Difference between revisions of "2021 AIME I Problems/Problem 9"
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Revision as of 15:10, 12 March 2021
Problem
Let be an isosceles trapezoid with and Suppose that the distances from to the lines and are and respectively. Let be the area of Find
Solution 1
Construct your isosceles trapezoid. Let, for simplicity, , , and . Extend the sides and mark the intersection as . Following what the question states, drop a perpendicular from to labeling the foot as . Drop another perpendicular from to , calling the foot . Lastly, drop a perpendicular from to , labeling it . In addition, drop a perpendicular from to calling its foot .
--DIAGRAM COMING SOON--
Start out by constructing a triangle congruent to with its side of length on line . This works because all isosceles triangles are cyclic and as a result, .
Notice that by AA similarity. We are given that and by symmetry we can deduce that . As a result, . This gives us that .
The question asks us along the lines of finding the area, , of the trapezoid . We look at the area of and notice that it can be represented as . Substituting , we solve for , getting .
Now let us focus on isosceles triangle , where . Since, is an altitude from to of an isosceles triangle, must be equal to . Since and , we can solve to get that and .
We must then set up equations using the Pythagorean Theorem, writing everything in terms of , , and . Looking at right triangle we get Looking at right triangle we get Now rearranging and solving, we get two equation Those are convenient equations as which gives us After some "smart" calculation, we get that .
Notice that the question asks for , and by applying the trapezoid area formula. Fortunately, this is just , and plugging in the value of , we get that .
~Math_Genius_164
Solution 2(LOC and Trig)
Call AD and BC . Draw diagonal AC and call the foot of the perpendicular from B to AC . Call the foot of the perpendicular from A to line BC F, and call the foot of the perpindicular from A to DC H. Triangles CBG and CAF are similar, and we get that = Therefore, . It then follows that triangles ABF and ADH are similar. Using similar triangles, we can then find that . Using the Law of Cosine on ABC, We can find that the cosine of angle ABC is . Since angles ABF and ADH are equivalent and supplementary to angle ABC, we know that the cosine of angle ADH is 1/3. It then follows that . Then it can be found that the area is . Multiplying this by , the answer is . -happykeeper
See also
2021 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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