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Difference between revisions of "2004 AMC 12B Problems/Problem 14"

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== Solution ==
 
== Solution ==
  
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=== Solution 1 ===
 
The triangle <math>ABC</math> is clearly a right triangle, its area is <math>\frac{5\cdot 12}2 = 30</math>. If we knew the areas of triangles <math>AMJ</math> and <math>BNK</math>, we could subtract them to get the area of the pentagon.
 
The triangle <math>ABC</math> is clearly a right triangle, its area is <math>\frac{5\cdot 12}2 = 30</math>. If we knew the areas of triangles <math>AMJ</math> and <math>BNK</math>, we could subtract them to get the area of the pentagon.
  
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Finally, the area of the pentagon is <math>30 - \frac{30}{169} - \frac{1920}{169} = \boxed{\frac{240}{13}}</math>.
 
Finally, the area of the pentagon is <math>30 - \frac{30}{169} - \frac{1920}{169} = \boxed{\frac{240}{13}}</math>.
  
 +
==== Solution 1a ====
 +
 +
Observe that all of the triangles in the problem are right triangles and similar with a ratio of 5-12-13. The largest triangle <math>\Delta ABC</math> has area <math>\dfrac{5\cdot12}2=30</math>. <math>\Delta AMJ</math> is linearly scaled down from <math>\Delta ABC</math> by a factor of <math>\dfrac{5-4}{13}=\dfrac1{13}</math> (as we can see from comparing the two hypotenuses), while <math>\Delta NBK</math> is scaled by a factor of <math>\dfrac{12-4}{13}=\dfrac8{13}</math>. The area we desire is the combined area of <math>\Delta AMJ</math> and <math>\Delta NBK</math> subtracted from the area of <math>\Delta ABC</math>, which is
 +
 +
<cmath>30-30\left(\dfrac1{13}\right)^2-30\left(\dfrac8{13}\right)^2=30\left(1-\left(\dfrac1{13}\right)^2-\left(\dfrac8{13}\right)^2\right)=30\left(1-\dfrac{65}{169}\right)=30\left(\dfrac8{13}\right)=\boxed{\textbf{(D) }\dfrac{240}{13}}.</cmath>
 +
 +
~Technodoggo
 +
 +
=== Solution 2 ===
 +
 +
Split the pentagon along a different diagonal as follows:
 +
 +
<asy>
 +
unitsize(0.5cm);
 +
defaultpen(0.8);
 +
pair C=(0,0), A=(0,5), B=(12,0), M=(0,4), N=(4,0);
 +
pair J=intersectionpoint(A--B, M--(M+rotate(90)*(B-A)) );
 +
pair K=intersectionpoint(A--B, N--(N+rotate(90)*(B-A)) );
 +
pair L=intersectionpoint(A--B, C--(C+rotate(90)*(B-A)) );
 +
draw( A--B--C--cycle );
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draw( M--J );
 +
draw( N--K );
 +
draw( M--N, dashed );
 +
label("$A$",A,NW);
 +
label("$B$",B,SE);
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label("$C$",C,SW);
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label("$M$",M,SW);
 +
label("$N$",N,S);
 +
label("$J$",J,NE);
 +
label("$K$",K,NE);
 +
label("$L$",L,NE);
 +
</asy>
 +
The area of the pentagon is then the sum of the areas of the resulting right triangle and trapezoid. As before, triangles <math>ABC</math>, <math>AMJ</math>, and <math>NBK</math> are all similar.
 +
 +
Since <math>BN=12-4=8</math>, <math>NK=\frac{5}{13}(8)=\frac{40}{13}</math> and <math>BK=\frac{12}{13}(8)=\frac{96}{13}</math>. Since <math>AM=5-4=1</math>, <math>JM=\frac{12}{13}</math> and <math>AJ=\frac{5}{13}</math>.
 +
 +
The trapezoid's height is therefore <math>13-\frac{5}{13}-\frac{96}{13}=\frac{68}{13}</math>, and its area is <math>\frac{1}{2}\left(\frac{68}{13}\right)\left(\frac{12}{13}+\frac{40}{13}\right)=\frac{34}{13}(4)=\frac{136}{13}</math>.
