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 "2000 AMC 12 Problems/Problem 19"

m
(→‎Solution)
Line 5: Line 5:
  
 
== Solution ==
 
== Solution ==
−
The answer is exactly 3, choice <math>\mathrm{(C)}</math>.
+
The answer is exactly <math>3</math>, choice <math>\mathrm{(C)}</math>.
−
We can use either Heron's on the larger triangle to calculate the height of ADE or we can decompose ABC into two right triangles of 5-12-13 and 9-12-15 to notice that height is 12. Then angle bisector theorem leads directly to the base of ADE, which leads to an area of 3.
+
We can find the area of triangle <math>ADE</math> by using the simple formula <math>\frac{bh}{2}</math>. Dropping an altitude from <math>A</math>, we see that it has length <math>12</math> ( we can split the large triangle into a <math>9-12-15</math> and a <math>5-12-13</math> triangle). Then we can apply the Angle Bisector theorem on triangle <math>ABC</math> to solve for <math>BE</math>. Solving <math>\frac{13}{BE}=\frac{15}{14-BE}</math>, we get that <math>BE=\frac{13}{2}</math>. <math>D</math> is the midpoint of <math>BC</math> so <math>BD=7</math>. Thus we get the base of triangle ADE, DE, to be <math>\frac{1}{2}</math> units long. Applying the formula <math>\frac{bh}{2}</math>, we get <math>\frac{12*\frac{1}{2}}{2}=3</math>.
  
 
== See also ==
 
== See also ==

Revision as of 23:39, 16 August 2015

Problem

In triangle $ABC$, $AB = 13$, $BC = 14$, $AC = 15$. Let $D$ denote the midpoint of $\overline{BC}$ and let $E$ denote the intersection of $\overline{BC}$ with the bisector of angle $BAC$. Which of the following is closest to the area of the triangle $ADE$?

$\text {(A)}\ 2 \qquad \text {(B)}\ 2.5 \qquad \text {(C)}\ 3 \qquad \text {(D)}\ 3.5 \qquad \text {(E)}\ 4$

Solution

The answer is exactly $3$, choice $\mathrm{(C)}$. We can find the area of triangle $ADE$ by using the simple formula $\frac{bh}{2}$. Dropping an altitude from $A$, we see that it has length $12$ ( we can split the large triangle into a $9-12-15$ and a $5-12-13$ triangle). Then we can apply the Angle Bisector theorem on triangle $ABC$ to solve for $BE$. Solving $\frac{13}{BE}=\frac{15}{14-BE}$, we get that $BE=\frac{13}{2}$. $D$ is the midpoint of $BC$ so $BD=7$. Thus we get the base of triangle ADE, DE, to be $\frac{1}{2}$ units long. Applying the formula $\frac{bh}{2}$, we get $\frac{12*\frac{1}{2}}{2}=3$.

See also

2000 AMC 12 (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_12_Problems/Problem_19&oldid=71659"