Difference between revisions of "2019 AIME II Problems/Problem 1"
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Let <math>a = \angle{CAB}</math>. By Law of Cosines, | Let <math>a = \angle{CAB}</math>. By Law of Cosines, | ||
<cmath>\cos a = \frac{17^2+9^2-10^2}{2*9*17} = \frac{15}{17}</cmath> | <cmath>\cos a = \frac{17^2+9^2-10^2}{2*9*17} = \frac{15}{17}</cmath> | ||
− | <cmath>\sin a = \sqrt{1-\cos a} = \frac{8}{17}</cmath> | + | <cmath>\sin a = \sqrt{1-\cos^2 a} = \frac{8}{17}</cmath> |
<cmath>\tan a = \frac{8}{15}</cmath> | <cmath>\tan a = \frac{8}{15}</cmath> | ||
<cmath>A = \frac{1}{2}* 9*\frac{9}{2}\tan a = \frac{54}{5}</cmath> | <cmath>A = \frac{1}{2}* 9*\frac{9}{2}\tan a = \frac{54}{5}</cmath> |
Revision as of 18:18, 11 March 2020
Problem
Two different points, and , lie on the same side of line so that and are congruent with , , and . The intersection of these two triangular regions has area , where and are relatively prime positive integers. Find .
Solution
- Diagram by Brendanb4321
Extend to form a right triangle with legs and such that is the hypotenuse and connect the points so
that you have a rectangle. The base of the rectangle will be . Now, let be the intersection of and . This means that and are with ratio . Set up a proportion, knowing that the two heights add up to 8. We will let be the height from to , and be the height of .
This means that the area is . This gets us
-Solution by the Math Wizard, Number Magician of the Second Order, Head of the Council of the Geometers
Solution 2
Using the diagram in Solution 1, let be the intersection of and . We can see that angle is in both and . Since and are congruent by AAS, we can then state and . It follows that and . We can now state that the area of is the area of the area of . Using Heron's formula, we compute the area of . Using the Law of Cosines on angle , we obtain
(For convenience, we're not going to simplify.)
Applying the Law of Cosines on yields This means . Next, apply Heron's formula to get the area of , which equals after simplifying. Subtracting the area of from the area of yields the area of , which is , giving us our answer, which is -Solution by flobszemathguy
Solution 3 (Very quick)
- Diagram by Brendanb4321 extended by Duoquinquagintillion
Begin with the first step of solution 1, seeing is the hypotenuse of a triangle and calling the intersection of and point . Next, notice is the hypotenuse of an triangle. Drop an altitude from with length , so the other leg of the new triangle formed has length . Notice we have formed similar triangles, and we can solve for .
So has area And - Solution by Duoquinquagintillion
Solution 4
Let . By Law of Cosines, And
- by Mathdummy
See Also
2019 AIME II (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
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.