Difference between revisions of "2004 AMC 10B Problems/Problem 22"
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<math> \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} </math> | <math> \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} </math> | ||
− | ==Solution== | + | ==Solution 1== |
− | This is | + | <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 <math>(0,0)</math> and the other two vertices are at <math>(12,0)</math> and <math>(0,5)</math>. | ||
As this is a right triangle, the center of the circumcircle is in the middle of the hypotenuse, at <math>(6,2.5)</math>. | As this is a right triangle, the center of the circumcircle is in the middle of the hypotenuse, at <math>(6,2.5)</math>. | ||
− | The radius <math>r</math> of the inscribed circle can be computed using the well-known identity <math>\frac{rP}2=S</math>, where <math>S</math> is the area of the triangle and <math>P</math> its perimeter. In our case, <math>S=5\cdot 12 | + | The radius <math>r</math> of the inscribed circle can be computed using the well-known identity <math>\frac{rP}2=S</math>, where <math>S</math> is the area of the triangle and <math>P</math> its perimeter. In our case, <math>S=\frac{5\cdot 12}{2}=30</math> and <math>P=5+12+13=30</math>. Thus, <math>r=2</math>. As the inscribed circle touches both legs, its center must be at <math>(r,r)=(2,2)</math>. |
The distance of these two points is then <math>\sqrt{ (6-2)^2 + (2.5-2)^2 } = \sqrt{16.25} = \sqrt{\frac{65}4} = \boxed{\frac{\sqrt{65}}2}</math>. | The distance of these two points is then <math>\sqrt{ (6-2)^2 + (2.5-2)^2 } = \sqrt{16.25} = \sqrt{\frac{65}4} = \boxed{\frac{\sqrt{65}}2}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | We directly apply Euler’s Theorem, which states that if the circumcenter is <math>O</math> and the incenter <math>I</math>, and the inradius is <math>r</math> and the circumradius is <math>R</math>, then | ||
+ | <cmath>OI^2=R(R-2r)</cmath> | ||
+ | |||
+ | We can see that this is a right triangle, and hence has area <math>30</math>. We then find the inradius with the formula <math>A=rs</math>, where <math>s</math> denotes semiperimeter. We easily see that <math>s=15</math>, so <math>r=2</math>. | ||
+ | |||
+ | We now find the circumradius with the formula <math>A=\frac{abc}{4R}</math>. Solving for <math>R</math> gives <math>R=\frac{13}{2}</math>. | ||
+ | |||
+ | Substituting all of this back into our formula gives: | ||
+ | <cmath>OI^2= \frac{65}{4}</cmath> | ||
+ | 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> | ||
+ | |||
+ | ==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 == |
Revision as of 23:48, 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?
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
Solution 4
A triangle with sides and must be right. Let the right angle be at so that and . The circumcenter must be the midpoint of , so that Let be the incenter and be the point where is tangent to the incircle. Since is a right triangle. Therefore, to find , it suffices to find and . The area of triangle is equal to the semiperimeter times the inradius, . This allows us to set up an equation involving : Solving, we get . Now it only remains to find . To start, note that , where is the semiperimeter and is the length of . Simplifing, we get , so is on the same side of as , and Therefore,
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 |
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