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Difference between revisions of "2000 AMC 12 Problems/Problem 19"

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== Solution 1==
 
== Solution 1==
 
The answer is exactly <math>3</math>, choice <math>\mathrm{(C)}</math>.
 
The answer is exactly <math>3</math>, choice <math>\mathrm{(C)}</math>.
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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 <math>ADE, DE</math>, 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>.
+
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 <math>ADE, DE</math>, 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>.
  
 
==Solution 2==
 
==Solution 2==

Revision as of 21:15, 30 December 2016

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 1

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 $\frac{12*\frac{1}{2}}{2}=3$.

Solution 2

The area of $ADE$ is $\frac{DE\cdot h}{2}=\frac{DE}{BC} \cdot \frac{BC\cdot h}{2}=\frac{DE}{BC}[ABC]$ where $h$ is the height of triangle $ABC$. Using Angle Bisector Theorem, we find $\frac{13}{BE}=\frac{15}{14-BE}$, which we solve to get $BE=\frac{13}{2}$. $D$ is the midpoint of $BC$ so $BD=7$. Thus we get the base of triangle $ADE, DE$, to be $\frac{1}{2}$ units long. We can now use Heron's Formula on $ABC$. \[s=\frac{AB+BC+AC}{2}=21\] \[[ABC]=\sqrt{(s)(s-AB)(s-BC)(s-AC)}=\sqrt{(21)(8)(7)(6)}=84\] \[\frac{DE}{BC}[ABC]=\frac{\frac{1}{2}}{14}\cdot 84=3\] Therefore, the answer is $\mathrm{C}$.

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

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