Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2003 AIME II Problems/Problem 11"

(→‎Solution 2)
m (→‎Solution 3)
 
(5 intermediate revisions by 4 users not shown)
Line 7: Line 7:
 
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=AM-AN</math>, and <math>[ABC]=\frac{24 \cdot 7} {2} =\frac{25 \cdot (CN)} {2}.</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=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 ACN,</math> we have  
+
By the [[Pythagorean Theorem]] in <math>\Delta BCN,</math> we have  
  
−
<math>AN=\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>  
+
<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 26: Line 26:
  
 
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===
Line 42: Line 44:
 
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},\, 2.5-2.4\sqrt{11})</math>.
+
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>.
Line 55: Line 58:
  
 
== 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}}
−
i got the mojo deals
 

Latest revision as of 18:47, 1 August 2023

Problem

Triangle $ABC$ is a right triangle with $AC = 7,$ $BC = 24,$ and right angle at $C.$ Point $M$ is the midpoint of $AB,$ and $D$ is on the same side of line $AB$ as $C$ so that $AD = BD = 15.$ Given that the area of triangle $CDM$ may be expressed as $\frac {m\sqrt {n}}{p},$ where $m,$ $n,$ and $p$ are positive integers, $m$ and $p$ are relatively prime, and $n$ is not divisible by the square of any prime, find $m + n + p.$

Solution

Solution 1

We use the Pythagorean Theorem on $ABC$ to determine that $AB=25.$

Let $N$ be the orthogonal projection from $C$ to $AB.$ Thus, $[CDM]=\frac{(DM)(MN)} {2}$, $MN=BN-BM$, and $[ABC]=\frac{24 \cdot 7} {2} =\frac{25 \cdot (CN)} {2}.$

From the third equation, we get $CN=\frac{168} {25}.$

By the Pythagorean Theorem in $\Delta BCN,$ we have

$BN=\sqrt{\left(\frac{24 \cdot 25} {25}\right)^2-\left(\frac{24 \cdot 7} {25}\right)^2}=\frac{24} {25}\sqrt{25^2-7^2}=\frac{576} {25}.$

Thus, $MN=\frac{576} {25}-\frac{25} {2}=\frac{527} {50}.$

In $\Delta ADM$, we use the Pythagorean Theorem to get $DM=\sqrt{15^2-\left(\frac{25} {2}\right)^2}=\frac{5} {2} \sqrt{11}.$

Thus, $[CDM]=\frac{527 \cdot 5\sqrt{11}} {50 \cdot 2 \cdot 2}= \frac{527\sqrt{11}} {40}.$

Hence, the answer is $527+11+40=\boxed{578}.$

~ minor edits by kundusne000

Solution 2

By the Pythagorean Theorem in $\Delta AMD$, we get $DM=\frac{5\sqrt{11}} {2}$. Since $ABC$ is a right triangle, $M$ is the circumcenter and thus, $CM=\frac{25} {2}$. We let $\angle CMD=\theta$. By the Law of Cosines,

$24^2 = 2 \cdot (12.5)^2-2 \cdot (12.5)^2 * \cos (90+\theta).$

It follows that $\sin \theta = \frac{527} {625}$. Thus, $[CMD]=\frac{1} {2} (12.5) \left(\frac{5\sqrt{11}} {2}\right)\left(\frac{527} {625}\right)=\frac{527\sqrt{11}} {40}$.

Solution 3

Suppose $ABC$ is plotted on the cartesian plane with $C$ at $(0,0)$, $A$ at $(0,7)$, and $B$ at $(24,0)$. Then $M$ is at $(12,3.5)$. Since $\Delta ABD$ is isosceles, $MD$ is perpendicular to $AM$, and since $AM=12.5$ and $AD=15, MD=2.5\sqrt{11}$. The slope of $AM$ is $-\frac{7}{24}$ so the slope of $MD$ is $\frac{24}{7}$. Draw a vertical line through $M$ and a horizontal line through $D$. Suppose these two lines meet at $X$. then $MX=\frac{24}{7}DX$ so $MD=\frac{25}{7}DX=\frac{25}{24}MD$ by the pythagorean theorem. So $MX=2.4\sqrt{11}$ and $DX=.7\sqrt{11}$ so the coordinates of D are $(12-.7\sqrt{11},\, 3.5-2.4\sqrt{11})$. Since we know the coordinates of each of the vertices of $\Delta CMD$, we can apply the Shoelace Theorem to find the area of $\Delta CMD, \frac{527 \sqrt{11}}{40}$.

~ minor edit by kundusne000

Solution 4

Let $E$ be the intersection of lines $BC$ and $DM$. Since triangles $\Delta CME$ and $\Delta CMD$ share a side and height, the area of $\Delta CDM$ is equal to $\frac{DM}{EM}$ times the area of $\Delta CME$. By AA similarity, $\Delta EMB$ is similar to $\Delta ACB$, $\frac{EM}{AC}=\frac{MB}{CB}$. Solving yields $EM=\frac{175}{48}$. Using the same method but for $EB$ yields $EB=\frac{625}{48}$. As in previous solutions, by the Pythagorean Theorem, $DM=\frac{5\sqrt{11}}{2}$. So, $\frac{DM}{EM}=\frac{24\sqrt{11}}{35}$. Now, since we know both $CB$ and $EB$, we can find that $CE=\frac{527}{48}$. The height of $\Delta CME$ is the length from point $M$ to $CB$. Since $M$ is the midpoint of $AB$, the height is just $\frac{1}{2}\cdot7=\frac{7}{2}$. Using this, we can find that the area of $\Delta CMD$ is $\frac{1}{2}\cdot\frac{7}{2}\cdot\frac{527}{48}\cdot\frac{24\sqrt{11}}{35}=\frac{527\sqrt{11}}{40}$, giving our answer of $\boxed{578}$.

Solution by someonenumber011.

See also

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
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_II_Problems/Problem_11&oldid=196332"