2018 AIME I Problems/Problem 4
Contents
[hide]- 1 Problem 4
- 2 Solution (Easiest Law of Cosines)
- 3 Solution 1 (No Trig)
- 4 Solution 2 (Easy Similar Triangles)
- 5 Solution 3 (Algebra w/ Law of Cosines)
- 6 Solution 4 (Coordinates)
- 7 Solution 5 (Law of Cosines)
- 8 Solution 6
- 9 Solution 7 (Area into Similar Triangles)
- 10 Solution 8 (Easiest way- Coordinates without bash)
- 11 Solution 9 one second accurate solve(1 variable equation)
- 12 Solution 10 (Law of Sines)
- 13 Solution 11 (Trigonometry)
- 14 Solution 12 (Double Angle Identity)
- 15 Video Solution
- 16 See Also
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 (Easiest Law of Cosines)
We apply Law of Cosines on twice (one from and one from ),
Solving for in both equations, we get
Note that this strategy of applying LOC on congruent or supplementary angles is very common. It's also how Stewart's Theorem is derived.
-RootThreeOverTwo, edits by epiconan
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
Diagram edited by Afly
Solution 2 (Easy Similar Triangles)
We start by adding a few points to the diagram. Call the midpoint of , and the midpoint of . (Note that and are altitudes of their respective triangles). We also call . Since triangle is isosceles, , and . Since , and . Since is a right triangle, .
Since and , triangles and are similar by Angle-Angle similarity. Using similar triangle ratios, we have . and because there are triangles in the problem. Call . Then , , and . Thus . Our ratio now becomes . Solving for gives us . Since is a height of the triangle , , or . Solving the equation gives us , so our answer is .
Solution 3 (Algebra w/ Law of Cosines)
As in the diagram, let . Consider point on the diagram shown above. Our goal is to be able to perform Pythagorean Theorem on , and . Let . Therefore, it is trivial to see that (leave everything squared so that it cancels nicely at the end). By Pythagorean Theorem on Triangle , we know that . Finally, we apply Law of Cosines on Triangle . We know that . Therefore, we get that . We can now do our final calculation: After some quick cleaning up, we get Therefore, our answer is .
~awesome1st
Solution 4 (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 5 (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 6
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 7 (Area into Similar Triangles)
After calling and , we see we have length ratios in terms of , which motivates area ratios. We look at the area of triangle in two ways in order to find (perpendicular from to ), and then use similar triangles to find .
Using area ratios, . (To find the total area , drop the altitudes from to , and call the foot of the altitude . By the 6-8-10 triangle, the height is and the area of is .)
The second way of finding the area of triangle is . The base is , and is the height. Therefore,
Now we have in terms of , we use the similar triangles and and set up the proportion
Solution 8 (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
Solution 9 one second accurate solve(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
- It is good to remember that doubling the smallest angle of a 3-4-5 triangle gives the larger (not right) angle in a 7-24-25 triangle.
Solution 10 (Law of Sines)
Let's label and . Using isosceles triangle properties and the triangle angle sum equation, we get Solving, we find .
Relabelling our triangle, we get . Dropping an altitude from to and using the Pythagorean theorem, we find . Using the sine area formula, we see . Plugging in our sine angle cofunction identity, , we get .
Now, using the Law of Sines on , we get After applying numerous trigonometric and algebraic tricks, identities, and simplifications, such as and , we find .
Therefore, our answer is .
~Tiblis
Solution 11 (Trigonometry)
We start by labelling a few angles (all of them in degrees). Let . Also let . By sine rule in we get Using sine rule in , we get . Hence we get . Hence . Therefore, our answer is
Alternatively, use sine rule in . (It’s easier)
~Prabh1512
Solution 12 (Double Angle Identity)
We let . Then, angle is and so is angle . We note that . We drop an altitude from to , and we call the foot . We note that . Using the double angle identity, we have Therefore, We now use the Pythagorean Theorem, which gives . Rearranging and simplifying, this becomes . Using the quadratic formula, this is . We take out a from the square root and make it a outside of the square root to make it simpler. We end up with . We note that this must be less than 10 to ensure that is positive. Therefore, we take the minus, and we get
~john0512
Video Solution
https://www.youtube.com/watch?v=iE8paW_ICxw
https://youtu.be/dI6uZ67Ae2s ~yofro
See Also
2018 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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