Difference between revisions of "2018 AIME I Problems/Problem 4"
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− | == Even Faster Law of Cosines== | + | == Even Faster Law of Cosines(1 variable equation)== |
Doing law of cosines we know that <math>\cos A</math> is <math>\frac{7}{25}.</math> Dropping the perpendicular from <math>D</math> to <math>AE</math> we get that <cmath>\frac{10-x}{2}=\frac{7x}{25}.</cmath> | Doing law of cosines we know that <math>\cos A</math> is <math>\frac{7}{25}.</math> Dropping the perpendicular from <math>D</math> to <math>AE</math> we get that <cmath>\frac{10-x}{2}=\frac{7x}{25}.</cmath> |
Revision as of 12:37, 19 February 2019
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 via 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
Solution 6 (Easiest way- Coordinates without bash)
Let , and . From there, we know that , so line is . Hence, for some , and so . Now, notice that by symmetry, , so . Because , we now have , which simplifies to , so , and . It follows that , and our answer is .
-Stormersyle
Even Faster Law of Cosines(1 variable equation)
Doing law of cosines we know that is Dropping the perpendicular from to we get that Solving for we get so our answer is .
-harsha12345
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.