2021 AIME II Problems/Problem 2
Equilateral triangle has side length . Point lies on the same side of line as such that . The line through parallel to line intersects sides and at points and , respectively. Point lies on such that is between and , is isosceles, and the ratio of the area of to the area of is . Find .
Solution 1 (Area Formulas for Triangles)
By angle chasing, we conclude that is a triangle, and is a triangle.
Let It follows that and By the side-length ratios in we have and
Let the brackets denote areas. We have and
We set up and solve an equation for Since it is clear that Therefore, we take the positive square root for both sides:
We express the areas of and in terms of in order to solve for
We let Because is isosceles and is equilateral,
Let the height of be and the height of be Then we have that and
Now we can find and in terms of Because we are given that This allows us to use the sin formula for triangle area: the area of is Similarly, because the area of is
Now we can make an equation: To make further calculations easier, we scale everything down by (while keeping the same variable names, so keep that in mind). Thus Because we scaled down everything by the actual value of is
Solution 3 (Pretty Straightforward)
So, If is isosceles, it means that .
In , , Hence (because )
Now, as we know that the ratio of the areas of and is
Substituting the values, we get
Hence, . Solving this, we easily get
We have taken , Hence,
Solution 4 (Similar Triangles)
Since is isosceles, , and since is equilateral, . Thus, , and since these triangles share an altitude, they must have the same area.
Drop perpendiculars from and to line ; call the meeting points and , respectively. is clearly congruent to both and , and thus each of these new triangles has the same area as . But we can "slide" over to make it adjacent to , thus creating an equilateral triangle whose area has a ratio of when compared to (based on our conclusion from the first paragraph). Since these triangles are both equilateral, they are similar, and since the area ratio reduces to , the ratio of their sides must be . So, because and represent sides of these triangles, and they add to , must equal two-fifths of , or .
Video Solution by Interstigation (Similar Triangles)
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