Difference between revisions of "1993 AIME Problems/Problem 15"
m |
m (→Solution) |
||
| (12 intermediate revisions by 6 users not shown) | |||
| Line 3: | Line 3: | ||
== Solution == | == 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> | ||
| + | |||
| + | 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>. | ||
| + | |||
| + | 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>. | ||
| + | |||
| + | |||
| + | |||
| + | 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>. | ||
| + | ------------------------ | ||
| + | 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 == | ||
{{AIME box|year=1993|num-b=14|after=Last question}} | {{AIME box|year=1993|num-b=14|after=Last question}} | ||
| + | [[Category:Intermediate Geometry Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 20:39, 21 September 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. 
