Difference between revisions of "2013 AIME II Problems/Problem 13"
(→Solution 5 (Barycentric Coordinates)) |
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After drawing the figure, we suppose <math>BD=a</math>, so that <math>CD=3a</math>, <math>AC=4a</math>, and <math>AE=ED=b</math>. | After drawing the figure, we suppose <math>BD=a</math>, so that <math>CD=3a</math>, <math>AC=4a</math>, and <math>AE=ED=b</math>. | ||
− | Using | + | Using Law of Cosines for <math>\triangle AEC</math> and <math>\triangle CED</math>,we get |
<cmath>b^2+7-2\sqrt{7}\cdot \cos(\angle CED)=9a^2\qquad (1)</cmath> | <cmath>b^2+7-2\sqrt{7}\cdot \cos(\angle CED)=9a^2\qquad (1)</cmath> | ||
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So, <math>(1)+(2)</math>, we get<cmath>2b^2+14=25a^2. \qquad (3)</cmath> | So, <math>(1)+(2)</math>, we get<cmath>2b^2+14=25a^2. \qquad (3)</cmath> | ||
− | Using | + | Using Law of Cosines in <math>\triangle ACD</math>, we get |
<cmath>b^2+9a^2-2\cdot 2b\cdot 3a\cdot \cos(\angle ADC)=16a^2</cmath> | <cmath>b^2+9a^2-2\cdot 2b\cdot 3a\cdot \cos(\angle ADC)=16a^2</cmath> | ||
Line 18: | Line 18: | ||
So, <cmath>\cos(\angle ADC)=\frac{7a^2-4b^2}{12ab}.\qquad (4)</cmath> | So, <cmath>\cos(\angle ADC)=\frac{7a^2-4b^2}{12ab}.\qquad (4)</cmath> | ||
− | Using | + | Using Law of Cosines in <math>\triangle EDC</math> and <math>\triangle EDB</math>, we get |
<cmath>b^2+9a^2-2\cdot 3a\cdot b\cdot \cos(\angle ADC)=7\qquad (5)</cmath> | <cmath>b^2+9a^2-2\cdot 3a\cdot b\cdot \cos(\angle ADC)=7\qquad (5)</cmath> | ||
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Using <math>(3)</math> and <math>(7)</math>, we can solve <math>a=1</math> and <math>b=\frac{\sqrt{22}}{2}</math>. | Using <math>(3)</math> and <math>(7)</math>, we can solve <math>a=1</math> and <math>b=\frac{\sqrt{22}}{2}</math>. | ||
− | Finally, we use | + | Finally, we use Law of Cosines for <math>\triangle ADB</math>, |
<cmath>4(\frac{\sqrt{22}}{2})^2+1+2\cdot\2(\frac{\sqrt{22}}{2})\cdot \cos(ADC)=AB^2</cmath> | <cmath>4(\frac{\sqrt{22}}{2})^2+1+2\cdot\2(\frac{\sqrt{22}}{2})\cdot \cos(ADC)=AB^2</cmath> |
Revision as of 22:32, 29 December 2017
Contents
[hide]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 Law of Cosines for and
,we get
So,
, we get
Using Law of Cosines in , we get
So,
Using Law of Cosines in and
, we get
, and according to
, we can get
Using and
, we can solve
and
.
Finally, we use Law of Cosines 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
.
Solution 4
(Thanks to writer of Solution 2)
Let . Then
and
. Also, let
. Using Stewart's Theorem on
gives us the equation
or, after simplifying,
. We use Stewart's again on
:
, which becomes
. Substituting
, we see that
, or
. Then
.
We now use Law of Cosines on .
. Plugging in for
and
,
, so
. Using the Pythagorean trig identity
,
, so
.
, and our answer is
.
Solution 5 (Barycentric Coordinates)
Let ABC be the reference triangle, with ,
, and
. We can easily calculate
and subsequently
. Using distance formula on
and
gives
But we know that , so we can substitute and now we have two equations and two variables. So we can clear the denominators and prepare to cancel a variable:
Then we add the equations to get
Then plugging gives and
. Then the height from
is
, and the area is
and our answer is
.
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.