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Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 9"

(Created page with "==Problem== In <math>\triangle ABC</math>, let <math>D</math> be on <math>\overline{AB}</math> such that <math>AD=DC</math>. If <math>\angle ADC=2\angle ABC</math>, <math>AD=1...")
 
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In <math>\triangle ABC</math>, let <math>D</math> be on <math>\overline{AB}</math> such that <math>AD=DC</math>. If <math>\angle ADC=2\angle ABC</math>, <math>AD=13</math>, and <math>BC=10</math>, find <math>AC.</math>
 
In <math>\triangle ABC</math>, let <math>D</math> be on <math>\overline{AB}</math> such that <math>AD=DC</math>. If <math>\angle ADC=2\angle ABC</math>, <math>AD=13</math>, and <math>BC=10</math>, find <math>AC.</math>
  
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==Solution==
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==Solution 1==
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asdf
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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> ~samrocksnature
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==Solution 2 (Stewart's Theorem)==
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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
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==See also==
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#[[2021 JMPSC Invitationals Problems|Other 2021 JMPSC Invitationals Problems]]
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#[[2021 JMPSC Invitationals Answer Key|2021 JMPSC Invitationals Answer Key]]
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#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
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{{JMPSC Notice}}

Latest revision as of 18:10, 11 July 2021

Problem

In $\triangle ABC$, let $D$ be on $\overline{AB}$ such that $AD=DC$. If $\angle ADC=2\angle ABC$, $AD=13$, and $BC=10$, find $AC.$

Solution 1

From the fact that $AD=DB$ and $\angle ADC = 2\angle ABC,$ we find that $\triangle ABC$ is a right triangle with a right angle at $C;$ thus by the Pythagorean Theorem we obtain $AC=\boxed{24}.$ ~samrocksnature

Solution 2 (Stewart's Theorem)

Note that $\angle BDC = 180-x$, which means $\angle DCB = \angle DBC$ and $AD=DB=DC=13$. Now, Stewart's Theorem dictates $x^2 \cdot 13 = 7488$, yielding $AC=x=\boxed{24}$ ~Geometry285

See also

  1. Other 2021 JMPSC Invitationals Problems
  2. 2021 JMPSC Invitationals Answer Key
  3. All JMPSC Problems and Solutions

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