Difference between revisions of "2004 AMC 10A Problems/Problem 20"
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==Solution 1== | ==Solution 1== | ||
− | Since triangle <math>BEF</math> is equilateral, <math>EA=FC</math>, and <math>EAB</math> and <math>FCB</math> are <math> | + | Since triangle <math>BEF</math> is equilateral, <math>EA=FC</math>, and <math>EAB</math> and <math>FCB</math> are <math>HL</math> congruent. Thus, triangle <math>DEF</math> is an isosceles right triangle. So we let <math>DE=x</math>. Thus <math>EF=EB=FB=x\sqrt{2}</math>. If we go angle chasing, we find out that <math>\angle AEB=75^{\circ}</math>, thus <math>\angle ABE=15^{\circ}</math>. <math>\frac{AE}{EB}=\sin{15^{\circ}}=\frac{\sqrt{6}-\sqrt{2}}{4}</math>. Thus <math>\frac{AE}{x\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{4}</math>, or <math>AE=\frac{x(\sqrt{3}-1)}{2}</math>. Thus <math>AB=\frac{x(\sqrt{3}+1)}{2}</math>, and <math>[ABE]=\frac{x^2}{4}</math>, and <math>[DEF]=\frac{x^2}{2}</math>. Thus the ratio of the areas is <math>\boxed{\mathrm{(D)}\ 2}</math> |
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==Solution 2 (Non-trig) == | ==Solution 2 (Non-trig) == | ||
WLOG, let the side length of <math>ABCD</math> be 1. Let <math>DE = x</math>. It suffices that <math>AE = 1 - x</math>. Then triangles <math>ABE</math> and <math>CBF</math> are congruent by HL, so <math>CF = AE</math> and <math>DE = DF</math>. We find that <math>BE = EF = x \sqrt{2}</math>, and so, by the Pythagorean Theorem, we have | WLOG, let the side length of <math>ABCD</math> be 1. Let <math>DE = x</math>. It suffices that <math>AE = 1 - x</math>. Then triangles <math>ABE</math> and <math>CBF</math> are congruent by HL, so <math>CF = AE</math> and <math>DE = DF</math>. We find that <math>BE = EF = x \sqrt{2}</math>, and so, by the Pythagorean Theorem, we have | ||
<math>(1 - x)^2 + 1 = 2x^2.</math> This yields <math>x^2 + 2x = 2</math>, so <math>x^2 = 2 - 2x</math>. Thus, the desired ratio of areas is | <math>(1 - x)^2 + 1 = 2x^2.</math> This yields <math>x^2 + 2x = 2</math>, so <math>x^2 = 2 - 2x</math>. Thus, the desired ratio of areas is | ||
− | <cmath>\ | + | <cmath>\dfrac{\dfrac{x^2}{2}}{\dfrac{1-x}{2}} = \dfrac{x^2}{1 - x} = \boxed{\text{(D) }2}.</cmath> |
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− | ==Solution | + | ==Solution 3 (System of Equations)== |
Assume <math>AB=1</math>. Then, <math>FC</math> is <math>x</math> and <math>ED</math> is <math>1-x</math>. We see that using <math>HL</math>, <math>FCB</math> is congruent to EAB. Using Pythagoras of triangles <math>FCB</math> and <math>FDE</math> we get <math>2{(1-x)}^2=x^2+1</math>. Expanding, we get <math>2x^2-4x+2=x^2+1</math>. Simplifying gives <math>x^2-4x+1=0</math> solving using completing the square (or other methods) gives 2 answers: <math>2-\sqrt{3}</math> and <math>2+\sqrt{3}</math>. Because <math>x < 1</math>, <math>x=2-\sqrt{3}</math>. Using the areas, the answer is <math>\boxed{\text{(D) }2}</math> | Assume <math>AB=1</math>. Then, <math>FC</math> is <math>x</math> and <math>ED</math> is <math>1-x</math>. We see that using <math>HL</math>, <math>FCB</math> is congruent to EAB. Using Pythagoras of triangles <math>FCB</math> and <math>FDE</math> we get <math>2{(1-x)}^2=x^2+1</math>. Expanding, we get <math>2x^2-4x+2=x^2+1</math>. Simplifying gives <math>x^2-4x+1=0</math> solving using completing the square (or other methods) gives 2 answers: <math>2-\sqrt{3}</math> and <math>2+\sqrt{3}</math>. Because <math>x < 1</math>, <math>x=2-\sqrt{3}</math>. Using the areas, the answer is <math>\boxed{\text{(D) }2}</math> | ||
− | ==Solution | + | ==Solution 4== |