 +
 +
Triangle <math>MCN</math> has area <math>\frac{1}{2}(4)(4)=8</math>, and the total area is <math>\frac{104+136}{13}=\boxed{\frac{240}{13}}</math>.
 +
 +
=== Solution 3 ===
 +
Right triangle <math>ABC</math> has area <math>1/2 \cdot AC \cdot BC = 1/2 \cdot 5 \cdot 12 = 30</math>.
 +
 +
We are given that <math>AC = 5</math> and <math>CM = 4</math>, so <math>AM = AC - CM = 1</math>. Right triangles <math>AMJ</math> and <math>ABC</math> are similar, with hypotenuses 1 and 13, respectively, so triangle <math>AMJ</math> has area
 +
<cmath>\left( \frac{1}{13} \right)^2 \cdot 30.</cmath>
 +
Also, we are given that <math>BC = 12</math> and <math>CN = 4</math>, so <math>BN = BC - CN = 8</math>. Right triangles <math>NBK</math> and <math>ABC</math> are similar, with hypotenuses 8 and 13, respectively, so triangle <math>NBK</math> has area
 +
<cmath>\left( \frac{8}{13} \right)^2 \cdot 30.</cmath>
 +
Therefore, the area of pentagon <math>CMJKN</math> is
 +
<cmath>30 - \left( \frac{1}{13} \right)^2 \cdot 30 - \left( \frac{8}{13} \right)^2 \cdot 30 = \boxed{\frac{240}{13}}.</cmath>
 +
 +
~math31415926535
 +
 +
===Solution 4===
 +
Because triangle ABC, triangle NBK, and triangle AMJ are similar right triangles whose hypotenuses are in the ratio 13 : 8 : 1, their areas are in
 +
the ratio 169 : 64 : 1. The area of triangle ABC is 1/2 (12)(5) = 30, so the areas of triangle NBK and triangle AMJ are (64/169) (30) and (1/169)(30), respectively. Thus the area of pentagon CMJKN is (1 −  64/169 - 1/169)(30)= <math>\boxed{\mathbf{(D)}240/13} </math>
 +
 +
Credit to http://billingswest.billings.k12.mt.us/math/AMC%201012/AMC%2012%20work%20sheets/2004%20AMC%2012B%20ws-15.pdf for Solution 3.
 +
 +
===Solution 5===
 +
We fake-solve. Draw a perpendicular line to side <math>CB</math> that goes through <math>N</math>. Similarly, draw a perpendicular line to side <math>AC</math> that goes through <math>M</math>. Let them intersect at <math>O</math>. It follows that quadrilateral <math>MONC</math> is a square with side <math>4</math> and thus area <math>16</math>. Now, draw a perpendicular from <math>J</math> to <math>MO</math> and let them intersect at <math>Z</math>. There is a obvious pair of congruent triangles. Fill in the gap. Hence, we see that the area of the pentagon is definitely greater than <math>16</math>. How much greater? Well, if we let the intersection of <math>AB</math> and <math>NO</math> be <math>X</math>, we can approximate that <math>\triangle NXK</math> has an area greater than <math>2</math>. So we place bets on <math>\boxed{\mathbf{(D)}240/13} </math> which is correct!
 +
 +
This would be much easier to show and more convincing with a diagram but idk how to upload GeoGebra to AoPSwiki.
 +
 +
Solution by franzliszt
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC12 box|year=2004|ab=B|num-b=13|num-a=15}}
 
{{AMC12 box|year=2004|ab=B|num-b=13|num-a=15}}
 +
{{MAA Notice}}

Latest revision as of 20:16, 30 October 2024

Problem

In $\triangle ABC$, $AB=13$, $AC=5$, and $BC=12$. Points $M$ and $N$ lie on $AC$ and $BC$, respectively, with $CM=CN=4$. Points $J$ and $K$ are on $AB$ so that $MJ$ and $NK$ are perpendicular to $AB$. What is the area of pentagon $CMJKN$?

