus by the Pythagorean Theorem we obtain <math>AC=\boxed{24}.</math> | From the fact that <math>AD=DB</math> and <math>\angle ADC = 2\angle ABC,</math> we find that <math>\triangle ABC</math> is a right triangle with a right angle at <math>C;</math> thus by the Pythagorean Theorem we obtain <math>AC=\boxed{24}.</math> | ||
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| + | ==Solution== | ||
| + | Note that <math>\angle BDC = 180-x</math>, which means <math>\angle DCB = \angle DBC</math> and <math>AD=DB=DC=13</math>. Now, Stewart's Theorem dictates <math>x^2 \cdot 13 = 7488</math>, yielding <math>AC=x=\boxed{24}</math> ~Geometry285 | ||
==See also== | ==See also== | ||
Revision as of 17:10, 11 July 2021
Contents
Problem
In 
, let 
be on 
such that 
. If 
, 
, and 
, find 
Solution
From the fact that 
and 
we find that is a right triangle with a right angle at 
thus by the Pythagorean Theorem we obtain 
Solution
Note that 
, which means 
and 
. Now, Stewart's Theorem dictates 
, yielding 
~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. 
