1991 AIME Problems/Problem 12
Problem
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 .
Contents
Solution
Solution 1
Let be the center of the rhombus. Via parallel sides and alternate interior angles, we see that the opposite triangles are congruent (, ). Quickly we realize that is also the center of the rectangle.
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 .
Solution 2
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
Solution 3
The triangles are isosceles, and similar (because they have ).
Hence .
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 .
Solution 4
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.
Solution 5
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 .
Solution 6
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,
Solution 7
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
See also
1991 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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