[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,5), B=(12,0), M=(0,4), N=(4,0); pair J=intersectionpoint(A--B, M--(M+rotate(90)*(B-A)) ); pair K=intersectionpoint(A--B, N--(N+rotate(90)*(B-A)) ); draw( A--B--C--cycle ); draw( M--J ); draw( N--K ); label("$A$",A,NW); label("$B$",B,SE); label("$C$",C,SW); label("$M$",M,SW); label("$N$",N,S); label("$J$",J,NE); label("$K$",K,NE); [/asy]

$\mathrm{(A)}\ 15 \qquad \mathrm{(B)}\ \frac{81}{5} \qquad \mathrm{(C)}\ \frac{205}{12} \qquad \mathrm{(D)}\ \frac{240}{13} \qquad \mathrm{(E)}\ 20$

Solution

Solution 1

The triangle $ABC$ is clearly a right triangle, its area is $\frac{5\cdot 12}2 = 30$. If we knew the areas of triangles $AMJ$ and $BNK$, we could subtract them to get the area of the pentagon.

Draw the height $CL$ from $C$ onto $AB$. As $AB=13$ and the area is $30$, we get $CL=\frac{60}{13}$. The situation is shown in the picture below:

[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,5), B=(12,0), M=(0,4), N=(4,0); pair J=intersectionpoint(A--B, M--(M+rotate(90)*(B-A)) ); pair K=intersectionpoint(A--B, N--(N+rotate(90)*(B-A)) ); pair L=intersectionpoint(A--B, C--(C+rotate(90)*(B-A)) ); draw( A--B--C--cycle ); draw( M--J ); draw( N--K ); draw( C--L, dashed ); label("$A$",A,NW); label("$B$",B,SE); label("$C$",C,SW); label("$M$",M,SW); label("$N$",N,S); label("$J$",J,NE); label("$K$",K,NE); label("$L$",L,NE); [/asy]

Now note that the triangles $ABC$, $AMJ$, $ACL$, $CBL$ and $NBK$ all have the same angles and therefore they are similar. We already know some of their sides, and we will use this information to compute their areas. Note that if two polygons are similar with ratio $k$, their areas have ratio $k^2$. We will use this fact repeatedly. Below we will use $[XYZ]$ to denote the area of the triangle $XYZ$.

We have $\frac{CL}{BC} = \frac{60/13}{12} = \frac 5{13}$, hence $[ACL] = \frac{ 25[ABC] }{169} = \frac{750}{169}$.

Also, $\frac{CL}{AC} = \frac{60/13}5 = \frac{12}{13}$, hence $[CBL] = \frac{ 144[ABC] }{169} = \frac{4320}{169}$.

Now for the smaller triangles:

We know that $\frac{AM}{AC}=\frac 15$, hence $[AMJ] = \frac{[ACL]}{25} = \frac{30}{169}$.

Similarly, $\frac{BN}{BC}=\frac 8{12} = \frac 23$, hence $[NBK] = \frac{4[CBL]}9 = \frac{1920}{169}$.

Finally, the area of the pentagon is $30 - \frac{30}{169} - \frac{1920}{169} = \boxed{\frac{240}{13}}$.

Solution 1a

Observe that all of the triangles in the problem are right triangles and similar with a ratio of 5-12-13. The largest triangle $\Delta ABC$ has area $\dfrac{5\cdot12}2=30$. $\Delta AMJ$ is linearly scaled down from $\Delta ABC$ by a factor of $\dfrac{5-4}{13}=\dfrac1{13}$ (as we can see from comparing the two hypotenuses), while $\Delta NBK$ is scaled by a factor of $\dfrac{12-4}{13}=\dfrac8{13}$. The area we desire is the combined area of $\Delta AMJ$ and $\Delta NBK$ subtracted from the area of $\Delta ABC$, which is

\[30-30\left(\dfrac1{13}\right)^2-30\left(\dfrac8{13}\right)^2=30\left(1-\left(\dfrac1{13}\right)^2-\left(\dfrac8{13}\right)^2\right)=30\left(1-\dfrac{65}{169}\right)=30\left(\dfrac8{13}\right)=\boxed{\textbf{(D) }\dfrac{240}{13}}.\]

~Technodoggo

Solution 2

Split the pentagon along a different diagonal as follows:

