2018 AIME I Problems/Problem 4
- 1 Problem 4
- 2 Solution 0
- 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 (Fastest via Law of Cosines)
- 10 Solution 8 (Easiest way- Coordinates without bash)
- 11 Solution 9 Even Faster Law of Cosines(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
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 .
By the Law of Cosines, we have: Let be midpoint of , then So, our answer is .
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 (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 .
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
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
Solution 7 (Fastest via Law of Cosines)
We can have 2 Law of Cosines applied on (one from and one from ),
Solving for in both equations, we get
and , so the answer is
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 .
Solution 9 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 .
- 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 .
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)
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
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