Difference between revisions of "2013 AMC 10B Problems/Problem 16"
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From the solutions above, we know that the sides CP and AP are 3 and 4 respectively because of the properties of medians that divide cevians into 1:2 ratios. We can then proceed to use the heron's formula on the middle triangle EPD and get the area of EPD as 3/2, (its simple computation really, nothing large). Then we can find the areas of the remaining triangles based on side and ratio length of the bases. | From the solutions above, we know that the sides CP and AP are 3 and 4 respectively because of the properties of medians that divide cevians into 1:2 ratios. We can then proceed to use the heron's formula on the middle triangle EPD and get the area of EPD as 3/2, (its simple computation really, nothing large). Then we can find the areas of the remaining triangles based on side and ratio length of the bases. | ||
+ | |||
+ | ==Solution 5== | ||
+ | We know that <math>[AEDC]=\frac{3}{4}[ABC]</math>, and <math>[ABC]=3[PAC]</math> using median properties. So Now we try to find <math>[PAC]</math>. Since <math>\triangle PAC\sim \triangle PDE</math>, then the side lengths of <math>\triangle PAC</math> are twice as long as <math>\triangle PDE</math> since <math>D</math> and <math>E</math> are midpoints. Therefore, <math>\frac{[PAC]}{[PDE]}=2^2=4</math>. It suffices to compute <math>[PDE]</math>. Notice that <math>(1.5, 2, 2.5)</math> is a Pythagorean Triple, so <math>[PDE]=\frac{1.5\times 2}{2}=1.5</math>. This implies <math>[PAC]=1.5\cdot 4=6</math>, and then <math>[ABC]=3\cdot 6=18</math>. Finally, <math>[AEDC]=\frac{3}{4}\times 18=\boxed{13.5}</math>. | ||
+ | |||
+ | ~CoolJupiter | ||
== See also == | == See also == |
Revision as of 09:11, 31 August 2020
Problem
In triangle , medians
and
intersect at
,
,
, and
. What is the area of
?
Solution 1
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
. Since the diagonals of quadrilaterals
,
and
, are perpendicular, the area of
is
Solution 2
Note that triangle is a right triangle, and that the four angles (angles
and
) 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
Solution 3
From the solution above, we can find that the lengths of the diagonals are and
. Now, since the diagonals of AEDC are perpendicular, we use the area formula to find that the total area is
Solution 4
From the solutions above, we know that the sides CP and AP are 3 and 4 respectively because of the properties of medians that divide cevians into 1:2 ratios. We can then proceed to use the heron's formula on the middle triangle EPD and get the area of EPD as 3/2, (its simple computation really, nothing large). Then we can find the areas of the remaining triangles based on side and ratio length of the bases.
Solution 5
We know that , and
using median properties. So Now we try to find
. Since
, then the side lengths of
are twice as long as
since
and
are midpoints. Therefore,
. It suffices to compute
. Notice that
is a Pythagorean Triple, so
. This implies
, and then
. Finally,
.
~CoolJupiter
See also
2013 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 10 Problems and Solutions |
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