Difference between revisions of "2004 AMC 10B Problems/Problem 22"

 
Line 80: Line 80:
 
</asy>
 
</asy>
 
Construct <math>\triangle{ABC}</math> such that <math>AB=5</math>, <math>AC=12</math>, and <math>BC=13</math>. Since this is a pythagorean triple, <math>\angle{A}=90</math>. By a property of circumcircles and right triangles, the circumcenter, <math>G</math>, lies on the midpoint of <math>\overline{BC}</math>, so <math>BG=\frac{13}{2}</math>. Turning to the incircle, we find that the inradius is <math>2</math>, using the formula <math>A=rs</math>, where <math>A</math> is the area of the triangle, <math>r</math> is the inradius, and <math>s</math> is the semiperimeter. We then denote the incenter <math>I</math>, along with the points of tangency <math>D</math>, <math>E</math>, and <math>F</math>. Because <math>\angle{IDA}=\angle{IEA}=90</math> by a property of tangency, <math>\angle{EID}=90</math>, and so <math>IDAE</math> is a square. Then, since <math>IE=2</math>, <math>AD=2</math>. As <math>AB=5</math>, <math>BD=3</math>, and because <math>\triangle{BID}\cong\triangle{BIF}</math> by HL, <math>BD=BF=3</math>. Therefore, <math>FG=\frac{7}{2}</math>. Because <math>IF=2</math>, pythagorean theorem gives <math>IG=\boxed{\frac{\sqrt{65}}{2}}</math>
 
Construct <math>\triangle{ABC}</math> such that <math>AB=5</math>, <math>AC=12</math>, and <math>BC=13</math>. Since this is a pythagorean triple, <math>\angle{A}=90</math>. By a property of circumcircles and right triangles, the circumcenter, <math>G</math>, lies on the midpoint of <math>\overline{BC}</math>, so <math>BG=\frac{13}{2}</math>. Turning to the incircle, we find that the inradius is <math>2</math>, using the formula <math>A=rs</math>, where <math>A</math> is the area of the triangle, <math>r</math> is the inradius, and <math>s</math> is the semiperimeter. We then denote the incenter <math>I</math>, along with the points of tangency <math>D</math>, <math>E</math>, and <math>F</math>. Because <math>\angle{IDA}=\angle{IEA}=90</math> by a property of tangency, <math>\angle{EID}=90</math>, and so <math>IDAE</math> is a square. Then, since <math>IE=2</math>, <math>AD=2</math>. As <math>AB=5</math>, <math>BD=3</math>, and because <math>\triangle{BID}\cong\triangle{BIF}</math> by HL, <math>BD=BF=3</math>. Therefore, <math>FG=\frac{7}{2}</math>. Because <math>IF=2</math>, pythagorean theorem gives <math>IG=\boxed{\frac{\sqrt{65}}{2}}</math>
 +
 +
==Solution 4==
 +
A triangle with sides <math>5,12, </math> and <math>13</math> must be right. Let the right angle be at <math>A</math> so that <math>AB=5,AC=12, </math> and <math>BC=13</math>. The circumcenter <math>O</math> must be the midpoint of <math>BC</math>, so that <math>BO=CO=\frac{13}{2}.</math> Let <math>I</math> be the incenter and <math>D</math> be the point where <math>BC</math> is tangent to the incircle. Since <math>ID \perp DO, \triangle IDO</math> is a right triangle. Therefore, to find <math>IO</math>, it suffices to find <math>ID</math> and <math>DO</math>. The area of triangle <math>ABC</math> is equal to the semiperimeter times the inradius, <math>r</math>. This allows us to set up an equation involving <math>r</math>: <cmath>\frac{5 \cdot 12}{2}= r \cdot \frac{5+12+13}{2}.</cmath> Solving, we get <math>r=2</math>. Now it only remains to find <math>DO</math>. To start, note that <math>BD=s-b</math>, where <math>s</math> is the semiperimeter and <math>b</math> is the length of <math>AC</math>. Simplifing, we get <math>BD=3<\frac{13}{2}</math>, so <math>D</math> is on the same side of <math>O</math> as <math>B</math>, and <math>DO=BO-BD=\frac{13}{2}-3=\frac{7}{2}.</math> Therefore, <math>IO=\sqrt{(\frac{7}{2})^2+2^2}=\sqrt{\frac{65}{4}}=\boxed{(D)\frac{\sqrt{65}}{2}}.</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 22:48, 25 October 2021

Problem

A triangle with sides of 5, 12, and 13 has both an inscribed and a circumscribed circle. What is the distance between the centers of those circles?

$\mathrm{(A) \ } \frac{3\sqrt{5}}{2} \qquad \mathrm{(B) \ } \frac{7}{2} \qquad \mathrm{(C) \ } \sqrt{15} \qquad \mathrm{(D) \ } \frac{\sqrt{65}}{2} \qquad \mathrm{(E) \ } \frac{9}{2}$

Solution 1

[asy] import geometry;  unitsize(0.6 cm);  pair A, B, C, D, E, F, I, O;  A = (5^2/13,5*12/13); B = (0,0); C = (13,0); I = incenter(A,B,C); D = (I + reflect(B,C)*(I))/2; E = (I + reflect(C,A)*(I))/2; F = (I + reflect(A,B)*(I))/2; O = (B + C)/2;  draw(A--B--C--cycle); draw(incircle(A,B,C)); draw(I--D); draw(I--E); draw(I--F); draw(I--O);  label("$A$", A, N); label("$B$", B, SW); label("$C$", C, SE); dot("$D$", D, S); dot("$E$", E, NE); dot("$F$", F, NW); dot("$I$", I, N); dot("$O$", O, S); [/asy] This is a right triangle. Pick a coordinate system so that the right angle is at $(0,0)$ and the other two vertices are at $(12,0)$ and $(0,5)$.

