1991 AIME Problems/Problem 12
Rhombus is inscribed in rectangle so that vertices , , , and are interior points on sides , , , and , respectively. It is given that , , , and . Let , in lowest terms, denote the perimeter of . Find .
By the Pythagorean Theorem, we can solve for a side of the rhombus; . Since the diagonals of a rhombus are perpendicular bisectors, we have that . Also, , so quadrilateral is cyclic. By Ptolemy's Theorem, .
By similar logic, we have is a cyclic quadrilateral. Let , . The Pythagorean Theorem gives us . Ptolemy’s Theorem gives us . Since the diagonals of a rectangle are equal, , and . Solving for , we get . Substituting into ,
We reject because then everything degenerates into squares, but the condition that gives us a contradiction. Thus , and backwards solving gives . The perimeter of is , and .
From above, we have and . Returning to note that Hence, by similarity. From here, it's clear that Similarly, Therefore, the perimeter of rectangle is
The length of could be found easily from the area of :
From the right triangle we have . We could have also defined a similar formula: , and then we found , the segment is tangent to the circles with diameters .
The perimeter is .
For convenience, let . Since the opposite triangles are congruent we have that , and therefore . Let , then we have , or . Expanding with the formula , and since we have , we can solve for . The rest then follows similarily from above.
We can just find coordinates of the points. After drawing a picture, we can see 4 congruent right triangles with sides of , namely triangles and .
Let the points of triangle be . Let point be on , such that and . Triangle can be split into two similar 3-4-5 right triangles, and . By the Pythagorean Theorem, point is away from point . Repeating the process, if we break down triangle into two more similar triangles, we find that point is at .
By reflecting point over point , we get point . By reflecting point over point , we get point . Thus, the perimeter is equal to , making the final answer .
We can just use areas. Let and . . Also, we can add up the areas of all 8 right triangles and let that equal the total area of the rectangle, . This gives . Solving this system of equation gives , , from which it is straightforward to find the answer, . Thus,
We will bash with trigonometry.
Firstly, by Pythagoras Theorem, . We observe that . Thus, if we drop an altitude from to to point , it will have length . In particular, since we form a 7-24-25 triangle.
Now, . Thus, since , we get that . Now, by the Pythagorean Theorem, .
Using the same idea, . Thus, since .
Now, we can finish. We know . We also know . Thus, our perimeter is
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