2019 AIME I Problems/Problem 3
Contents
- 1 Problem
- 2 Diagram
- 3 Solution 1
- 4 Solution 2
- 5 Solution 3 (Easiest, uses only basic geometry too)
- 6 Solution 4
- 7 Solution 5 (Official MAA)
- 8 Solution 6 (Using simple trigonometry)
- 9 Video Solution #1(Complementary Area Counting?)
- 10 Video Solution
- 11 Video Solution 2
- 12 Video Solution 3
- 13 See Also
Problem
In , , , and . Points and lie on , points and lie on , and points and lie on , with . Find the area of hexagon .
Diagram
Solution 1
We know the area of the hexagon to be . Since , we know that is a right triangle. Thus the area of is . Another way to compute the area is Then the area of . Preceding in a similar fashion for , the area of is . Since , the area of . Thus our desired answer is
Solution 2
Let be the origin. Noticing that the triangle is a 3-4-5 right triangle, we can see that , and . Using the shoelace theorem, the area is . Shoelace theorem:Suppose the polygon has vertices , , ... , , listed in clockwise order. Then the area of is
You can also go counterclockwise order, as long as you find the absolute value of the answer.
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Solution 3 (Easiest, uses only basic geometry too)
Note that has area and is a -- right triangle. Then, by similar triangles, the altitude from to has length and the altitude from to has length , so , meaning that . -Stormersyle
Solution 4
Knowing that has area and is a -- triangle, we can find the area of the smaller triangles , , and and subtract them from to obtain our answer. First off, we know has area since it is a right triangle. To the find the areas of and , we can use Law of Cosines () to find the lengths of and , respectively. Computing gives and . Now, using Heron's Formula, we find and . Adding these and subtracting from , we get -Starsher
Solution 5 (Official MAA)
Triangle is a right triangle with are . Each of and shares an angle with . Because the area of a triangle with sides and included angle is it follows that the areas of and are each where and are the lengths of the sides of adjacent to the shared angle. Thus the sum of the areas of and is Therefore has area .
Solution 6 (Using simple trigonometry)
Let's say that angle APF is 'u'. Then, we know that sin(u) is 20/25 = 4/5. Therefore, the area of APF is 0.5*5*5*sin(u)=10. Now, let's do the same thing with triangle BQC. If we name angle BQC as 'w', we know that sin(w) is 15/25 = 3/5. Therefore, the area of BQC is 0.5*5*5*sin(w)=7.5. Since triangle ERD is a right-angled isosceles triangle, the area of ERD is 12.5. In conclusion, (area of the hexagon) = tri(PRQ)-{tri(APF)+tri(BQC)+tri(ERD)}=150-(10+7.5+12.5)=120
Video Solution #1(Complementary Area Counting?)
https://youtu.be/JQdad7APQG8?t=417
Video Solution
https://www.youtube.com/watch?v=4jOfXNiQ6WM
Video Solution 2
https://youtu.be/TSKcjht8Rfk?t=941
~IceMatrix
Video Solution 3
~Shreyas S
See Also
2019 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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.