Difference between revisions of "1994 AIME Problems/Problem 10"
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== Solution 3 == | == Solution 3 == | ||
− | Using similar right triangles, we identify that <math>CD = \sqrt{AD \cdot BD}</math>. Let <math>AD</math> be <math>29 \cdot k^2</math>, to avoid too many radicals, getting <math>CD = k \cdot 29^2</math>. Next we know that <math>AC = \sqrt{AB \cdot AD}</math> and that <math>BC = \sqrt{AB \cdot BD}</math>. Applying the logic with the established values of k, we get <math>AC = 29k \cdot \sqrt{29^2 + k^2}</math> and <math>BC = 29^2 \cdot \sqrt{29^2 + k^2}</math>. Next we look to the integer requirement. Since k is both outside and inside square roots, we know it must be an integer to keep all sides as integers. Let <math>y</math> be \sqrt{29^2 + k^2}, thus <math>29^2 = y^2 - k^2</math>, and <math>29^2 = (y + | + | Using similar right triangles, we identify that <math>CD = \sqrt{AD \cdot BD}</math>. Let <math>AD</math> be <math>29 \cdot k^2</math>, to avoid too many radicals, getting <math>CD = k \cdot 29^2</math>. Next we know that <math>AC = \sqrt{AB \cdot AD}</math> and that <math>BC = \sqrt{AB \cdot BD}</math>. Applying the logic with the established values of k, we get <math>AC = 29k \cdot \sqrt{29^2 + k^2}</math> and <math>BC = 29^2 \cdot \sqrt{29^2 + k^2}</math>. Next we look to the integer requirement. Since <math>k</math> is both outside and inside square roots, we know it must be an integer to keep all sides as integers. Let <math>y</math> be <math>\sqrt{29^2 + k^2}</math>, thus <math>29^2 = y^2 - k^2</math>, and <math>29^2 = (y + k)(y - k)</math>. Since <math>29</math> is prime, and <math>k</math> cannot be zero, we find <math>k = 420</math> and <math>y = 421</math> as the smallest integers satisfying this quadratic Diophantine equation. Then, since <math>cos B</math> = <math>\frac{29}{\sqrt{29^2 + k^2}}</math>. Plugging in we get <math>cos B = \frac{29}{421}</math>, thus our answer is <math>\boxed{450}</math>. |
== See also == | == See also == |
Latest revision as of 16:23, 26 August 2018
Problem
In triangle angle
is a right angle and the altitude from
meets
at
The lengths of the sides of
are integers,
and
, where
and
are relatively prime positive integers. Find
Solution 1
Since , we have
. It follows that
and
, so
and
are in the form
and
, respectively, where x is an integer.
By the Pythagorean Theorem, we find that , so
. Letting
, we obtain after dividing through by
,
. As
, the pairs of factors of
are
; clearly
, so
. Then,
.
Thus, , and
.
Solution 2
We will solve for using
, which gives us
. By the Pythagorean Theorem on
, we have
. Trying out factors of
, we can either guess and check or just guess to find that
and
(The other pairs give answers over 999). Adding these, we have
and
, and our answer is
.
Solution 3
Using similar right triangles, we identify that . Let
be
, to avoid too many radicals, getting
. Next we know that
and that
. Applying the logic with the established values of k, we get
and
. Next we look to the integer requirement. Since
is both outside and inside square roots, we know it must be an integer to keep all sides as integers. Let
be
, thus
, and
. Since
is prime, and
cannot be zero, we find
and
as the smallest integers satisfying this quadratic Diophantine equation. Then, since
=
. Plugging in we get
, thus our answer is
.
See also
1994 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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.