Difference between revisions of "2018 AIME I Problems/Problem 4"
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(→Solution 5 (Fastest [Law of Cosines])) |
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Solving for <math>cos A</math> in both equations, we get | Solving for <math>cos A</math> in both equations, we get | ||
− | <math>cos A = \frac{(10-x)^2}{(2)(10-x)(x)}</math> and <math>cos A = \frac{7}{25} \implies \frac{(10-x^2 | + | <math>cos A = \frac{(10-x)^2}{(2)(10-x)(x)}</math> and <math>cos A = \frac{7}{25} \implies \frac{(10-x)^2}{(2)(10-x)(x)} = \frac{7}{25} \implies x = \frac{250}{39}</math>, so the answer is <math>\boxed {289}</math> |
~RootThreeOverTwo | ~RootThreeOverTwo |
Revision as of 00:54, 13 May 2018
Contents
[hide]Problem 4
In and . Point lies strictly between and on and point lies strictly between and on so that . Then can be expressed in the form , where and are relatively prime positive integers. Find .
Solution 1 (No Trig)
We draw the altitude from to to get point . We notice that the triangle's height from to is 8 because it is a Right Triangle. To find the length of , we let represent and set up an equation by finding two ways to express the area. The equation is , which leaves us with . We then solve for the length , which is done through pythagorean theorm and get = . We can now see that is a Right Triangle. Thus, we set as , and yield that . Now, we can see = . Solving this equation, we yield , or . Thus, our final answer is . ~bluebacon008
Solution 2 (Coordinates)
Let , , and . Then, let be in the interval and parametrically define and as and respectively. Note that , so . This means that However, since is extraneous by definition, ~ mathwiz0803
Solution 3 (Law of Cosines)
As shown in the diagram, let denote . Let us denote the foot of the altitude of to as . Note that can be expressed as and is a triangle . Therefore, and . Before we can proceed with the Law of Cosines, we must determine . Using LOC, we can write the following statement: Thus, the desired answer is ~ blitzkrieg21
Solution 4
In isosceles triangle, draw the altitude from onto . Let the point of intersection be . Clearly, , and hence .
Now, we recognise that the perpendicular from onto gives us two -- triangles. So, we calculate and
. And hence,
Inspecting gives us Solving the equation gives
~novus677
Solution 5 (Fastest [Law of Cosines])
We can have 2 Law of Cosines applied on (one from and one from ),
and
Solving for in both equations, we get
and , so the answer is
~RootThreeOverTwo
See Also
2018 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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.