Difference between revisions of "2013 AIME II Problems/Problem 13"
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Also, as <math>LE:BE=5:3</math>, we know that <math>BL=\frac{8}{5}\cdot 3=\frac{24}{5}</math>. Furthermore, as <math>\triangle YLA\sim \triangle XCA</math>, and as <math>AL:LC=1:4</math>, we know that <math>LY=\frac{h}{5}</math> and <math>AY=\frac{b}{10}</math>, so <math>YB=\frac{9b}{10}</math>. Therefore, by the Pythagorean Theorem on <math>\triangle BLY</math>, we get <cmath>\frac{81b^2}{100}+\frac{h^2}{25}=\frac{576}{25}.</cmath> | Also, as <math>LE:BE=5:3</math>, we know that <math>BL=\frac{8}{5}\cdot 3=\frac{24}{5}</math>. Furthermore, as <math>\triangle YLA\sim \triangle XCA</math>, and as <math>AL:LC=1:4</math>, we know that <math>LY=\frac{h}{5}</math> and <math>AY=\frac{b}{10}</math>, so <math>YB=\frac{9b}{10}</math>. Therefore, by the Pythagorean Theorem on <math>\triangle BLY</math>, we get <cmath>\frac{81b^2}{100}+\frac{h^2}{25}=\frac{576}{25}.</cmath> | ||
Solving this system of equations yields <math>b=2\sqrt{7}</math> and <math>h=3</math>. Therefore, the area of the triangle is <math>3\sqrt{7}</math>, giving us an answer of <math>\boxed{010}</math>. | Solving this system of equations yields <math>b=2\sqrt{7}</math> and <math>h=3</math>. Therefore, the area of the triangle is <math>3\sqrt{7}</math>, giving us an answer of <math>\boxed{010}</math>. | ||
+ | |||
+ | === Solution 3 === | ||
+ | Let the coordinates of A, B and C be (-a, 0), (a, 0) and (0, h) respectively. | ||
+ | Then <math>D = (\frac{3a}{4}, \frac{h}{4})</math> and <math>E = (-\frac{a}{8},\frac{h}{8}).</math> | ||
+ | <math>EC^2 = 7</math> implies <math>a^2 + 49h^2 = 448</math>; <math>EB^2 = 9</math> implies <math>81a^2 + h^2 = 576.</math> | ||
+ | Solve this system of equations simultaneously, <math>a=\sqrt{7}</math> and <math>h=3</math>. | ||
+ | Area of the triangle is ah = <math>3\sqrt{7}</math>, giving us an answer of <math>\boxed{010}</math>. | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2013|n=II|num-b=12|num-a=14}} | {{AIME box|year=2013|n=II|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 23:52, 2 August 2015
Problem 13
In ,
, and point
is on
so that
. Let
be the midpoint of
. Given that
and
, the area of
can be expressed in the form
, where
and
are positive integers and
is not divisible by the square of any prime. Find
.
Solution
Solution 1
After drawing the figure, we suppose , so that
,
, and
.
Using cosine law for and
,we get
So,
, we get
Using cosine law in , we get
So,
Using cosine law in and
, we get
, and according to
, we can get
Using and
, we can solve
and
.
Finally, we use cosine law for ,
then , so the height of this
is
.
Then the area of is
, so the answer is
.
Solution 2
Let be the foot of the altitude from
with other points labelled as shown below.
Now we proceed using mass points. To balance along the segment
, we assign
a mass of
and
a mass of
. Therefore,
has a mass of
. As
is the midpoint of
, we must assign
a mass of
as well. This gives
a mass of
and
a mass of
.
Now let be the base of the triangle, and let
be the height. Then as
, and as
, we know that
Also, as
, we know that
. Therefore, by the Pythagorean Theorem on
, we know that
Also, as , we know that
. Furthermore, as
, and as
, we know that
and
, so
. Therefore, by the Pythagorean Theorem on
, we get
Solving this system of equations yields
and
. Therefore, the area of the triangle is
, giving us an answer of
.
Solution 3
Let the coordinates of A, B and C be (-a, 0), (a, 0) and (0, h) respectively.
Then and
implies
;
implies
Solve this system of equations simultaneously,
and
.
Area of the triangle is ah =
, giving us an answer of
.
See Also
2013 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.