Difference between revisions of "2019 AMC 10A Problems/Problem 21"
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We have that <math>\triangle{BDC}</math> is a <math>3-4-5</math> triangle, so, as in Solution 1, <math>BD = 9</math>. | We have that <math>\triangle{BDC}</math> is a <math>3-4-5</math> triangle, so, as in Solution 1, <math>BD = 9</math>. | ||
From this, we know that <math>\overline{BI}=9-r</math>. | From this, we know that <math>\overline{BI}=9-r</math>. | ||
− | Since AB is tangent to circle <math>I</math>, we also know <math>IEB</math> is a right triangle. <math>\triangle{BIE}</math> and <math>\triangle{BDA}</math> share angle <math>DBA</math>, so <math>\triangle{BIE} \sim \triangle{BDA}</math> since they have two equal angles. | + | Since <math>AB</math> is tangent to circle <math>I</math>, we also know <math>IEB</math> is a right triangle. <math>\triangle{BIE}</math> and <math>\triangle{BDA}</math> share angle <math>DBA</math>, so <math>\triangle{BIE} \sim \triangle{BDA}</math> since they have two equal angles. |
Hence, by this similarity, <math>\dfrac{9-r}{5}=\dfrac{r}{4}</math>. Cross-multiplying, we get <math>36-4r =5r</math>, which gives <math>r=4</math>. | Hence, by this similarity, <math>\dfrac{9-r}{5}=\dfrac{r}{4}</math>. Cross-multiplying, we get <math>36-4r =5r</math>, which gives <math>r=4</math>. | ||
We now take another cross section of the sphere, perpendicular to the plane of the triangle. | We now take another cross section of the sphere, perpendicular to the plane of the triangle. |
Revision as of 21:33, 19 April 2019
- The following problem is from both the 2019 AMC 10A #21 and 2019 AMC 12A #18, so both problems redirect to this page.
Contents
Problem
A sphere with center has radius . A triangle with sides of length and is situated in space so that each of its sides is tangent to the sphere. What is the distance between and the plane determined by the triangle?
Diagram
3D: Plane through triangle:
Solution 1
The triangle is placed on the sphere so that its three sides are tangent to the sphere. The cross-section of the sphere created by the plane of the triangle is also the incircle of the triangle. To find the inradius, use . The area of the triangle can be found by drawing an altitude from the vertex between sides with length to the midpoint of the side with length . The Pythagorean triple - - allows us easily to determine that the base is and the height is . The formula can also be used to find the area of the triangle as , while the semiperimeter is simply . After plugging into the equation, we thus get , so the inradius is . Now, let the distance between and the triangle be . Choose a point on the incircle and denote it by . The distance is , because it is just the radius of the sphere. The distance from point to the center of the incircle is , because it is the radius of the incircle. By using the Pythagorean Theorem, we thus find .
Solution 2
As in Solution 1, we note that by the Pythagorean Theorem, the height of the triangle is , and that the three sides of the triangle are tangent to the sphere, so the circle in the cross-section of the sphere is the incenter of the triangle.
Recall that the inradius is the intersection of the angle bisectors. To find the inradius of the incircle, we use the Angle Bisector Theorem. Since we know that (the height) is equal to , (the inradius) is . From here, the problem can be solved in the same way as in Solution 1. The answer is .
Solution 3 (similar triangles)
First, we label a few points:
We have that is a triangle, so, as in Solution 1, . From this, we know that . Since is tangent to circle , we also know is a right triangle. and share angle , so since they have two equal angles. Hence, by this similarity, . Cross-multiplying, we get , which gives . We now take another cross section of the sphere, perpendicular to the plane of the triangle.
Using the Pythagorean Theorem, we find that the distance from the center to the plane is .
solution by woofle628 and GeniusKid1221
Solution 4 (educated guess)
Test all the answer choices by plugging them into the expression to find the inradius of the triangle. Seeing that only gives an integer inradius, we pick .
See Also
2019 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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 |
2019 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.