Difference between revisions of "2012 AMC 12A Problems/Problem 18"
(→See Also) |
(→Solution 2) |
||
Line 5: | Line 5: | ||
<math> \textbf{(A)}\ 15\qquad\textbf{(B)}\ 5+\sqrt{26}+3\sqrt{3}\qquad\textbf{(C)}\ 3\sqrt{26}\qquad\textbf{(D)}\ \frac{2}{3}\sqrt{546}\qquad\textbf{(E)}\ 9\sqrt{3} </math> | <math> \textbf{(A)}\ 15\qquad\textbf{(B)}\ 5+\sqrt{26}+3\sqrt{3}\qquad\textbf{(C)}\ 3\sqrt{26}\qquad\textbf{(D)}\ \frac{2}{3}\sqrt{546}\qquad\textbf{(E)}\ 9\sqrt{3} </math> | ||
− | == Solution == | + | == Solution 1 == |
Inscribe circle <math>C</math> of radius <math>r</math> inside triangle <math>ABC</math> so that it meets <math>AB</math> at <math>Q</math>, <math>BC</math> at <math>R</math>, and <math>AC</math> at <math>S</math>. Note that angle bisectors of triangle <math>ABC</math> are concurrent at the center <math>O</math>(also <math>I</math>) of circle <math>C</math>. Let <math>x=QB</math>, <math>y=RC</math> and <math>z=AS</math>. Note that <math>BR=x</math>, <math>SC=y</math> and <math>AQ=z</math>. Hence <math>x+z=27</math>, <math>x+y=25</math>, and <math>z+y=26</math>. Subtracting the last 2 equations we have <math>x-z=-1</math> and adding this to the first equation we have <math>x=13</math>. | Inscribe circle <math>C</math> of radius <math>r</math> inside triangle <math>ABC</math> so that it meets <math>AB</math> at <math>Q</math>, <math>BC</math> at <math>R</math>, and <math>AC</math> at <math>S</math>. Note that angle bisectors of triangle <math>ABC</math> are concurrent at the center <math>O</math>(also <math>I</math>) of circle <math>C</math>. Let <math>x=QB</math>, <math>y=RC</math> and <math>z=AS</math>. Note that <math>BR=x</math>, <math>SC=y</math> and <math>AQ=z</math>. Hence <math>x+z=27</math>, <math>x+y=25</math>, and <math>z+y=26</math>. Subtracting the last 2 equations we have <math>x-z=-1</math> and adding this to the first equation we have <math>x=13</math>. | ||
Line 12: | Line 12: | ||
Since the radius of circle <math>O</math> is perpendicular to <math>BC</math> at <math>R</math>, we have by the pythagorean theorem <math>BO^2=BI^2=r^2+x^2=56+169=225</math> so that <math>BI=15</math>. | Since the radius of circle <math>O</math> is perpendicular to <math>BC</math> at <math>R</math>, we have by the pythagorean theorem <math>BO^2=BI^2=r^2+x^2=56+169=225</math> so that <math>BI=15</math>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | 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)\:6.}</math> | ||
+ | |||
+ | -Solution by '''arowaaron''' | ||
== See Also == | == See Also == |
Revision as of 18:57, 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 25, 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.