Difference between revisions of "2021 Fall AMC 12B Problems/Problem 24"
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==Solution 1== | ==Solution 1== | ||
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<b>Claim:</b> <math>\triangle ADC \sim \triangle ABE.</math> | <b>Claim:</b> <math>\triangle ADC \sim \triangle ABE.</math> | ||
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~Steven Chen (www.professorchenedu.com) | ~Steven Chen (www.professorchenedu.com) | ||
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+ | ==Solution 3 (Stewart Bash)== | ||
+ | By the Inscribed Angle Theorem and the definition of angle bisectors note that<cmath>\angle ABD=\angle ABC=\angle AEC\ \text{and}\ \angle BAD=\angle DAC=\angle EAC</cmath>so <math>\triangle ABD\sim\triangle AEC</math>. Therefore <math>\frac{AB}{AD}=\frac{AE}{AC}\rightarrow AB\cdot AC=AD\cdot AE</math>. By PoP, we can also express <math>AD\cdot AE</math> as <math>AB\cdot AF,</math> so <math>AB\cdot AC=AB\cdot AF\rightarrow AC=AF=20</math> and <math>BF=20-AB=20-11=9</math>. Let <math>CF=x</math>. Applying Stewart’s theorem on <math>\triangle ACF</math> with cevian <math>\overrightarrow{CB},</math> we have | ||
+ | <cmath>\begin{align*} | ||
+ | 11\cdot 9\cdot 20+24\cdot 20\cdot 24&=11x^{2}+20\cdot 9\cdot 20 \ | ||
+ | 1980+11{,}520&=11x^{2}+3600 \ | ||
+ | 13{,}500&=11x^{2}+3600 \ | ||
+ | 11x^{2}&=9900 \ | ||
+ | x^{2}&=900 \ | ||
+ | x&=\boxed{\textbf{(C)} ~30}. | ||
+ | \end{align*}</cmath> | ||
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+ | ~Punxsutawney Phil | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2021 Fall|ab=B|num-a=25|num-b=23}} | {{AMC12 box|year=2021 Fall|ab=B|num-a=25|num-b=23}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 23:39, 27 November 2021
Problem
Triangle has side lengths , and . The bisector of intersects in point , and intersects the circumcircle of in point . The circumcircle of intersects the line in points and . What is ?
Solution 1
Claim:
Proof: Note that and meaning that our claim is true by AA similarity.
Because of this similarity, we have that by Power of a Point. Thus,
Now, note that and plug into Law of Cosines to find the angle's cosine:
So, we observe that we can use Law of Cosines again to find :
- kevinmathz
Solution 3
This solution is based on this figure: Image:2021_AMC_12B_(Nov)_Problem_24,_sol.png
Denote by the circumcenter of . Denote by the circumradius of .
In , following from the law of cosines, we have
For , we have The fourth equality follows from the property that , , are concyclic. The fifth and the ninth equalities follow from the property that , , , are concyclic.
Because bisects , following from the angle bisector theorem, we have
Hence, .
In , following from the law of cosines, we have and
Hence, and . Hence, .
Now, we are ready to compute whose expression is given in Equation (2). We get .
Now, we can compute whose expression is given in Equation (1). We have .
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
Solution 3 (Stewart Bash)
By the Inscribed Angle Theorem and the definition of angle bisectors note thatso . Therefore . By PoP, we can also express as so and . Let . Applying Stewart’s theorem on with cevian we have
~Punxsutawney Phil
See Also
2021 Fall AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.