Difference between revisions of "2018 AIME I Problems/Problem 4"
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− | We draw the altitude from <math>B</math> to <math>\overline{AC}</math> to get point <math>F</math>. We notice that the triangle's height from <math>A</math> to <math>\overline{BC}</math> is 8 because it is a <math>3-4-5</math> Right Triangle. To find the length of <math>\overline{BF}</math>, we let <math>h</math> represent <math>\overline{BF}</math> and set up an equation by finding two ways to express the area. The equation is <math>(8)(12)=(10)(h)</math>, which leaves us with <math>h=9.6</math>. We then solve for the length <math>\overline{AF}</math>, which is done through pythagorean theorm and get <math>\overline{AF}</math> = <math>2.8</math>. We can now see that <math>\triangle | + | We draw the altitude from <math>B</math> to <math>\overline{AC}</math> to get point <math>F</math>. We notice that the triangle's height from <math>A</math> to <math>\overline{BC}</math> is 8 because it is a <math>3-4-5</math> Right Triangle. To find the length of <math>\overline{BF}</math>, we let <math>h</math> represent <math>\overline{BF}</math> and set up an equation by finding two ways to express the area. The equation is <math>(8)(12)=(10)(h)</math>, which leaves us with <math>h=9.6</math>. We then solve for the length <math>\overline{AF}</math>, which is done through pythagorean theorm and get <math>\overline{AF}</math> = <math>2.8</math>. We can now see that <math>\triangle AFB</math> is a <math>7-24-25</math> Right Triangle. Thus, we set <math>\overline{AG}</math> as <math>5-</math><math>\tfrac{x}{2}</math>, and yield that <math>\overline{AD}</math> <math>=</math> <math>(</math> <math>5-</math> <math>\tfrac{x}{2}</math> <math>)</math> <math>(</math> <math>\tfrac{25}{7}</math> <math>)</math>. Now, we can see <math>x</math> = <math>(</math> <math>5-</math> <math>\tfrac{x}{2}</math> <math>)</math> <math>(</math> <math>\tfrac{25}{7}</math> <math>)</math>. Solving this equation, we yield <math>39x=250</math>, or <math>x=</math> <math>\tfrac{250}{39}</math>. Thus, our final answer is <math>250+39=\boxed{289}</math>. |
~bluebacon008 | ~bluebacon008 | ||
Revision as of 14:39, 24 February 2021
Contents
- 1 Problem 4
- 2 Solution 1
- 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 Video Solution
- 15 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 1
. 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 .
~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 (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 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 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
- 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
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 | |
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.