Difference between revisions of "1993 AIME Problems/Problem 15"
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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]] |
------------------------ | ------------------------ | ||
Edit by phoenixfire: | Edit by phoenixfire: | ||
| Line 54: | Line 54: | ||
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> | 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 Yiyj1 | + | ~minor edit by [[Yiyj1]] |
| − | Note: We can also just right it as <math>RS=\frac{|b-a|(a+b-c)}{2c}</math> since <math>a+b-c \leq 0</math> by the triangle inequality. ~Yiyj1 | + | Note: We can also just right it as <math>RS=\frac{|b-a|(a+b-c)}{2c}</math> since <math>a+b-c \leq 0</math> by the triangle inequality. ~[[Yiyj1]] |
== See also == | == See also == | ||
Revision as of 18:37, 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
![[asy] unitsize(48); pair A,B,C,H; A=(8,0); B=origin; C=(3,4); H=(3,0); draw(A--B--C--cycle); draw(C--H); label("$A$",A,SE); label("$B$",B,SW); label("$C$",C,N); label("$H$",H,NE); draw(circle((2,1),1)); pair [] x=intersectionpoints(C--H,circle((2,1),1)); dot(x[0]); label("$S$",x[0],SW); draw(circle((4.29843788128,1.29843788128),1.29843788128)); pair [] y=intersectionpoints(C--H,circle((4.29843788128,1.29843788128),1.29843788128)); dot(y[0]); label("$R$",y[0],NE); label("$1993$",(1.5,2),NW); label("$1994$",(5.5,2),NE); label("$1995$",(4,0),S); [/asy]](http://latex.artofproblemsolving.com/b/8/b/b8b732c641481c3b1635527f52f3125316ac3f3d.png)
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
![\[RS=\dfrac{|b-a|}{2c}|a+b-c|\]](http://latex.artofproblemsolving.com/b/9/b/b9b574252807c16d09785459431180ff371816fc.png)
Over here 
, so using the formula gives ![\[RS = \dfrac{|1994 - 1993|}{2 \cdot 1995}|1993 + 1994 - 1995| = \dfrac{1 \cdot 1992}{2(1995)} = \frac{332}{665}.\]](http://latex.artofproblemsolving.com/1/a/2/1a2e6264ee2aa1a4702409db23e487f928eb97c0.png)
~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. 
