Difference between revisions of "1994 AIME Problems/Problem 2"
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:<math>AB^2 - 16 AB - 240 = 0</math> | :<math>AB^2 - 16 AB - 240 = 0</math> | ||
− | The [[quadratic formula]] shows that the answer is <math>\frac{16 \pm \sqrt{16^2 + 4 \cdot 240}}{2} = 8 \pm \sqrt{304}</math>. Discard the negative root, so our answer is <math>8 + 304 = 312</math>. | + | The [[quadratic formula]] shows that the answer is <math>\frac{16 \pm \sqrt{16^2 + 4 \cdot 240}}{2} = 8 \pm \sqrt{304}</math>. Discard the negative root, so our answer is <math>8 + 304 = \boxed{312}</math>. |
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/nPVDavMoG9M?t=32 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
== See also == | == See also == |
Latest revision as of 03:53, 23 January 2023
Problem
A circle with diameter of length 10 is internally tangent at
to a circle of radius 20. Square
is constructed with
and
on the larger circle,
tangent at
to the smaller circle, and the smaller circle outside
. The length of
can be written in the form
, where
and
are integers. Find
.
Note: The diagram was not given during the actual contest.
Solution
Call the center of the larger circle . Extend the diameter
to the other side of the square (at point
), and draw
. We now have a right triangle, with hypotenuse of length
. Since
, we know that
. The other leg,
, is just
.
Apply the Pythagorean Theorem:
The quadratic formula shows that the answer is . Discard the negative root, so our answer is
.
Video Solution by OmegaLearn
https://youtu.be/nPVDavMoG9M?t=32
~ pi_is_3.14
See also
1994 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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.