Difference between revisions of "2004 AMC 10A Problems/Problem 20"
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<math> \mathrm{(A) \ } \frac{4}{3} \qquad \mathrm{(B) \ } \frac{3}{2} \qquad \mathrm{(C) \ } \sqrt{3} \qquad \mathrm{(D) \ } 2 \qquad \mathrm{(E) \ } 1+\sqrt{3} </math> | <math> \mathrm{(A) \ } \frac{4}{3} \qquad \mathrm{(B) \ } \frac{3}{2} \qquad \mathrm{(C) \ } \sqrt{3} \qquad \mathrm{(D) \ } 2 \qquad \mathrm{(E) \ } 1+\sqrt{3} </math> | ||
− | ==Solution | + | ==Solution 1== |
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Since triangle <math>BEF</math> is equilateral, <math>EA=FC</math>, and <math>EAB</math> and <math>FCB</math> are <math>SAS</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> | Since triangle <math>BEF</math> is equilateral, <math>EA=FC</math>, and <math>EAB</math> and <math>FCB</math> are <math>SAS</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> | ||
− | ==Solution | + | ==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>\frac{\frac{x^2}{2}}{\frac{1-x}{2}} = \frac{x^2}{1 - x} = \boxed{\text{(D) }2}.</cmath> | <cmath>\frac{\frac{x^2}{2}}{\frac{1-x}{2}} = \frac{x^2}{1 - x} = \boxed{\text{(D) }2}.</cmath> | ||
− | ==Solution | + | ==Solution 3== |
<math>\bigtriangleup BEF</math> is equilateral, so <math>\angle EBF = 60^{\circ}</math>, and <math>\angle EBA = \angle FBC</math> so they must each be <math>15^{\circ}</math>. Then let <math>BE=EF=FB=1</math>, which gives <math>EA=\sin{15^{\circ}}</math> and <math>AB=\cos{15^{\circ}}</math>. | <math>\bigtriangleup BEF</math> is equilateral, so <math>\angle EBF = 60^{\circ}</math>, and <math>\angle EBA = \angle FBC</math> so they must each be <math>15^{\circ}</math>. Then let <math>BE=EF=FB=1</math>, which gives <math>EA=\sin{15^{\circ}}</math> and <math>AB=\cos{15^{\circ}}</math>. | ||
The area of <math>\bigtriangleup ABE</math> is then <math>\frac{1}{2}\sin{15^{\circ}}\cos{15^{\circ}}=\frac{1}{4}\sin{30^{\circ}}=\frac{1}{8}</math>. | The area of <math>\bigtriangleup ABE</math> is then <math>\frac{1}{2}\sin{15^{\circ}}\cos{15^{\circ}}=\frac{1}{4}\sin{30^{\circ}}=\frac{1}{8}</math>. | ||
<math>\bigtriangleup DEF</math> is an isosceles right triangle with hypotenuse 1, so <math>DE=DF=\frac{1}{\sqrt{2}}</math> and therefore its area is <math>\frac{1}{2}\left(\frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}}\right)=\frac{1}{4}</math>. | <math>\bigtriangleup DEF</math> is an isosceles right triangle with hypotenuse 1, so <math>DE=DF=\frac{1}{\sqrt{2}}</math> and therefore its area is <math>\frac{1}{2}\left(\frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}}\right)=\frac{1}{4}</math>. | ||
The ratio of areas is then <math>\frac{\frac{1}{4}}{\frac{1}{8}}=\framebox{(D) 2}</math> | The ratio of areas is then <math>\frac{\frac{1}{4}}{\frac{1}{8}}=\framebox{(D) 2}</math> | ||
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+ | ==Solution 4(system of equations)== | ||
+ | Assume AB=1 then FC is x ED is <math>1-x</math> then we see that using HL FCB is congruent is EAB. Using Pythagoras of triangles FCB and FDE 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> then <math>x=2-\sqrt{3}</math> then using the areas we get the answer to be D | ||
==Solution 5== | ==Solution 5== | ||
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 2. Choice D | 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 2. Choice D |
Revision as of 19:38, 1 August 2020
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
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
is equilateral, so , and so they must each be . Then let , which gives and . The area of is then . is an isosceles right triangle with hypotenuse 1, so and therefore its area is . The ratio of areas is then
Solution 4(system of equations)
Assume AB=1 then FC is x ED is then we see that using HL FCB is congruent is EAB. Using Pythagoras of triangles FCB and FDE we get . Expanding we get . Simplifying gives solving using completing the square(or other methods) gives 2 answers and because then then using the areas we get the answer to be D
Solution 5
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 2. Choice D