Difference between revisions of "2017 AMC 10B Problems/Problem 21"
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− | + | ==Problem== | |
+ | In <math>\triangle ABC</math>, <math>AB=6</math>, <math>AC=8</math>, <math>BC=10</math>, and <math>D</math> is the midpoint of <math>\overline{BC}</math>. What is the sum of the radii of the circles inscribed in <math>\triangle ADB</math> and <math>\triangle ADC</math>? | ||
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+ | <math>\textbf{(A)}\ \sqrt{5}\qquad\textbf{(B)}\ \frac{11}{4}\qquad\textbf{(C)}\ 2\sqrt{2}\qquad\textbf{(D)}\ \frac{17}{6}\qquad\textbf{(E)}\ 3</math> | ||
==Solution== | ==Solution== | ||
− | + | 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>, 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>. | |
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+ | ==Video Solution== | ||
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+ | https://youtu.be/EfKFDwTDRjs | ||
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==See Also== | ==See Also== | ||
{{AMC10 box|year=2017|ab=B|num-b=20|num-a=22}} | {{AMC10 box|year=2017|ab=B|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 16:01, 12 July 2020
Contents
Problem
In , , , , and is the midpoint of . What is the sum of the radii of the circles inscribed in and ?
Solution
We note that by the converse of the Pythagorean Theorem, is a right triangle with a right angle at . Therefore, , and . Since , the inradius of is , and the inradius of is . Adding the two together, we have .
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 |
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