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

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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 $\mathrm{C}$ .

Solution 3

$[asy] unitsize(0.5cm); pair A,B,C,D,E; B=(0,0); C=(14,0); A=intersectionpoint(arc(B,13,0,90),arc(C,15,90,180)); draw(A--B--C--cycle); D=(7,0); E=(6.5,0); draw(A--E); draw(A--D); label("$A$",A,N); label("$B$",B,S); label("$C$",C,S); label("$E$",E,NW); label("$D$",D,NE); label("$13$",A--B,NW); label("$15$",A--C,NE); label("$14$",B--C,S); [/asy]$

Let's find the area of $\Delta ABC$ by Heron,

$s=\frac{a+b+c}{2}\\\\s=\frac{14+15+13}{2}\to\boxed{s=21}$

Then,

$A^2=s(s-a)(s-b)(s-c)\\\\A^2=21(21-14)(21-15)(21-13)\\\\\boxed{A=84}$

Knowing that D is the midpoint of BC, thus the area of $\Delta ADC=\Delta ABD$ , which is half the area of $\Delta ABC$ , so $\Delta ADC=\Delta ABD=42$ .

By Angle Bisector Theorem we know that:

$\frac{13}{BE}=\frac{15}{CE}$

$\boxed{CE=14-BE}$

$\frac{13}{BE}=\frac{15}{14-BE}$

$\boxed{BE=6.5}\Rightarrow CE=7.5$

Also, we know that:

$\frac{A_{\Delta ADE}}{A_{\Delta ABC}}=\frac{ED}{BC}$

And, we can easily see that $DE=0.5$ , so,

$\frac{A_{\Delta ADE}}{84}=\frac{0.5}{14}$

$\boxed{A_{\Delta ADE}=3}$

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
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