Difference between revisions of "2004 AMC 10B Problems/Problem 22"
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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> | ||
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+ | ==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?
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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