Difference between revisions of "2003 AIME II Problems/Problem 11"
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Since triangles <math>\Delta CME</math> and <math>\Delta CMD</math> share a side and height, the area of <math>\Delta CDM</math> is equal to <math>\frac{DM}{EM}</math> times the area of <math>\Delta CEM</math>. | Since triangles <math>\Delta CME</math> and <math>\Delta CMD</math> share a side and height, the area of <math>\Delta CDM</math> is equal to <math>\frac{DM}{EM}</math> times the area of <math>\Delta CEM</math>. | ||
By AA similarity, <math>\Delta EMB</math> is similar to <math>\Delta ACB</math>, <math>\frac{EM}{AC}=\frac{MB}{CB}</math>. Solving yields <math>EM=\frac{175}{48}</math>. Using the same method but for <math>EB</math> yields <math>EB=\frac{625}{48}</math>. As in previous solutions, by the [[Pythagorean Theorem]], <math>DM=\frac{5\sqrt{11}}{2}</math>. So, <math>\frac{DM}{EM}=\frac{24\sqrt{11}}{35}</math>. | By AA similarity, <math>\Delta EMB</math> is similar to <math>\Delta ACB</math>, <math>\frac{EM}{AC}=\frac{MB}{CB}</math>. Solving yields <math>EM=\frac{175}{48}</math>. Using the same method but for <math>EB</math> yields <math>EB=\frac{625}{48}</math>. As in previous solutions, by the [[Pythagorean Theorem]], <math>DM=\frac{5\sqrt{11}}{2}</math>. So, <math>\frac{DM}{EM}=\frac{24\sqrt{11}}{35}</math>. | ||
− | Now, since we know both <math>CB</math> and <math>EB</math>, we can find that <math>CE=\frac{527}{48}. The height of < | + | Now, since we know both <math>CB</math> and <math>EB</math>, we can find that <math>CE=\frac{527}{48}</math>. The height of <math>\Delta CME</math> is the length from point <math>M</math> to <math>CB</math>. Since <math>M</math> is the midpoint of <math>AB</math>, the height is just <math>\frac{1}{2}\cdot(7)=\frac{7}{2}. Using this, we can find that the area of </math>\Delta CMD<math> is </math>\frac{1}{2}\cdot\frac{7}{2}\cdot\frac{527}{48}\cdot\frac{24\sqrt{11}}{35}=\frac{527\sqrt{11}}{40}$, giving our answer of \boxed{578}. |
Solution by someonenumber011. | Solution by someonenumber011. |
Revision as of 17:32, 26 December 2019
Contents
Problem
Triangle is a right triangle with and right angle at Point is the midpoint of and is on the same side of line as so that Given that the area of triangle may be expressed as where and are positive integers, and are relatively prime, and is not divisible by the square of any prime, find
Solution
Solution 1
We use the Pythagorean Theorem on to determine that
Let be the orthogonal projection from to Thus, , , and
From the third equation, we get
By the Pythagorean Theorem in we have
Thus,
In , we use the Pythagorean Theorem to get
Thus,
Hence, the answer is
Solution 2
By the Pythagorean Theorem in , we get . Since is a right triangle, is the circumcenter and thus, . We let . By the Law of Cosines,
It follows that . Thus, .
Solution 3
Suppose is plotted on the cartesian plane with at , at , and at . Then is at . Since is isosceles, is perpendicular to , and since and . The slope of is so the slope of is . Draw a vertical line through and a horizontal line through . Suppose these two lines meet at . then so by the pythagorean theorem. So and so the coordinates of D are . Since we know the coordinates of each of the vertices of , we can apply the Shoelace Theorem to find the area of .
Solution 4
Since triangles and share a side and height, the area of is equal to times the area of . By AA similarity, is similar to , . Solving yields . Using the same method but for yields . As in previous solutions, by the Pythagorean Theorem, . So, . Now, since we know both and , we can find that . The height of is the length from point to . Since is the midpoint of , the height is just \Delta CMD\frac{1}{2}\cdot\frac{7}{2}\cdot\frac{527}{48}\cdot\frac{24\sqrt{11}}{35}=\frac{527\sqrt{11}}{40}$, giving our answer of \boxed{578}.
Solution by someonenumber011.
See also
2003 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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