Difference between revisions of "2013 AIME II Problems/Problem 13"
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In <math>\triangle ABC</math>, <math>AC = BC</math>, and point <math>D</math> is on <math>\overline{BC}</math> so that <math>CD = 3\cdot BD</math>. Let <math>E</math> be the midpoint of <math>\overline{AD}</math>. Given that <math>CE = \sqrt{7}</math> and <math>BE = 3</math>, the area of <math>\triangle ABC</math> can be expressed in the form <math>m\sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers and <math>n</math> is not divisible by the square of any prime. Find <math>m+n</math>. | In <math>\triangle ABC</math>, <math>AC = BC</math>, and point <math>D</math> is on <math>\overline{BC}</math> so that <math>CD = 3\cdot BD</math>. Let <math>E</math> be the midpoint of <math>\overline{AD}</math>. Given that <math>CE = \sqrt{7}</math> and <math>BE = 3</math>, the area of <math>\triangle ABC</math> can be expressed in the form <math>m\sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers and <math>n</math> is not divisible by the square of any prime. Find <math>m+n</math>. | ||
==Solution== | ==Solution== | ||
+ | |||
+ | ==Stewart's Solid Start== | ||
+ | Draw a good diagram. This involves paper and ruler. We can set <math>AE=ED=m</math>. Set <math>BD=k</math>, therefore <math>CD=3k, AC=4k</math>. Thereafter, by Stewart's Theorem on <math>\triangle ACD</math> and cevian <math>CE</math>, we get <math>2m^2+14=25k^2</math>. Also apply Stewart's Theorem on <math>\triangle CEB</math> with cevian <math>DE</math>. After simplification, <math>2m^2=17-6k^2</math>. Therefore, <math>k=1, m=\frac{\sqrt{22}}{2}</math>. Finally, note that (using [] for area) <math>[CED]=[CAE]=3[EDB]=3[AEB]=\frac{3}{8}[ABC]</math>, because of base-ratios. Using Heron's Formula on <math>\triangle EDB</math>, as it is simplest, we see that <math>[ABC]=3\sqrt{7}</math>, so your answer is <math>10</math>. Every step was straightforward and by adopting the simplest steps, we solved the problem quickly. | ||
=== Solution 1 === | === Solution 1 === |
Revision as of 20:52, 28 February 2018
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
Stewart's Solid Start
Draw a good diagram. This involves paper and ruler. 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 . Every step was straightforward and by adopting the simplest steps, we solved the problem quickly.
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 .
Solution 6 (Desperate for points)
Note that uniquely determines . Assume WTMLOG (without too much loss of generality) that . Then, the 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.