Difference between revisions of "1987 AIME Problems/Problem 15"
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== Messy Trig Solution == | == Messy Trig Solution == | ||
− | Let <math>\theta</math> be the smaller angle in the triangle. Then the sum of shorter and longer leg is <math>\sqrt{441}(2+\tan{\theta}+\cot{\theta})</math>. We observe that the short leg has length <math>\sqrt{441}(1+\tan{\theta}) = \sqrt{440}(\sec{\theta}+\sin{\theta})</math>. Grouping and squaring, we get | + | Let <math>\theta</math> be the smaller angle in the triangle. Then the sum of shorter and longer leg is <math>\sqrt{441}(2+\tan{\theta}+\cot{\theta})</math>. We observe that the short leg has length <math>\sqrt{441}(1+\tan{\theta}) = \sqrt{440}(\sec{\theta}+\sin{\theta})</math>. Grouping and squaring, we get <math>\sqrt{\frac{440}{441}} = \frac{\sin{\theta}+\cos{\theta}}{1+\sin{\theta}\cos{\theta}}</math>. Squaring and using the double angle identity for sine, we get, <math>110(\sin{2\theta})^2 + \sin{2*\theta} - 1 = 0</math>. Solving, we get <math>\sin{2*\theta} = \frac{1}{10}</math>. Now to find <math>\tan{theta}</math>, we find <math>\cos{2*\theta}</math> using the Pythagorean |
+ | Identity, and then use the tangent double angle identity. Thus, <math>\tan{\theta} = 10-3\sqrt{11}</math>. Substituting into the original sum, | ||
+ | we get <math>\boxed{462}</math>. | ||
== See also == | == See also == |
Revision as of 18:33, 20 January 2020
Problem
Squares and
are inscribed in right triangle
, as shown in the figures below. Find
if area
and area
.
Solution
Because all the triangles in the figure are similar to triangle , it's a good idea to use area ratios. In the diagram above,
Hence,
and
. Additionally, the area of triangle
is equal to both
and
Setting the equations equal and solving for ,
. Therefore,
. However,
is equal to the area of triangle
! This means that the ratio between the areas
and
is
, and the ratio between the sides is
. As a result,
. We now need
to find the value of
, because
.
Let denote the height to the hypotenuse of triangle
. Notice that
. (The height of
decreased by the corresponding height of
) Thus,
. Because
,
.
Easy Trig Solution
Let . Now using the 1st square,
and
. Using the second square,
. We have
, or
Rearranging and letting
gives us
We take the positive root, so
, which means
.
Messy Trig Solution
Let be the smaller angle in the triangle. Then the sum of shorter and longer leg is
. We observe that the short leg has length
. Grouping and squaring, we get
. Squaring and using the double angle identity for sine, we get,
. Solving, we get
. Now to find
, we find
using the Pythagorean
Identity, and then use the tangent double angle identity. Thus,
. Substituting into the original sum,
we get
.
See also
1987 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.