Difference between revisions of "2013 AMC 10B Problems/Problem 16"
(→Solution 2) |
(→Problem) |
||
Line 3: | Line 3: | ||
<math>\qquad\textbf{(A)}13\qquad\textbf{(B)}13.5\qquad\textbf{(C)}14\qquad\textbf{(D)}14.5\qquad\textbf{(E)}</math> | <math>\qquad\textbf{(A)}13\qquad\textbf{(B)}13.5\qquad\textbf{(C)}14\qquad\textbf{(D)}14.5\qquad\textbf{(E)}</math> | ||
+ | <asy> | ||
+ | pair A,B,C,D,E,P; | ||
+ | A=(0,0); | ||
+ | B=(80,0); | ||
+ | C=(20,40); | ||
+ | D=(50,30); | ||
+ | E=(40,0); | ||
+ | P=(33.3,13.3); | ||
+ | draw(A--B); | ||
+ | draw(B--C); | ||
+ | draw(A--C); | ||
+ | draw(C--E); | ||
+ | draw(A--D); | ||
+ | draw(D--E); | ||
+ | dot(A); | ||
+ | dot(B); | ||
+ | dot(C); | ||
+ | dot(D); | ||
+ | dot(E); | ||
+ | dot(P); | ||
+ | label("A",A,NNW); | ||
+ | label("B",B,NNE); | ||
+ | label("C",C,ENE); | ||
+ | label("D",D,ESE); | ||
+ | label("E",E,SSE); | ||
+ | label("P",P,SSE); | ||
+ | </asy> | ||
==Solution== | ==Solution== |
Revision as of 20:31, 22 February 2013
Problem
In triangle , medians
and
intersect at
,
,
, and
. What is the area of
?
Solution
Let us use mass points:
Assign mass
. Thus, because
is the midpoint of
,
also has a mass of
. Similarly,
has a mass of
.
and
each have a mass of
because they are between
and
and
and
respectively. Note that the mass of
is twice the mass of
, so AP must be twice as long as
. PD has length
, so
has length
and
has length
. Similarly,
is twice
and
, so
and
. Now note that triangle
is a
right triangle with the right angle
. This means that the quadrilateral
is a kite. The area of a kite is half the product of the diagonals,
and
. Recall that they are
and
respectively, so the area of
is
Solution 2
Note that triangle is a right triangle, and that the four angles that have point
are all right angles. Using the fact that the centroid (
) divides each median in a
ratio,
and
. Quadrilateral
is now just four right triangles. The area is