As this is a right triangle, the center of the circumcircle is in the middle of the hypotenuse, at $(6,2.5)$.

The radius $r$ of the inscribed circle can be computed using the well-known identity $\frac{rP}2=S$, where $S$ is the area of the triangle and $P$ its perimeter. In our case, $S=\frac{5\cdot 12}{2}=30$ and $P=5+12+13=30$. Thus, $r=2$. As the inscribed circle touches both legs, its center must be at $(r,r)=(2,2)$.

The distance of these two points is then $\sqrt{ (6-2)^2 + (2.5-2)^2 } = \sqrt{16.25} = \sqrt{\frac{65}4} = \boxed{\frac{\sqrt{65}}2}$.

Solution 2

We directly apply Euler’s Theorem, which states that if the circumcenter is $O$ and the incenter $I$, and the inradius is $r$ and the circumradius is $R$, then \[OI^2=R(R-2r)\]

We can see that this is a right triangle, and hence has area $30$. We then find the inradius with the formula $A=rs$, where $s$ denotes semiperimeter. We easily see that $s=15$, so $r=2$.

We now find the circumradius with the formula $A=\frac{abc}{4R}$. Solving for $R$ gives $R=\frac{13}{2}$.

Substituting all of this back into our formula gives: \[OI^2= \frac{65}{4}\] So, $OI=\frac{\sqrt{65}}{2}\implies \boxed{D}$

Solution 3

[asy] size(15cm); draw((0,0)--(0,5), linewidth(2));  draw((0,0)--(12,0), linewidth(2));  draw((12,0)--(0,5), linewidth(2));  draw((2,0)--(2,2), linewidth(2));  draw((2,2)--(2.770565628817799,3.8455976546592505), linewidth(2));  draw((2,2)--(6.023716614191289,2.4901180774202962), linewidth(2));  draw((2,2)--(0,5), linewidth(2));  draw((2,2)--(12,0), linewidth(2));  draw((0,2)--(2,2), linewidth(2));  label("$A$", (0.14164244785738467,0.25966489738837517), NE);  label("$B$", (0.14164244785738467,5.311734560831129), NE);  label("$C$", (12.120449134133493,0.3232129434694161), NE);  label("$D$", (0.14164244785738467,2.324976395022205), NE);  label("$E$", (2.111631876369636,0.3232129434694161), NE);  label("$F$", (2.9059824523826405,4.167869731372392), NE);  label("$G$", (6.146932802515699,2.801586740630012), NE);  label("$I$", (2.1, 2.1), NE); [/asy] Construct $\triangle{ABC}$ such that $AB=5$, $AC=12$, and $BC=13$. Since this is a pythagorean triple, $\angle{A}=90$. By a property of circumcircles and right triangles, the circumcenter, $G$, lies on the midpoint of $\overline{BC}$, so $BG=\frac{13}{2}$. Turning to the incircle, we find that the inradius is $2$, using the formula $A=rs$, where $A$ is the area of the triangle, $r$ is the inradius, and $s$ is the semiperimeter. We then denote the incenter $I$, along with the points of tangency $D$, $E$, and $F$. Because $\angle{IDA}=\angle{IEA}=90$ by a property of tangency, $\angle{EID}=90$, and so $IDAE$ is a square. Then, since $IE=2$, $AD=2$. As $AB=5$, $BD=3$, and because $\triangle{BID}\cong\triangle{BIF}$ by HL, $BD=BF=3$. Therefore, $FG=\frac{7}{2}$. Because $IF=2$, pythagorean theorem gives $IG=\boxed{\frac{\sqrt{65}}{2}}$

Solution 4

A triangle with sides $5,12,$ and $13$ must be right. Let the right angle be at $A$ so that $AB=5,AC=12,$ and $BC=13$. The circumcenter $O$ must be the midpoint of $BC$, so that $BO=CO=\frac{13}{2}.$ Let $I$ be the incenter and $D$ be the point where $BC$ is tangent to the incircle. Since $ID \perp DO, \triangle IDO$ is a right triangle. Therefore, to find $IO$, it suffices to find $ID$ and $DO$. The area of triangle $ABC$ is equal to the semiperimeter times the inradius, $r$. This allows us to set up an equation involving $r$: \[\frac{5 \cdot 12}{2}= r \cdot \frac{5+12+13}{2}.\] Solving, we get $r=2$. Now it only remains to find $DO$. To start, note that $BD=s-b$, where $s$ is the semiperimeter and $b$ is the length of $AC$. Simplifing, we get $BD=3<\frac{13}{2}$, so $D$ is on the same side of $O$ as $B$, and $DO=BO-BD=\frac{13}{2}-3=\frac{7}{2}.$ Therefore, $IO=\sqrt{(\frac{7}{2})^2+2^2}=\sqrt{\frac{65}{4}}=\boxed{(D)\frac{\sqrt{65}}{2}}.$

See also

2004 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS