Difference between revisions of "2012 AMC 12A Problems/Problem 18"
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− | We can use mass points and Stewart's to solve this problem. Because we are looking at the Incenter we then label <math>A</math> with a mass of 25, <math>B</math> with <math>26</math>, and <math>C</math> with <math>27</math>. We also label where the angle bisectors intersect the opposite side <math>A'</math>, <math>B'</math>, and <math>C'</math> correspondingly. It follows then that point <math>B'</math> has mass 52. Which means that <math>\overline{BB'}</math> is split into a <math>2:1</math> ratio. We can then use Stewart's to find <math>\overline{BB'}</math>. So we have <math>25^2\frac{27}{2} + 27^2\frac{25}{2} = \frac{25 \cdot 26 \cdot 27}{4} + 26\overline{BB'}^2</math>. Solving we get <math>\overline{BB'} = \frac{45}{2}</math>. Plugging it in we get <math>\overline{BI} = 15</math>. Therefore the answer is <math>\boxed{(E)\:15.}</math> | + | We can use mass points and Stewart's to solve this problem. Because we are looking at the Incenter we then label <math>A</math> with a mass of <math>25</math>, <math>B</math> with <math>26</math>, and <math>C</math> with <math>27</math>. We also label where the angle bisectors intersect the opposite side <math>A'</math>, <math>B'</math>, and <math>C'</math> correspondingly. It follows then that point <math>B'</math> has mass 52. Which means that <math>\overline{BB'}</math> is split into a <math>2:1</math> ratio. We can then use Stewart's to find <math>\overline{BB'}</math>. So we have <math>25^2\frac{27}{2} + 27^2\frac{25}{2} = \frac{25 \cdot 26 \cdot 27}{4} + 26\overline{BB'}^2</math>. Solving we get <math>\overline{BB'} = \frac{45}{2}</math>. Plugging it in we get <math>\overline{BI} = 15</math>. Therefore the answer is <math>\boxed{(E)\:15.}</math> |
-Solution by '''arowaaron''' | -Solution by '''arowaaron''' |
Revision as of 19:02, 16 July 2017
Contents
[hide]Problem
Triangle has , , and . Let denote the intersection of the internal angle bisectors of . What is ?
Solution 1
Inscribe circle of radius inside triangle so that it meets at , at , and at . Note that angle bisectors of triangle are concurrent at the center (also ) of circle . Let , and . Note that , and . Hence , , and . Subtracting the last 2 equations we have and adding this to the first equation we have .
By Heron's formula for the area of a triangle we have that the area of triangle is . On the other hand the area is given by . Then so that .
Since the radius of circle is perpendicular to at , we have by the pythagorean theorem so that .
Solution 2
We can use mass points and Stewart's to solve this problem. Because we are looking at the Incenter we then label with a mass of , with , and with . We also label where the angle bisectors intersect the opposite side , , and correspondingly. It follows then that point has mass 52. Which means that is split into a ratio. We can then use Stewart's to find . So we have . Solving we get . Plugging it in we get . Therefore the answer is
-Solution by arowaaron
See Also
2012 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.