− | First, since <math>\bigtriangleup BEF</math> is equilateral and <math>ABCD</math> is a square, by the Hypothenuse Leg Theorem, <math>\bigtriangleup ABE</math> is congruent to <math>\bigtriangleup CBF</math>. Then, assume length <math>AB = BC = x</math> and length <math>DE = DF = y</math>, then <math>AE = FC = x - y</math>. <math>\bigtriangleup BEF</math> is equilateral, so <math>EF = EB</math> and <math>EB^2 = EF^2</math>, it is given that <math>ABCD</math> is a square and <math>\bigtriangleup DEF</math> and <math>\bigtriangleup ABE</math> are right triangles. Then we use the Pythagorean theorem to prove that <math>AB^2 + AE^2 = EB^2</math> and since we know that <math>EB^2 = EF^2</math> and <math>EF^2 = DE^2 + DF^2</math>, which means <math>AB^2 + AE^2 = DE^2 + DF^2</math>. Now we plug in the variables and the equation becomes <math>x^2 + (x | + | First, since <math>\bigtriangleup BEF</math> is equilateral and <math>ABCD</math> is a square, by the Hypothenuse Leg Theorem, <math>\bigtriangleup ABE</math> is congruent to <math>\bigtriangleup CBF</math>. Then, assume length <math>AB = BC = x</math> and length <math>DE = DF = y</math>, then <math>AE = FC = x - y</math>. <math>\bigtriangleup BEF</math> is equilateral, so <math>EF = EB</math> and <math>EB^2 = EF^2</math>, it is given that <math>ABCD</math> is a square and <math>\bigtriangleup DEF</math> and <math>\bigtriangleup ABE</math> are right triangles. Then we use the Pythagorean theorem to prove that <math>AB^2 + AE^2 = EB^2</math> and since we know that <math>EB^2 = EF^2</math> and <math>EF^2 = DE^2 + DF^2</math>, which means <math>AB^2 + AE^2 = DE^2 + DF^2</math>. Now we plug in the variables and the equation becomes <math>x^2 + (x-y)^2 = 2y^2</math>, expand and simplify and you get <math>2x^2 - 2xy = y^2</math>. We want the ratio of area of <math>\bigtriangleup DEF</math> to <math>\bigtriangleup ABE</math>. Expressed in our variables, the ratio of the area is <math>\frac{y^2}{x^2 - xy}</math> and we know <math>2x^2 - 2xy = y^2</math>, so the ratio must be <math>2</math>. So, the answer is <math>\boxed{\text{(D) }2}</math> |
==Video Solution== | ==Video Solution== |
Latest revision as of 12:59, 1 November 2024
Contents
Problem
Points and are located on square so that is equilateral. What is the ratio of the area of to that of ?
Solution 1
Since triangle is equilateral, , and and are congruent. Thus, triangle is an isosceles right triangle. So we let . Thus . If we go angle chasing, we find out that , thus . . Thus , or . Thus , and , and . Thus the ratio of the areas is
z
Solution 2 (Non-trig)
WLOG, let the side length of be 1. Let . It suffices that . Then triangles and are congruent by HL, so and . We find that , and so, by the Pythagorean Theorem, we have This yields , so . Thus, the desired ratio of areas is
Solution 3 (System of Equations)
Assume . Then, is and is . We see that using , is congruent to EAB. Using Pythagoras of triangles and we get . Expanding, we get . Simplifying gives solving using completing the square (or other methods) gives 2 answers: and . Because , . Using the areas, the answer is
Solution 4
First, since is equilateral and is a square, by the Hypothenuse Leg Theorem, is congruent to . Then, assume length and length , then . is equilateral, so and , it is given that is a square and and are right triangles. Then we use the Pythagorean theorem to prove that and since we know that and , which means . Now we plug in the variables and the equation becomes , expand and simplify and you get . We want the ratio of area of to . Expressed in our variables, the ratio of the area is and we know , so the ratio must be . So, the answer is
Video Solution
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Video Solution by TheBeautyofMath
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See also
2004 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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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