Difference between revisions of "2017 AMC 10B Problems/Problem 19"
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==Problem== | ==Problem== | ||
− | Let <math>ABC</math> be an equilateral triangle. Extend side <math>\overline{AB}</math> beyond <math>B</math> to a point <math>B'</math> so that <math>BB'= | + | Let <math>ABC</math> be an equilateral triangle. Extend side <math>\overline{AB}</math> beyond <math>B</math> to a point <math>B'</math> so that <math>BB'=3 \cdot AB</math>. Similarly, extend side <math>\overline{BC}</math> beyond <math>C</math> to a point <math>C'</math> so that <math>CC'=3 \cdot BC</math>, and extend side <math>\overline{CA}</math> beyond <math>A</math> to a point <math>A'</math> so that <math>AA'=3 \cdot CA</math>. What is the ratio of the area of <math>\triangle A'B'C'</math> to the area of <math>\triangle ABC</math>? |
<math>\textbf{(A)}\ 9:1\qquad\textbf{(B)}\ 16:1\qquad\textbf{(C)}\ 25:1\qquad\textbf{(D)}\ 36:1\qquad\textbf{(E)}\ 37:1</math> | <math>\textbf{(A)}\ 9:1\qquad\textbf{(B)}\ 16:1\qquad\textbf{(C)}\ 25:1\qquad\textbf{(D)}\ 36:1\qquad\textbf{(E)}\ 37:1</math> | ||
− | |||
− | ===Solution 1=== | + | == Solution == |
− | Note that by symmetry, <math>\triangle A'B'C'</math> is also equilateral. Therefore, we only need to find one of the sides of <math>A'B'C'</math> to determine the area ratio. WLOG, let <math>AB = BC = CA = 1</math>. Therefore, <math>BB' = 3</math> and <math>BC' = 4</math>. Also, <math>\angle B'BC' = 120^{\circ}</math>, so by the Law of Cosines, <math>B'C' = \sqrt{37}</math>. Therefore, the answer is <math>(\sqrt{37})^2 : 1^2 = \boxed{\textbf{(E) } 37 : 1}</math> | + | <asy> |
+ | size(12cm); | ||
+ | dot((0,0)); | ||
+ | dot((2,0)); | ||
+ | dot((1,1.732)); | ||
+ | dot((-6,0)); | ||
+ | dot((5,-5.196)); | ||
+ | dot((4,6.928)); | ||
+ | draw((0,0)--(2,0)--(1,1.732)--cycle); | ||
+ | draw((-6,0)--(5,-5.196)--(4,6.928)--cycle); | ||
+ | draw((-6,0)--(0,0)); | ||
+ | draw((5,-5.196)--(2,0)); | ||
+ | draw((4,6.928)--(1,1.732)); | ||
+ | |||
+ | </asy> | ||
+ | ===Solution 1 (Uses Trig) === | ||
+ | Note that by symmetry, <math>\triangle A'B'C'</math> is also equilateral. Therefore, we only need to find one of the sides of <math>A'B'C'</math> to determine the area ratio. WLOG, let <math>AB = BC = CA = 1</math>. Therefore, <math>BB' = 3</math> and <math>BC' = 4</math>. Also, <math>\angle B'BC' = 120^{\circ}</math>, so by the Law of Cosines, <math>B'C' = \sqrt{37}</math>. Therefore, the answer is <math>(\sqrt{37})^2 : 1^2 = \boxed{\textbf{(E) } 37:1}</math> | ||
===Solution 2 === | ===Solution 2 === | ||
− | As mentioned in the first solution, <math>\triangle A'B'C'</math> is equilateral. WLOG, let <math>AB=2</math>. Let <math>D</math> be on the line passing through <math>AB</math> such that <math>A'D</math> is perpendicular to <math>AB</math>. Note that <math>\triangle A'DA</math> is a 30-60-90 with right angle at <math>D</math>. Since <math>AA'=6</math>, <math>AD=3</math> and <math>A'D=3\sqrt{3}</math>. So we know that <math>DB'=11</math>. Note that <math>\triangle A'DB'</math> is a right triangle with right angle at <math>D</math>. So by the Pythagorean theorem, we find <math>A'B'= \sqrt{(3\sqrt{3})^2 + 11^2} = 2\sqrt{37}.</math> Therefore, the answer is <math>(2\sqrt{37})^2 : 2^2 = \boxed{\textbf{(E) } 37 : 1}</math>. | + | As mentioned in the first solution, <math>\triangle A'B'C'</math> is equilateral. WLOG, let <math>AB=2</math>. Let <math>D</math> be on the line passing through <math>AB</math> such that <math>A'D</math> is perpendicular to <math>AB</math>. Note that <math>\triangle A'DA</math> is a <math>30-60-90</math> with right angle at <math>D</math>. Since <math>AA'=6</math>, <math>AD=3</math> and <math>A'D=3\sqrt{3}</math>. So we know that <math>DB'=11</math>. Note that <math>\triangle A'DB'</math> is a right triangle with right angle at <math>D</math>. So by the Pythagorean theorem, we find <math>A'B'= \sqrt{(3\sqrt{3})^2 + 11^2} = 2\sqrt{37}.</math> Therefore, the answer is <math>(2\sqrt{37})^2 : 2^2 = \boxed{\textbf{(E) } 37:1}</math>. |
