Difference between revisions of "2003 AIME II Problems/Problem 11"
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== Solution == | == Solution == | ||
+ | ===Solution 1=== | ||
+ | |||
We use the [[Pythagorean Theorem]] on <math>ABC</math> to determine that <math>AB=25.</math> | We use the [[Pythagorean Theorem]] on <math>ABC</math> to determine that <math>AB=25.</math> | ||
− | Let <math>N</math> be the orthogonal projection from <math>C</math> to <math>AB.</math> Thus, <math>[CDM]=\frac{(DM)(MN)} {2}</math>, <math>MN= | + | Let <math>N</math> be the orthogonal projection from <math>C</math> to <math>AB.</math> Thus, <math>[CDM]=\frac{(DM)(MN)} {2}</math>, <math>MN=BN-BM</math>, and <math>[ABC]=\frac{24 \cdot 7} {2} =\frac{25 \cdot (CN)} {2}.</math> |
From the third equation, we get | From the third equation, we get | ||
<math>CN=\frac{168} {25}.</math> | <math>CN=\frac{168} {25}.</math> | ||
− | By the [[Pythagorean Theorem]] in <math>\Delta | + | By the [[Pythagorean Theorem]] in <math>\Delta BCN,</math> we have |
− | <math> | + | <math>BN=\sqrt{\left(\frac{24 \cdot 25} {25}\right)^2-\left(\frac{24 \cdot 7} {25}\right)^2}=\frac{24} {25}\sqrt{25^2-7^2}=\frac{576} {25}.</math> |
Thus, | Thus, | ||
Line 18: | Line 20: | ||
In <math>\Delta ADM</math>, we use the [[Pythagorean Theorem]] to get | In <math>\Delta ADM</math>, we use the [[Pythagorean Theorem]] to get | ||
− | <math>DM=\sqrt{15^2-(\frac{25} {2})^2}=\frac{5} {2} \sqrt{11}.</math> | + | <math>DM=\sqrt{15^2-\left(\frac{25} {2}\right)^2}=\frac{5} {2} \sqrt{11}.</math> |
Thus, | Thus, | ||
− | <math>[CDM]=\frac{527 \cdot 5\sqrt{11}} {50 \cdot 2 \cdot 2}=\frac{527\sqrt{11}} {40}.</math> | + | <math>[CDM]=\frac{527 \cdot 5\sqrt{11}} {50 \cdot 2 \cdot 2}= \frac{527\sqrt{11}} {40}.</math> |
Hence, the answer is <math>527+11+40=\boxed{578}.</math> | Hence, the answer is <math>527+11+40=\boxed{578}.</math> | ||
+ | |||
+ | ~ minor edits by kundusne000 | ||
+ | |||
+ | ===Solution 2=== | ||
+ | |||
+ | By the [[Pythagorean Theorem]] in <math>\Delta AMD</math>, we get <math>DM=\frac{5\sqrt{11}} {2}</math>. Since <math>ABC</math> is a right triangle, <math>M</math> is the circumcenter and thus, <math>CM=\frac{25} {2}</math>. We let <math>\angle CMD=\theta</math>. By the [[Law of Cosines]], | ||
+ | |||
+ | <math>24^2 = 2 \cdot (12.5)^2-2 \cdot (12.5)^2 * \cos (90+\theta).</math> | ||
+ | |||
+ | It follows that <math>\sin \theta = \frac{527} {625}</math>. Thus, | ||
+ | <math>[CMD]=\frac{1} {2} (12.5) \left(\frac{5\sqrt{11}} {2}\right)\left(\frac{527} {625}\right)=\frac{527\sqrt{11}} {40}</math>. | ||
+ | |||
+ | ===Solution 3=== | ||
+ | |||
+ | Suppose <math>ABC</math> is plotted on the [[cartesian plane]] with <math>C</math> at <math>(0,0)</math>, <math>A</math> at <math>(0,7)</math>, and <math>B</math> at <math>(24,0)</math>. | ||
+ | Then <math>M</math> is at <math>(12,3.5)</math>. Since <math>\Delta ABD</math> is isosceles, <math>MD</math> is perpendicular to <math>AM</math>, and since <math>AM=12.5</math> and <math>AD=15, MD=2.5\sqrt{11}</math>. | ||
+ | The slope of <math>AM</math> is <math>-\frac{7}{24}</math> so the slope of <math>MD</math> is <math>\frac{24}{7}</math>. | ||
+ | Draw a vertical line through <math>M</math> and a horizontal line through <math>D</math>. Suppose these two lines meet at <math>X</math>. then <math>MX=\frac{24}{7}DX</math> so <math>MD=\frac{25}{7}DX=\frac{25}{24}MD</math> by the pythagorean theorem. | ||
+ | So <math>MX=2.4\sqrt{11}</math> and <math>DX=.7\sqrt{11}</math> so the coordinates of D are <math>(12-.7\sqrt{11},\, 3.5-2.4\sqrt{11})</math>. | ||
+ | Since we know the coordinates of each of the vertices of <math>\Delta CMD</math>, we can apply the [[Shoelace Theorem]] to find the area of <math>\Delta CMD, \frac{527 \sqrt{11}}{40}</math>. | ||
+ | |||
+ | ~ minor edit by kundusne000 | ||
+ | |||
+ | ===Solution 4=== | ||
+ | |||
+ | Let <math>E</math> be the intersection of lines <math>BC</math> and <math>DM</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 CME</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}</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}\cdot7=\frac{7}{2}</math>. 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}</math>, giving our answer of <math>\boxed{578}</math>. | ||
+ | |||
+ | Solution by someonenumber011. | ||
== See also == | == See also == | ||
+ | Video Solution from Khan Academy: https://www.youtube.com/watch?v=smtrrefmC40 | ||
{{AIME box|year=2003|n=II|num-b=10|num-a=12}} | {{AIME box|year=2003|n=II|num-b=10|num-a=12}} | ||
+ | |||
+ | [[Category: Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 18:47, 1 August 2023
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
~ minor edits by kundusne000
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 .
~ minor edit by kundusne000
Solution 4
Let be the intersection of lines and . 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 . Using this, we can find that the area of is , giving our answer of .
Solution by someonenumber011.
See also
Video Solution from Khan Academy: https://www.youtube.com/watch?v=smtrrefmC40
2003 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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