Difference between revisions of "2013 AIME II Problems/Problem 13"
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Let ABC be the reference triangle, with <math>A=(1,0,0)</math>, <math>B=(0,1,0)</math>, and <math>C=(0,0,1)</math>. We can easily calculate <math>D=(0,\frac{3}{4},\frac{1}{4})</math> and subsequently <math>E=(\frac{1}{2},\frac{3}{8},\frac{1}{8})</math>. Using distance formula on <math>\overline{EC}=(\frac{1}{2},\frac{3}{8},-\frac{7}{8})</math> and <math>\overline{EB}=(\frac{1}{2},-\frac{5}{8},\frac{1}{8})</math> gives | Let ABC be the reference triangle, with <math>A=(1,0,0)</math>, <math>B=(0,1,0)</math>, and <math>C=(0,0,1)</math>. We can easily calculate <math>D=(0,\frac{3}{4},\frac{1}{4})</math> and subsequently <math>E=(\frac{1}{2},\frac{3}{8},\frac{1}{8})</math>. Using distance formula on <math>\overline{EC}=(\frac{1}{2},\frac{3}{8},-\frac{7}{8})</math> and <math>\overline{EB}=(\frac{1}{2},-\frac{5}{8},\frac{1}{8})</math> gives | ||
− | + | ||
− | \begin{align} | + | \begin{align*} |
\begin{cases} | \begin{cases} | ||
7=|EC|^2= | 7=|EC|^2= | ||
9=|EB|^2= | 9=|EB|^2= | ||
\end{cases} | \end{cases} | ||
− | \end{align} | + | \end{align*} |
− | + | $ | |
==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 15:01, 18 March 2017
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 .
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 .
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
\begin{align*}
\begin{cases}
7=|EC|^2=
9=|EB|^2=
\end{cases}
\end{align*}
$
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.