Difference between revisions of "2006 AMC 10A Problems/Problem 17"
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==Solution 3== | ==Solution 3== | ||
We see that if we draw a line to <math>BZ</math> it is half the width of the rectangle so that length would be <math>1</math>, and the resulting triangle is a <math>45-45-90</math> so using the Pythagorean Theorem we can get that each side is <math>\sqrt{\frac{1^2}{2}}</math> so the area of the middle square would be <math>(\sqrt{\frac{1^2}{2}})^2=(\sqrt{\frac{1}{2}})^2=\frac{1}{2}</math> which is our answer. | We see that if we draw a line to <math>BZ</math> it is half the width of the rectangle so that length would be <math>1</math>, and the resulting triangle is a <math>45-45-90</math> so using the Pythagorean Theorem we can get that each side is <math>\sqrt{\frac{1^2}{2}}</math> so the area of the middle square would be <math>(\sqrt{\frac{1^2}{2}})^2=(\sqrt{\frac{1}{2}})^2=\frac{1}{2}</math> which is our answer. | ||
+ | ==Solution 4== | ||
+ | Solution: | ||
+ | Since <math>B</math> and <math>C</math> are trisection points and <math>AC = 2</math>, we see that <math>AD = 3</math>. Also, <math>AC = AH</math>, so triangle <math>ACH</math> is a right isosceles triangle, i.e. <math>\angle ACH = \angle AHC = 45^\circ</math>. By symmetry, triangles <math>AFH</math>, <math>DEG</math>, and <math>BED</math> are also right isosceles triangles. Therefore, <math>\angle WAD = \angle WDA = 45^\circ</math>, which means triangle <math>AWD</math> is also a right isosceles triangle. Also, triangle <math>AXC</math> is a right isosceles triangle. | ||
+ | |||
+ | Then <math>AW = AD/\sqrt{2} = 3/\sqrt{2}</math>, and <math>AX = AC/\sqrt{2} = 2/\sqrt{2}</math>. Hence, <math>XW = AW - AX = 3/\sqrt{2} - 2/\sqrt{2} = 1/\sqrt{2}</math>. | ||
+ | |||
+ | By symmetry, quadrilateral <math>WXYZ</math> is a square, so its area is | ||
+ | <cmath>XW^2 = \left( \frac{1}{\sqrt{2}} \right)^2 = \boxed{\frac{1}{2}}.</cmath> | ||
+ | |||
+ | ~made by AoPS HW-put here by qkddud~ | ||
== See Also == | == See Also == |
Revision as of 21:14, 6 August 2019
Problem
In rectangle , points and trisect , and points and trisect . In addition, , and . What is the area of quadrilateral shown in the figure?
Solution
Solution 1
It is not difficult to see by symmetry that is a square. Draw . Clearly . Then is isosceles, and is a . Hence , and .
There are many different similar ways to come to the same conclusion using different 45-45-90 triangles.
Solution 2
Draw the lines as shown above, and count the squares. There are 12, so we have .
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
Solution: 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 HW-put here by qkddud~
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 |
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