Difference between revisions of "2006 AMC 10A Problems/Problem 17"
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~made by AoPS (somewhere) -put here by qkddud~ | ~made by AoPS (somewhere) -put here by qkddud~ | ||
+ | |||
+ | == Solution 5 (Proof Bash) == | ||
+ | |||
+ | <asy> | ||
+ | size(7cm); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10); | ||
+ | pair A,B,C,D,E,F,G,H,W,X,Y,Z; | ||
+ | A=(0,2); B=(1,2); C=(2,2); D=(3,2); | ||
+ | H=(0,0); G=(1,0); F=(2,0); E=(3,0); | ||
+ | D('A',A, N); D('B',B,N); D('C',C,N); D('D',D,N); D('E',E,SE); D('F',F,SE); D('G',G,SW); D('H',H,SW); | ||
+ | D(A--F); D(B--E); D(D--G); D(C--H); | ||
+ | Z=IP(A--F, C--H); Y=IP(A--F, D--G); X=IP(B--E,D--G); W=IP(B--E,C--H); | ||
+ | D('W',W,1.6*N); D('X',X,1.6*plain.E); D('Y',Y,1.6*S); D('Z',Z,1.6*plain.W); | ||
+ | D(A--D--E--H--cycle); | ||
+ | </asy> | ||
+ | |||
+ | |||
+ | By symmetry, quadrilateral <math>WXYZ</math> is a square. | ||
+ | |||
+ | First step, proving that <math>\triangle BXD \sim \triangle BWC</math>. | ||
+ | |||
+ | We can tell quadrilateral <math>CDGH</math> is a parallelogram because <math>CD \parallel GH</math> and <math>CD \cong GH</math>. | ||
+ | |||
+ | By knowing that, we can say that <math>CW \parallel DX</math>. | ||
+ | |||
+ | Finally, we can now prove <math>\triangle BXD \sim \triangle BWC</math> by AA, with a ratio of 2:1. | ||
+ | |||
+ | Since <math>BD = DE = 2</math> and <math>\angle BDE = 90</math>. Then <math>\triangle BDE</math> is a 45-45-90 triangle. | ||
+ | |||
+ | This will make <math>\angle DBE = 45</math> making <math>\triangle BXD</math> and <math>\triangle BWC</math> a 45-45-90 triangle. | ||
+ | |||
+ | This will make, <math>BD = 2, BC=1, DX=BX=\sqrt 2, BW=WX=\frac{\sqrt 2}{2}</math>. Since <math>WX</math> is the length of the square, our answer will be <math>(\frac{\sqrt 2}{2})^2 = \frac{2}{4} = \boxed{\textbf{(A) }\frac{1}{2}}.</math> | ||
+ | |||
+ | ~ghfhgvghj10 | ||
==Video Solution by the Beauty of Math== | ==Video Solution by the Beauty of Math== |
Latest revision as of 23:21, 1 February 2023
Contents
Problem
In rectangle , points and trisect , and points and trisect . In addition, , and . What is the area of quadrilateral shown in the figure?
Solution 1
By symmetry, is a square.
Draw . , so is a . Hence , and .
There are many different similar ways to come to the same conclusion using different 45-45-90 triangles.
Solution 2
Drawing lines as shown above and piecing together the triangles, we see that is made up of squares congruent to . Hence .
Solution 3
We see that if we draw a line to it is half the width of the rectangle so that length would be , and the resulting triangle is a so using the Pythagorean Theorem we can get that each side is so the area of the middle square would be which is our answer.
Solution 4
Since and are trisection points and , we see that . Also, , so triangle is a right isosceles triangle, i.e. . By symmetry, triangles , , and are also right isosceles triangles. Therefore, , which means triangle is also a right isosceles triangle. Also, triangle is a right isosceles triangle.
Then , and . Hence, .
By symmetry, quadrilateral is a square, so its area is
~made by AoPS (somewhere) -put here by qkddud~
Solution 5 (Proof Bash)
By symmetry, quadrilateral is a square.
First step, proving that .
We can tell quadrilateral is a parallelogram because and .
By knowing that, we can say that .
Finally, we can now prove by AA, with a ratio of 2:1.
Since and . Then is a 45-45-90 triangle.
This will make making and a 45-45-90 triangle.
This will make, . Since is the length of the square, our answer will be
~ghfhgvghj10
Video Solution by the Beauty of Math
https://www.youtube.com/watch?v=GX33rxlJz7s
~IceMatrix
See Also
2006 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.