[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,5), B=(12,0), M=(0,4), N=(4,0); pair J=intersectionpoint(A--B, M--(M+rotate(90)*(B-A)) ); pair K=intersectionpoint(A--B, N--(N+rotate(90)*(B-A)) ); pair L=intersectionpoint(A--B, C--(C+rotate(90)*(B-A)) ); draw( A--B--C--cycle ); draw( M--J ); draw( N--K ); draw( M--N, dashed ); label("$A$",A,NW); label("$B$",B,SE); label("$C$",C,SW); label("$M$",M,SW); label("$N$",N,S); label("$J$",J,NE); label("$K$",K,NE); label("$L$",L,NE); [/asy] The area of the pentagon is then the sum of the areas of the resulting right triangle and trapezoid. As before, triangles $ABC$, $AMJ$, and $NBK$ are all similar.

Since $BN=12-4=8$, $NK=\frac{5}{13}(8)=\frac{40}{13}$ and $BK=\frac{12}{13}(8)=\frac{96}{13}$. Since $AM=5-4=1$, $JM=\frac{12}{13}$ and $AJ=\frac{5}{13}$.

The trapezoid's height is therefore $13-\frac{5}{13}-\frac{96}{13}=\frac{68}{13}$, and its area is $\frac{1}{2}\left(\frac{68}{13}\right)\left(\frac{12}{13}+\frac{40}{13}\right)=\frac{34}{13}(4)=\frac{136}{13}$.

Triangle $MCN$ has area $\frac{1}{2}(4)(4)=8$, and the total area is $\frac{104+136}{13}=\boxed{\frac{240}{13}}$.

Solution 3

Right triangle $ABC$ has area $1/2 \cdot AC \cdot BC = 1/2 \cdot 5 \cdot 12 = 30$.

We are given that $AC = 5$ and $CM = 4$, so $AM = AC - CM = 1$. Right triangles $AMJ$ and $ABC$ are similar, with hypotenuses 1 and 13, respectively, so triangle $AMJ$ has area \[\left( \frac{1}{13} \right)^2 \cdot 30.\] Also, we are given that $BC = 12$ and $CN = 4$, so $BN = BC - CN = 8$. Right triangles $NBK$ and $ABC$ are similar, with hypotenuses 8 and 13, respectively, so triangle $NBK$ has area \[\left( \frac{8}{13} \right)^2 \cdot 30.\] Therefore, the area of pentagon $CMJKN$ is \[30 - \left( \frac{1}{13} \right)^2 \cdot 30 - \left( \frac{8}{13} \right)^2 \cdot 30 = \boxed{\frac{240}{13}}.\]

~math31415926535

Solution 4

Because triangle ABC, triangle NBK, and triangle AMJ are similar right triangles whose hypotenuses are in the ratio 13 : 8 : 1, their areas are in the ratio 169 : 64 : 1. The area of triangle ABC is 1/2 (12)(5) = 30, so the areas of triangle NBK and triangle AMJ are (64/169) (30) and (1/169)(30), respectively. Thus the area of pentagon CMJKN is (1 − 64/169 - 1/169)(30)= $\boxed{\mathbf{(D)}240/13}$

Credit to http://billingswest.billings.k12.mt.us/math/AMC%201012/AMC%2012%20work%20sheets/2004%20AMC%2012B%20ws-15.pdf for Solution 3.

Solution 5

We fake-solve. Draw a perpendicular line to side $CB$ that goes through $N$. Similarly, draw a perpendicular line to side $AC$ that goes through $M$. Let them intersect at $O$. It follows that quadrilateral $MONC$ is a square with side $4$ and thus area $16$. Now, draw a perpendicular from $J$ to $MO$ and let them intersect at $Z$. There is a obvious pair of congruent triangles. Fill in the gap. Hence, we see that the area of the pentagon is definitely greater than $16$. How much greater? Well, if we let the intersection of $AB$ and $NO$ be $X$, we can approximate that $\triangle NXK$ has an area greater than $2$. So we place bets on $\boxed{\mathbf{(D)}240/13}$ which is correct!

This would be much easier to show and more convincing with a diagram but idk how to upload GeoGebra to AoPSwiki.

Solution by franzliszt

See Also

2004 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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

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