Difference between revisions of "2003 AIME II Problems/Problem 11"
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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, | ||
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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=== | ===Solution 2=== | ||
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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>. | 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. | 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},\, | + | 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>. | 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=== | ===Solution 4=== | ||
− | 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>. | + | 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>. | 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>. | 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>. | ||
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== 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]] | [[Category: Intermediate Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
− |
Latest revision as of 19:47, 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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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