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"

(→‎Solution 3)
(→‎Solution 3)
Line 37: Line 37:
 
label("$15$",A--C,NE);
 
label("$15$",A--C,NE);
 
label("$14$",B--C,S);
 
label("$14$",B--C,S);
 +
label("$6.5$",B--E,N);
 +
label("$7$",C--D,N);
 
</asy>
 
</asy>
  
Line 47: Line 49:
 
<math>A^2=s(s-a)(s-b)(s-c)\\\\A^2=21(21-14)(21-15)(21-13)\\\\\boxed{A=84}</math>
 
<math>A^2=s(s-a)(s-b)(s-c)\\\\A^2=21(21-14)(21-15)(21-13)\\\\\boxed{A=84}</math>
  
−
Knowing that D is the midpoint of BC, thus the area of <math>\Delta ADC=\Delta ABD</math>, which is half the area of <math>\Delta ABC</math>, so <math>\Delta ADC=\Delta ABD=42</math>.
+
Knowing that D is the midpoint of BC, then <math>BD=CD=7</math>.
  
 
By [[Angle Bisector Theorem]] we know that:
 
By [[Angle Bisector Theorem]] we know that:

Revision as of 16:27, 7 August 2017

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); label("$6.5$",B--E,N); label("$7$",C--D,N); [/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, then $BD=CD=7$.

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,

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

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

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=86933"