Difference between revisions of "2013 AIME II Problems/Problem 13"
m (→Solution 5 (Barycentric Coordinates)) |
(→Solutions) |
||
Line 138: | Line 138: | ||
Then plugging gives <math>b=4</math> and <math>c=2\sqrt{7}</math>. Then the height from <math>C</math> is <math>3</math>, and the area is <math>3\sqrt{7}</math> and our answer is <math>\boxed{010}</math>. | Then plugging gives <math>b=4</math> and <math>c=2\sqrt{7}</math>. Then the height from <math>C</math> is <math>3</math>, and the area is <math>3\sqrt{7}</math> and our answer is <math>\boxed{010}</math>. | ||
+ | |||
+ | ===Solution 7=== | ||
+ | Let <math>C=(0,0), A=(x,y),</math> and <math>B=(-x,y)</math>. | ||
+ | It is trivial to show that <math>D=\left(-\frac{3}{4}x,\frac{3}{4}y\right)</math> and <math>E=\left(\frac{1}{8}x,\frac{7}{8}y\right)</math>. Thus, since <math>BE=3</math> and <math>CE=\sqrt{7}</math>, we get that | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \left(\frac{1}{8}x\right)^2+\left(\frac{7}{8}y\right)^2&=7 \\ | ||
+ | \left(\frac{9}{8}x\right)^2+\left(\frac{1}{8}y\right)^2&=9 \\ | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Multiplying both equations by <math>64</math>, we get that | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | x^2+49y^2&=448 \\ | ||
+ | 81x^2+y^2&=576 \\ | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Solving these equations, we get that <math>x=\sqrt{7}</math> and <math>y=3</math>. | ||
+ | |||
+ | Thus, the area of <math>\triangle ABC</math> is <math>xy=3\sqrt{7}</math>, so our answer is <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 14:23, 24 November 2021
Contents
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
.
Solutions
Solution 1
We can set . Set
, therefore
. Thereafter, by Stewart's Theorem on
and cevian
, we get
. Also apply Stewart's Theorem on
with cevian
. After simplification,
. Therefore,
. Finally, note that (using [] for area)
, because of base-ratios. Using Heron's Formula on
, as it is simplest, we see that
, so your answer is
.
Solution 2
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 3
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 4
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 5
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
.
Note to writter: Couldn't we just use Heron's formula for after
is solved then noticing that
?
Solution 6 (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
.
Solution 7
Let and
.
It is trivial to show that
and
. Thus, since
and
, we get that
Multiplying both equations by , we get that
Solving these equations, we get that and
.
Thus, the area of is
, so 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.