Difference between revisions of "1993 AIME Problems/Problem 15"
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It can be shown that in any triangle with side lengths <math>n-1, n, n+1</math>, if you draw an altitude from the vertex to the side of <math>n+1</math>, and draw the incircles of the two right triangles, the distance between the two tangency points is simply <math>\frac{n-2}{2n+2)}=\frac{n-2}{2(n+1)}</math>. | It can be shown that in any triangle with side lengths <math>n-1, n, n+1</math>, if you draw an altitude from the vertex to the side of <math>n+1</math>, and draw the incircles of the two right triangles, the distance between the two tangency points is simply <math>\frac{n-2}{2n+2)}=\frac{n-2}{2(n+1)}</math>. | ||
− | Plugging in <math>n=1994</math> yields that the answer is <math>\frac{1992}{2(1995}</math>, which simplifies to <math>\frac{332}{665}</math> | + | Plugging in <math>n=1994</math> yields that the answer is <math>\frac{1992}{2(1995)}</math>, which simplifies to <math>\frac{332}{665}</math> |
~minor edit by Yiyj1 | ~minor edit by Yiyj1 |
Revision as of 18:36, 30 August 2023
Problem
Let be an altitude of
. Let
and
be the points where the circles inscribed in the triangles
and
are tangent to
. If
,
, and
, then
can be expressed as
, where
and
are relatively prime integers. Find
.
Solution
From the Pythagorean Theorem, , and
.
Subtracting those two equations yields .
After simplification, we see that , or
.
Note that .
Therefore we have that .
Therefore .
Now note that ,
, and
.
Therefore we have .
Plugging in and simplifying, we have
.
Edit by GameMaster402:
It can be shown that in any triangle with side lengths , if you draw an altitude from the vertex to the side of
, and draw the incircles of the two right triangles, the distance between the two tangency points is simply
.
Plugging in yields that the answer is
, which simplifies to
~minor edit by Yiyj1
Edit by phoenixfire:
It can further be shown for any triangle with sides that
Over here
, so using the formula gives
~minor edit by Yiyj1
Note: We can also just right it as since
by the triangle inequality. ~Yiyj1
See also
1993 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 |
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