Difference between revisions of "2004 AMC 12B Problems/Problem 14"
(→Problem) |
Heliootrope (talk | contribs) (added solution 2) |
||
Line 35: | Line 35: | ||
== Solution == | == Solution == | ||
+ | === 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. | ||
Line 75: | Line 76: | ||
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 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)) ); | ||
+ | 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 <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>. | ||
== 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}} |
Revision as of 22:01, 3 February 2013
Problem
In , , , and . Points and lie on and , respectively, with . Points and are on so that and are perpendicular to . What is the area of pentagon ?
Solution
Solution 1
The triangle is clearly a right triangle, its area is . If we knew the areas of triangles and , we could subtract them to get the area of the pentagon.
Draw the height from onto . As and the area is , we get . The situation is shown in the picture below:
Now note that the triangles , , , and 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 , their areas have ratio . We will use this fact repeatedly. Below we will use to denote the area of the triangle .
We have , hence .
Also, , hence .
Now for the smaller triangles:
We know that , hence .
Similarly, , hence .
Finally, the area of the pentagon is .
Solution 2
Split the pentagon along a different diagonal as follows:
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 ); 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); (Error compiling LaTeX. label("$L$",L,NE); ^ 0125ab1cc6f4632b43bedf8b6e324d871fc40be7.asy: 19.6: no matching function 'label(string, path(int k=<default>, pair A, pair B, real a=<default>, real b=<default>), pair)')
The area of the pentagon is then the sum of the areas of the resulting right triangle and trapezoid. As before, triangles , , and are all similar.
Since , and . Since , and .
The trapezoid's height is therefore , and its area is .
Triangle has area , and the total area is .
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 |