Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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...")
 
(→‎Solution)
 
(5 intermediate revisions by 2 users not shown)
Line 2: Line 2:
 
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>
  
−
==Solution==
+
==Solution 1==
−
asdf
+
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
 +
 
 +
==Solution 2 (Stewart's Theorem)==
 +
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==
 +
#[[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 17:10, 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

The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition. JMPSC.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_JMPSC_Invitationals_Problems/Problem_9&oldid=158125"