2019 AIME II Problems/Problem 1
Contents
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. (We know that is a , since is an .) 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
Solution 5
Because and , quadrilateral is cyclic. So, Ptolemy's theorem tells us that
From here, there are many ways to finish which have been listed above. If we let , then
Using Heron's formula on , we see that
Thus, our answer is . ~a.y.711
Solution 6
Let . Now consider , and if we find the coordinates of , by symmetry about , we can find the coordinates of D.
So let . So the following equations hold:
.
.
Solving by squaring both equations and then subtracting one from the other to eliminate , we get because is in the second quadrant.
Now by symmetry, .
So now you can proceed by finding the intersection and then calculating the area directly. We get .
~hastapasta
See Also
2019 AIME II (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
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