Difference between revisions of "2004 AMC 10A Problems/Problem 20"
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Points <math>E</math> and <math>F</math> are located on square <math>ABCD</math> so that <math>\triangle BEF</math> is equilateral. What is the ratio of the area of <math>\triangle DEF</math> to that of <math>\triangle ABE</math>? | Points <math>E</math> and <math>F</math> are located on square <math>ABCD</math> so that <math>\triangle BEF</math> is equilateral. What is the ratio of the area of <math>\triangle DEF</math> to that of <math>\triangle ABE</math>? | ||
− | <center> | + | <center> |
+ | <asy> | ||
+ | unitsize(3 cm); | ||
+ | |||
+ | pair A, B, C, D, E, F; | ||
+ | |||
+ | A = (0,0); | ||
+ | B = (1,0); | ||
+ | C = (1,1); | ||
+ | D = (0,1); | ||
+ | E = (0,Tan(15)); | ||
+ | F = (1 - Tan(15),1); | ||
+ | |||
+ | draw(A--B--C--D--cycle); | ||
+ | draw(B--E--F--cycle); | ||
+ | |||
+ | label("$A$", A, SW); | ||
+ | label("$B$", B, SE); | ||
+ | label("$C$", C, NE); | ||
+ | label("$D$", D, NW); | ||
+ | label("$E$", E, W); | ||
+ | label("$F$", F, N); | ||
+ | </asy> | ||
+ | </center> | ||
<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== |
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 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 | |
<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} = 2.</cmath> | + | <cmath>\frac{\frac{x^2}{2}}{\frac{1-x}{2}} = \frac{x^2}{1 - x} = \boxed{\text{(D) }2}.</cmath> |
+ | |||
+ | ==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>. | ||
+ | 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>. | ||
+ | The ratio of areas is then <math>\frac{\frac{1}{4}}{\frac{1}{8}}=\framebox{(D) 2}</math> | ||
+ | |||
+ | ==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== | ||
+ | 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 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/hwHIHRukYMk | ||
+ | |||
+ | Education, the Study of Everything | ||
+ | |||
+ | ==Video Solution by TheBeautyofMath== | ||
+ | https://youtu.be/BFKo9h8GhLY | ||
+ | |||
+ | ~IceMatrix | ||
==See also== | ==See also== | ||
− | |||
{{AMC10 box|year=2004|ab=A|num-b=19|num-a=21}} | {{AMC10 box|year=2004|ab=A|num-b=19|num-a=21}} | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
− | |||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 21:30, 29 August 2021
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
Video Solution
Education, the Study of Everything
Video Solution by TheBeautyofMath
~IceMatrix
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.