Difference between revisions of "1993 AIME Problems/Problem 15"
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− | From the [[Pythagorean Theorem]], <math>AH^2+CH^2=1994^2</math>, and <math>(1995-AH)^2+CH^2=1993^2 | + | From the [[Pythagorean Theorem]], <math>AH^2+CH^2=1994^2</math>, and <math>(1995-AH)^2+CH^2=1993^2</math>. |
− | + | Subtracting those two equations yields <math>AH^2-(1995-AH)^2=3987</math>. | |
− | <math>RS=\left| \frac{AH+CH-AC-CH-BH+BC}{2} \right|=\frac{|AH-BH-1994+1993|}{2}</math>. | + | After simplification, we see that <math>2*1995AH-1995^2=3987</math>, or <math>AH=\frac{1995}{2}+\frac{3987}{2*1995}</math>. |
+ | |||
+ | Note that <math>AH+BH=1995</math>. | ||
+ | |||
+ | Therefore we have that <math>BH=\frac{1995}{2}-\frac{3987}{2*1995}</math>. | ||
+ | |||
+ | Therefore <math>AH-BH=\frac{3987}{1995}</math>. | ||
+ | |||
+ | |||
+ | <math>\textbf{Note: }</math>An easier method is to use both facts together: | ||
+ | |||
+ | First, by the Pythagorean Theorem, we find that <math>1993^2-BH^2=1994^2-AH^2</math>, so <math>AH^2-BH^2=1994^2-1993^2=3987</math>. We know that <math>AH+BH=1995</math>, so by difference of squares and dividing we find that <math>AH-BH=\frac{3987}{1995}=\frac{1329}{665}</math>. ~eevee9406 | ||
+ | |||
+ | |||
+ | Now note that <math>RS=|HR-HS|</math>, <math>RH=\frac{AH+CH-AC}{2}</math>, and <math>HS=\frac{CH+BH-BC}{2}</math>. | ||
+ | |||
+ | Therefore we have <math>RS=\left| \frac{AH+CH-AC-CH-BH+BC}{2} \right|=\frac{|AH-BH-1994+1993|}{2}</math>. | ||
Plugging in <math>AH-BH</math> and simplifying, we have <math>RS=\frac{1992}{1995*2}=\frac{332}{665} \rightarrow 332+665=\boxed{997}</math>. | Plugging in <math>AH-BH</math> and simplifying, we have <math>RS=\frac{1992}{1995*2}=\frac{332}{665} \rightarrow 332+665=\boxed{997}</math>. | ||
+ | ------------------------ | ||
+ | Edit by GameMaster402: | ||
+ | |||
+ | 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> | ||
+ | |||
+ | ~minor edit by [[User: Yiyj1|Yiyj1]] | ||
+ | ------------------------ | ||
+ | Edit by phoenixfire: | ||
+ | |||
+ | It can further be shown for any triangle with sides <math>a=BC, b=CA, c=AB</math> that | ||
+ | <cmath>RS=\dfrac{|b-a|}{2c}|a+b-c|</cmath> | ||
+ | Over here <math>a=1993, b=1994, c=1995</math>, so using the formula gives <cmath>RS = \dfrac{|1994 - 1993|}{2 \cdot 1995}|1993 + 1994 - 1995| = \dfrac{1 \cdot 1992}{2(1995)} = \frac{332}{665}.</cmath> | ||
+ | |||
+ | ~minor edit by [[User: Yiyj1|Yiyj1]] | ||
+ | |||
+ | Note: We can also just right it as <math>RS=\frac{|b-a|(a+b-c)}{2c}</math> since <math>a+b-c \geq 0</math> by the triangle inequality. ~[[User: Yiyj1|Yiyj1]] | ||
== See also == | == See also == |
Latest revision as of 22:14, 30 January 2025
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 .
An easier method is to use both facts together:
First, by the Pythagorean Theorem, we find that , so
. We know that
, so by difference of squares and dividing we find that
. ~eevee9406
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.