Difference between revisions of "Mock AIME II 2012 Problems/Problem 9"
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Now, note that we have to find the coordinates of <math>E,F</math> and <math>D</math>. Assume WLOG that <math>A</math> is <math>(0,0)</math> and <math>B</math> be <math>(12,0)</math>. <math>E</math> is obviously <math>(3,0)</math>, <math>F</math> is going to be <math>(12+\frac{9+\sqrt{73}-12}{4}, \frac{\sqrt{219}-3\sqrt3}{4})</math> and the coordinates of <math>D</math> is going to be <math>(\frac{9+\sqrt{73}}{4}, \frac{\sqrt{219}-3\sqrt3}{4})</math>. Since <math>D</math> and <math>F</math> have the same <math>y</math> value, we can find <math>[EDF]</math> by using <math>\frac12*b*h</math>. The base is going to be equal to the distance from <math>D</math> to <math>F</math>, which is the same thing as <math>12+\frac{9+\sqrt{73}-12}{4}-\frac{9+\sqrt{73}}{4}=9=b</math>, and the height is the change in the <math>y</math> coordinate from <math>E</math> to <math>D</math>. This is the same thing as <math>\frac{\sqrt{219}-3\sqrt3}{4}-0=\frac{\sqrt{219}-3\sqrt3}{4}=h</math>. Hence, plugging these into <math>[EDF]=\frac12*b*h</math> give us <math>\frac12*9*\frac{\sqrt{219}-3\sqrt3}{4}=\frac{9\sqrt{219}-27\sqrt3}{8}</math>. The answer is thus <math>9+219+27+3+8=\boxed{266}</math>. | Now, note that we have to find the coordinates of <math>E,F</math> and <math>D</math>. Assume WLOG that <math>A</math> is <math>(0,0)</math> and <math>B</math> be <math>(12,0)</math>. <math>E</math> is obviously <math>(3,0)</math>, <math>F</math> is going to be <math>(12+\frac{9+\sqrt{73}-12}{4}, \frac{\sqrt{219}-3\sqrt3}{4})</math> and the coordinates of <math>D</math> is going to be <math>(\frac{9+\sqrt{73}}{4}, \frac{\sqrt{219}-3\sqrt3}{4})</math>. Since <math>D</math> and <math>F</math> have the same <math>y</math> value, we can find <math>[EDF]</math> by using <math>\frac12*b*h</math>. The base is going to be equal to the distance from <math>D</math> to <math>F</math>, which is the same thing as <math>12+\frac{9+\sqrt{73}-12}{4}-\frac{9+\sqrt{73}}{4}=9=b</math>, and the height is the change in the <math>y</math> coordinate from <math>E</math> to <math>D</math>. This is the same thing as <math>\frac{\sqrt{219}-3\sqrt3}{4}-0=\frac{\sqrt{219}-3\sqrt3}{4}=h</math>. Hence, plugging these into <math>[EDF]=\frac12*b*h</math> give us <math>\frac12*9*\frac{\sqrt{219}-3\sqrt3}{4}=\frac{9\sqrt{219}-27\sqrt3}{8}</math>. The answer is thus <math>9+219+27+3+8=\boxed{266}</math>. | ||
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+ | ==Solution 2== | ||
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+ | Let <math>S</math> be the area of the large triangle and let <math>x</math> be the area we are trying to find. We have that <math>\Delta ADE</math> is similar to <math>\Delta ABC</math> with a ratio of <math>1:4</math> while <math>\Delta CDF</math> is similar to <math>\Delta ABC</math> with a ratio of <math>3:4</math>. Then the area of <math>\Delta ADE = \frac{S}{16}</math> and the area of <math>\Delta CDE = \frac{9S}{16}</math>. Lastly, the area of <math>\Delta EBF = \frac{1}{2}EB*BF*\sin B = \frac{1}{2}* \frac{3AB}{4}*\frac{BC}{4}*\sin B = \frac{3}{16}S</math>. The area of <math>\Delta ABC</math> is equal to the sum of the areas of the four smaller triangles so <math>S= \frac{S}{16}+\frac{9S}{16}+\frac{3S}{16} + x</math>. Thus <math>x = \frac{3S}{16}</math>. | ||
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+ | It remains to find <math>S</math>. First we can use the Law of Cosines on <math>\Delta ABC</math> to find <math>12^2+BC^2 + 12BC = 400</math> which solves to <math>BC = -6 + 2\sqrt{73}</math>. Then <math>S = \frac{1}{2} AB* BC*\sin{120} = -18\sqrt3 + 6\sqrt{219}</math>. Then substituting yields <math>x =\frac{-27\sqrt 3 + 9 \sqrt{219}}{8}</math> so the answer is <math>\boxed{266}</math> |
Latest revision as of 13:04, 27 December 2012
Problem
In , , , and . and lie on , and , respectively. If , and , the area of can be expressed in the form where are all positive integers, and and do not have any perfect squares greater than as divisors. Find .
Solution
Here is a diagram (note that D should be on AC and F should be on BC): . Start out by finding . Remark that by the law of sines, . Therefore . We know because in , therefore is in the first quadrant. Use the Pythagorean Identity to give us .
Now, note that is the same thing as , from , therefore we have $\equal{} \frac{\sqrt3*\sqrt{73}}{20}-\frac{3\sqrt3}{20}=\frac{\sqrt3*\sqrt{73}-3\sqrt3}{20}$ (Error compiling LaTeX. Unknown error_msg).
Next, use the law of sines to give us . This gives us . This gives us .
Now, we use coordinates to find that the coordinates of , , , and . is going to be the point from adding point to create triangle as shown in this diagram (where is the right angle, is the angle, and is the angle):
Now, note that we have to find the coordinates of and . Assume WLOG that is and be . is obviously , is going to be and the coordinates of is going to be . Since and have the same value, we can find by using . The base is going to be equal to the distance from to , which is the same thing as , and the height is the change in the coordinate from to . This is the same thing as . Hence, plugging these into give us . The answer is thus .
Solution 2
Let be the area of the large triangle and let be the area we are trying to find. We have that is similar to with a ratio of while is similar to with a ratio of . Then the area of and the area of . Lastly, the area of . The area of is equal to the sum of the areas of the four smaller triangles so . Thus .
It remains to find . First we can use the Law of Cosines on to find which solves to . Then . Then substituting yields so the answer is