Difference between revisions of "2004 AMC 10B Problems/Problem 22"
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<cmath>OI^2= \frac{65}{4}</cmath> | <cmath>OI^2= \frac{65}{4}</cmath> | ||
So, <math>OI=\frac{\sqrt{65}}{2}\implies \boxed{D}</math> | So, <math>OI=\frac{\sqrt{65}}{2}\implies \boxed{D}</math> | ||
+ | |||
+ | ==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 <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> | ||
== See also == | == See also == |
Revision as of 13:01, 2 August 2021
Contents
[hide]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?
Solution 1
This is a right triangle. Pick a coordinate system so that the right angle is at
and the other two vertices are at
and
.
As this is a right triangle, the center of the circumcircle is in the middle of the hypotenuse, at .
The radius of the inscribed circle can be computed using the well-known identity
, where
is the area of the triangle and
its perimeter. In our case,
and
. Thus,
. As the inscribed circle touches both legs, its center must be at
.
The distance of these two points is then .
Solution 2
We directly apply Euler’s Theorem, which states that if the circumcenter is and the incenter
, and the inradius is
and the circumradius is
, then
We can see that this is a right triangle, and hence has area . We then find the inradius with the formula
, where
denotes semiperimeter. We easily see that
, so
.
We now find the circumradius with the formula . Solving for
gives
.
Substituting all of this back into our formula gives:
So,
Solution 3
Construct
such that
,
, and
. Since this is a pythagorean triple,
. By a property of circumcircles and right triangles, the circumcenter,
, lies on the midpoint of
, so
. Turning to the incircle, we find that the inradius is
, using the formula
, where
is the area of the triangle,
is the inradius, and
is the semiperimeter. We then denote the incenter
, along with the points of tangency
,
, and
. Because
by a property of tangency,
, and so
is a square. Then, since
,
. As
,
, and because
by HL,
. Therefore,
. Because
, pythagorean theorem gives
See also
2004 AMC 10B (Problems • Answer Key • Resources) | ||
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.