Difference between revisions of "2017 AMC 10B Problems/Problem 21"
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==Solution 1== | ==Solution 1== | ||
− | We note that by the converse of the Pythagorean Theorem, <math>\triangle ABC</math> is a right triangle with a right angle at <math>A</math>. Therefore, <math>AD = BD = CD = 5</math>, and <math>[ADB] = [ADC] = 12</math>. Since <math>A = rs,</math> we have <math>r = \frac As</math>, so the inradius of <math>\triangle ADB</math> is <math>\frac{12}{(5+5+6)/2} = \frac 32</math>, and the inradius of <math>\triangle ADC</math> is <math>\frac{12}{(5+5+8)/2} = \frac 43</math>. Adding the two together, we have <math>\boxed{\textbf{(D) } \frac{17}6}</math>. | + | We note that by the converse of the Pythagorean Theorem, <math>\triangle ABC</math> is a right triangle with a right angle at <math>A</math>. Also, the median to the hypotenuse will be half of the hypotenuse. Therefore, <math>AD = BD = CD = 5</math>, and <math>[ADB] = [ADC] = 12</math>. Since <math>A = rs,</math> we have <math>r = \frac As</math>, so the inradius of <math>\triangle ADB</math> is <math>\frac{12}{(5+5+6)/2} = \frac 32</math>, and the inradius of <math>\triangle ADC</math> is <math>\frac{12}{(5+5+8)/2} = \frac 43</math>. Adding the two together, we have <math>\boxed{\textbf{(D) } \frac{17}6}</math>. |
==Solution 2== | ==Solution 2== | ||
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~MrThinker | ~MrThinker | ||
− | ==Solution 3== | + | ==Solution 3 (Stewart's)== |
− | Applying | + | Applying [https://artofproblemsolving.com/wiki/index.php/Stewart%27s_theorem] gives us the length of <math>\overline{AD}.</math> Using that length, we can find the areas of triangles <math>\triangle ABD</math> and <math>\triangle ACD</math> by using Heron’s formula. We can use that area to find the inradius of the circles by the inradius formula <math>A=sr.</math> Therefore, we get <math>\boxed{\textbf{(D) }\frac{17}{6}}.</math> Although this solution works perfectly fine, it takes time and has room for error so apply Stewart’s and Heron’s with caution. |
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+ | ~peelybonehead | ||
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+ | Edited by ~Jadon_Jung | ||
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+ | ==Solution 4 (using Shoelace and general inradius)== | ||
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+ | First, start by plotting <math>\triangle{ABC}</math> on a grid; with B at (0,0), D at (5,0), and C at (10,0). | ||
+ | |||
+ | We can now solve for the position of point A. Let point A be (a,b). We can then set up the equation a^2 + b^2 = 36 and the equation (10-a)^2 + b^2 = 64. Solving for a and b, we get that a = <math>\frac{18}{5}</math> and b = <math>\frac{24}{5}</math>. We can then go ahead and use the Shoelace method to get the area of <math>\triangle{ABD}</math>, which ends up being 12. Since D is the midpoint of BC, <math>\triangle{ABD}</math> and <math>\triangle{ACD}</math> have the same area. | ||
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+ | Using the distance formula once again, AD has length 5, and now we can get the semiperimeter of <math>\triangle{ABD}</math> and <math>\triangle{ACD}</math>, which turns out to be 8 for <math>\triangle{ABD}</math> and 9 for <math>\triangle{ACD}</math>. Dividing Area by Semiperimeter and adding them, we get <math>\frac{3}{2}</math> + <math>\frac{4}{3}</math> = <math>\boxed{\textbf{(D) }\frac{17}{6}}.</math> | ||
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+ | ~BanSpeedrun | ||
==Video Solution== | ==Video Solution== |
Latest revision as of 16:39, 17 October 2024
Contents
Problem
In , , , , and is the midpoint of . What is the sum of the radii of the circles inscribed in and ?
Solution 1
We note that by the converse of the Pythagorean Theorem, is a right triangle with a right angle at . Also, the median to the hypotenuse will be half of the hypotenuse. Therefore, , and . Since we have , so the inradius of is , and the inradius of is . Adding the two together, we have .
Solution 2
We have Let be the radius of circle , and let be the radius of circle . We want to find .
We form 6 kites: , , , , , and . Since and are the midpoints of and , respectively, this means that , and .
Since is a kite, , and . The same applies to all kites in the diagram.
Now, we see that , and , thus is , making and isosceles. So, using the Pythagorean Theorem, and also using the Theorem. Hence, we know that .
Notice that the area of the kite (if the opposite angles are right) is , where and denoting each of the 2 congruent sides. This just simplifies to . Hence, we have
and
Solving for and , we find that and , so .
~MrThinker
Solution 3 (Stewart's)
Applying [1] gives us the length of Using that length, we can find the areas of triangles and by using Heron’s formula. We can use that area to find the inradius of the circles by the inradius formula Therefore, we get Although this solution works perfectly fine, it takes time and has room for error so apply Stewart’s and Heron’s with caution.
~peelybonehead
Edited by ~Jadon_Jung
Solution 4 (using Shoelace and general inradius)
First, start by plotting on a grid; with B at (0,0), D at (5,0), and C at (10,0).
We can now solve for the position of point A. Let point A be (a,b). We can then set up the equation a^2 + b^2 = 36 and the equation (10-a)^2 + b^2 = 64. Solving for a and b, we get that a = and b = . We can then go ahead and use the Shoelace method to get the area of , which ends up being 12. Since D is the midpoint of BC, and have the same area.
Using the distance formula once again, AD has length 5, and now we can get the semiperimeter of and , which turns out to be 8 for and 9 for . Dividing Area by Semiperimeter and adding them, we get + =
~BanSpeedrun
Video Solution
See Also
2017 AMC 10B (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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.