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Difference between revisions of "2004 Indonesia MO Problems/Problem 7"

(Solution to Problem 7)
 
(→‎See Also)
 
(One intermediate revision by one other user not shown)
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\end{align*}</cmath>
 
\end{align*}</cmath>
 
Thus, <math>x = \frac{a+b-c}{2},</math> so the diameter of the incircle is <math>a+b-c</math>.
 
Thus, <math>x = \frac{a+b-c}{2},</math> so the diameter of the incircle is <math>a+b-c</math>.
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==Solution 2==
 +
 +
The inradius of any triangle has <math>rs = A</math>, where <math>r</math> is the radius of the inscribed circle, <math>s</math> is the semi-perimeter of the triangle, and <math>A</math> is the area of the triangle. Thus,
 +
<cmath>\begin{align*}
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r = \frac{A}{s} = &\frac{ab/2}{(a+b+c)/2} = \frac{ab}{a+b+c} \\
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d = 2r &= \frac{2ab}{a+b+c}
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\end{align*}</cmath>
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 +
If we multiply both sides by the "conjugate" of <math>(a+b+c)</math>, we get
 +
<cmath>\begin{align*}
 +
d &= \frac{(2ab)(a+b-c)}{(a+b+c)(a+b-c)} \\
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d &= \frac{(2ab)(a+b-c)}{(a^2+2ab+b^2-c^2)} \\
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d &= \frac{(2ab)(a+b-c)}{a^2+2ab+b^2-(a^2+b^2)} \\
 +
d &= \frac{(2ab)(a+b-c)}{2ab} \\
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d &= a+b-c
 +
\end{align*}</cmath>
  
 
==See Also==
 
==See Also==
 
{{Indonesia MO box|year=2004|num-b=6|num-a=8}}
 
{{Indonesia MO box|year=2004|num-b=6|num-a=8}}
  
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[[Category:Intermediate Geometry Problems]]
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[[Category:Introductory Geometry Problems]]

Latest revision as of 14:37, 13 September 2024

Problem

Prove that in a triangle $ABC$ with $C$ as the right angle, where $a$ denote the side in front of angle $A$, $b$ denote the side in front of angle $B$, $c$ denote the side in front of angle $C$, the diameter of the incircle of $ABC$ equals to $a + b - c$.

Solution

[asy] draw((0,0)--(30,0)--(0,40)--cycle); draw(circle((10,10),10)); draw((0,10)--(10,10)--(10,0)); draw((10,10)--(18,16)); dot((0,0)); label("C",(0,0),SW); dot((30,0)); label("B",(30,0),SE); dot((0,40)); label("A",(0,40),NW); dot((0,10)); label("X",(0,10),W); dot((10,0)); label("Y",(10,0),S); dot((18,16)); label("Z",(18,16),NE); dot((10,10)); label("O",(10,10),N); draw((0,8)--(2,8)--(2,10)); draw((0,2)--(2,2)--(2,0)); draw((8,0)--(8,2)--(10,2)); [/asy] Let $O$ be the center of the circle, $X$ be the intersection of the incircle and $AC$, $Y$ be the intersection of the incircle and $BC$, and $Z$ be the intersection of the incircle and $AB$.


Note that $X$ and $Y$ are tangent points of the circle, so $OX \perp AC$ and $OY \perp BC.$ Since $XC \perp YC$, we know that $XOYC$ is a square, so $XO = OY = YC = XC.$


Let $x = XC = YC$, $y = AX$, and $z = BY.$ Since $X,Y,Z$ are tangent points to the incircle, we know that $y = AZ$ and $z = BZ.$ Thus, \begin{align*} x + z &= a \\ x + y &= b \\ y + z &= c \end{align*} Adding the three equations yields \begin{align*} 2x+2y+2z &= a+b+c \\ x+y+z &= \frac{a+b+c}{2} \end{align*} Thus, $x = \frac{a+b-c}{2},$ so the diameter of the incircle is $a+b-c$.

Solution 2

The inradius of any triangle has $rs = A$, where $r$ is the radius of the inscribed circle, $s$ is the semi-perimeter of the triangle, and $A$ is the area of the triangle. Thus, \begin{align*} r = \frac{A}{s} = &\frac{ab/2}{(a+b+c)/2} = \frac{ab}{a+b+c} \\ d = 2r &= \frac{2ab}{a+b+c} \end{align*}

If we multiply both sides by the "conjugate" of $(a+b+c)$, we get \begin{align*} d &= \frac{(2ab)(a+b-c)}{(a+b+c)(a+b-c)} \\ d &= \frac{(2ab)(a+b-c)}{(a^2+2ab+b^2-c^2)} \\ d &= \frac{(2ab)(a+b-c)}{a^2+2ab+b^2-(a^2+b^2)} \\ d &= \frac{(2ab)(a+b-c)}{2ab} \\ d &= a+b-c \end{align*}

See Also

2004 Indonesia MO (Problems)
Preceded by
Problem 6 		1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 Followed by
Problem 8
All Indonesia MO Problems and Solutions
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