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

Latest revision as of 16:37, 17 February 2024

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

$[asy] pair A,B,C,D,E; B=(0,0); C=(14,0); A=intersectionpoint(arc(B,13,0,90),arc(C,15,90,180)); draw(A--B--C--cycle); D=(7,0); E=(6.5,0); draw(A--E); draw(A--D); label("$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_{

Video Solution

https://youtu.be/A5oxZp3jem4

@@ Line 4: / Line 4: @@
 <math>\text {(A)}\ 2 \qquad \text {(B)}\ 2.5 \qquad \text {(C)}\ 3 \qquad \text {(D)}\ 3.5 \qquad \text {(E)}\ 4</math>
-== Solution ==
+== Solution 1==
-The area of <math>ADE = 3</math>\ \mathrm{(C)}$.
+The answer is exactly <math>3</math>, choice <math>\mathrm{(C)}</math>.
+We can find the area of triangle <math>ADE</math> by using the simple formula <math>\frac{bh}{2}</math>. Dropping an altitude from <math>A</math>, we see that it has length <math>12</math> (we can split the large triangle into a <math>9-12-15</math> and a <math>5-12-13</math> triangle). Then we can apply the [[Angle Bisector Theorem]] on triangle <math>ABC</math> to solve for <math>BE</math>. Solving <math>\frac{13}{BE}=\frac{15}{14-BE}</math>, we get that <math>BE=\frac{13}{2}</math>. <math>D</math> is the midpoint of <math>BC</math> so <math>BD=7</math>. Thus we get the base of triangle <math>ADE, DE</math>, to be <math>\frac{1}{2}</math> units long. Applying the formula <math>\frac{bh}{2}</math>, we get <math>\frac{12*\frac{1}{2}}{2}=3</math>.
+==Solution 2==
+The area of <math>ADE</math> is <math>\frac{DE\cdot h}{2}=\frac{DE}{BC} \cdot \frac{BC\cdot h}{2}=\frac{DE}{BC}[ABC]</math> where <math>h</math> is the height of triangle <math>ABC</math>. Using Angle Bisector Theorem, we find <math>\frac{13}{BE}=\frac{15}{14-BE}</math>, which we solve to get <math>BE=\frac{13}{2}</math>. <math>D</math> is the midpoint of <math>BC</math> so <math>BD=7</math>. Thus we get the base of triangle <math>ADE, DE</math>, to be <math>\frac{1}{2}</math> units long. We can now use [[Heron's Formula]] on <math>ABC</math>.
+<cmath>s=\frac{AB+BC+AC}{2}=21</cmath>
+<cmath>[ABC]=\sqrt{(s)(s-AB)(s-BC)(s-AC)}=\sqrt{(21)(8)(7)(6)}=84</cmath>
+<cmath>\frac{DE}{BC}[ABC]=\frac{\frac{1}{2}}{14}\cdot 84=3</cmath>
+Therefore, the answer is <math>\mathrm{C}</math>.
+==Solution 3==
+<asy>
+pair A,B,C,D,E;
+B=(0,0);
+C=(14,0);
+A=intersectionpoint(arc(B,13,0,90),arc(C,15,90,180));
+draw(A--B--C--cycle);
+D=(7,0);
+E=(6.5,0);
+draw(A--E);
+draw(A--D);
+label("$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 <math>\Delta ABC</math> by Heron,
+<math>s=\frac{a+b+c}{2}\\\\s=\frac{14+15+13}{2}\to\boxed{s=21}</math>
+Then,
+<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, then <math>BD=CD=7</math>.
+By [[Angle Bisector Theorem]] we know that:
+<math>\frac{13}{BE}=\frac{15}{CE}</math>
+<math>\boxed{CE=14-BE}</math>
+<math>\frac{13}{BE}=\frac{15}{14-BE}</math>
+<math>\boxed{BE=6.5}\Rightarrow CE=7.5</math>
+Also, we know that:
+<math>\frac{A_{\Delta ADE}}{A_{\Delta ABC}}=\frac{ED}{BC}</math>
+And, we can easily see that <math>DE=0.5</math>, so,
+$\frac{A_{
+== Video Solution ==
+https://youtu.be/A5oxZp3jem4
 == 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
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