Difference between revisions of "2000 AMC 12 Problems/Problem 19"
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== Solution == | == Solution == | ||
The answer is exactly <math>3</math>, choice <math>\mathrm{(C)}</math>. | 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 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>. | + | 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>. |
== See also == | == See also == |
Revision as of 00:40, 17 August 2015
Problem
In triangle ,
,
,
. Let
denote the midpoint of
and let
denote the intersection of
with the bisector of angle
. Which of the following is closest to the area of the triangle
?
Solution
The answer is exactly , choice
.
We can find the area of triangle
by using the simple formula
. Dropping an altitude from
, we see that it has length
( we can split the large triangle into a
and a
triangle). Then we can apply the Angle Bisector theorem on triangle
to solve for
. Solving
, we get that
.
is the midpoint of
so
. Thus we get the base of triangle
, to be
units long. Applying the formula
, we get
.
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.