+ | |||
+ | ===Solution 3 === | ||
+ | Let <math>AB=BC=CA=x</math>. We start by noting that we can just write <math>AB'</math> as just <math>AB+BB'=4AB</math>. | ||
+ | Similarly <math>BC'=4BC</math>, and <math>CA'=4CA</math>. We can evaluate the area of triangle <math>ABC</math> by simply using Heron's formula, | ||
+ | <math>[ABC]=\sqrt{\frac{3x}{2}\cdot {\Bigg(\frac{3x}{2}-x\Bigg)}^3}=\frac{x^2\sqrt{3}}{4}</math>. | ||
+ | Next in order to evaluate <math>A'B'C'</math> we need to evaluate the area of the larger triangles <math>AA'B',BB'C', \text{ and } CC'A'</math>. | ||
+ | In this solution we shall just compute <math>1</math> of these as the others are trivially equivalent. | ||
+ | In order to compute the area of <math>\Delta{AA'B'}</math> we can use the formula <math>[XYZ]=\frac{1}{2}xy\cdot\sin{z}</math>. | ||
+ | Since <math>ABC</math> is equilateral and <math>A</math>, <math>B</math>, <math>B'</math> are collinear, we already know <math>\angle{A'AB'}=180-60=120</math> | ||
+ | Similarly from above we know <math>AB'</math> and <math>A'A</math> to be <math>4x</math>, and <math>3x</math> respectively. Thus the area of <math>\Delta{AA'B'}</math> is <math>\frac{1}{2}\cdot 4x\cdot 3x \cdot \sin{120}=3x^2\cdot\sqrt{3}</math>. Likewise we can find <math>BB'C', \text{ and } CC'A'</math> to also be <math>3x^2\cdot\sqrt{3}</math>. | ||
+ | <math>[A'B'C']=[AA'B']+[BB'C']+[CC'A']+[ABC]=3\cdot3x^2\cdot\sqrt{3}+\frac{x^2\sqrt{3}}{4}=\sqrt{3}\cdot\Bigg(9x^2+\frac{x^2}{4}\Bigg)</math>. | ||
+ | Therefore the ratio of <math>[A'B'C']</math> to <math>[ABC]</math> is <math>\frac{\sqrt{3}\cdot\Bigg(9x^2+\frac{x^2}{4}\Bigg)}{\frac{x^2\sqrt{3}}{4}}=\boxed{\textbf{(E) } 37:1}</math> | ||
+ | |||
+ | ===Solution 4 (Elimination) === | ||
+ | Looking at the answer choices, we see that all but <math>{\textbf{(E)}}</math> has a perfect square in the ratio. With some intuition, we can guess that the sidelength of the new triangle formed is not an integer, thus we pick <math>\boxed{\textbf{(E) } 37:1}</math>. | ||
+ | |||
+ | Solution by sp1729 | ||
+ | |||
+ | ===Solution 5 (Barycentric Coordinates) === | ||
+ | We use barycentric coordinates wrt <math>\triangle ABC</math>, to which we can easily obtain that <math>A'=(4,0,-3)</math>, <math>B'=(-3,4,0)</math>, and <math>C'=(0,-3,4)</math>. Now, since the coordinates are homogenized (<math>-3+4=1</math>), we can directly apply the area formula to obtain that | ||
+ | <cmath>[A'B'C']=[ABC]\cdot\left| \begin{array}{ccc} 4 & 0 & -3 \\ -3 & 4 & 0 \\ 0 & -3 & 4 \end{array} \right| = (64-27)[ABC]=37[ABC],</cmath> | ||
+ | so the answer is <math>\boxed{\textbf{(E) } 37:1}</math> | ||
+ | |||
+ | ===Solution 6 (Area Comparison) === | ||
+ | First, comparing bases yields that <math>[BA'B']=3[AA'B]=9[ABC]\implies [AA'B']=12</math>. By congruent triangles, | ||
+ | <cmath>[AA'B']=[BB'C']=[CC'A']\implies [A'B'C']=(12+12+12+1)[ABC],</cmath> | ||
+ | so <math>[A'B'C']:[ABC]=\boxed{\textbf{(E) } 37:1}</math> | ||
+ | |||
+ | === Solution 7 (Quick Proportionality) === | ||
+ | Scale down the figure so that the area formulas for the <math>120^\circ</math> and equilateral triangles become proportional with proportionality constant equivalent to the product of the corresponding sides. By the proportionality, it becomes clear that the answer is <math>3\cdot4\cdot3+1\cdot1=37, \boxed{\text{E}}</math>. | ||
+ | |||
+ | ~ solution by mathchampion1 | ||
+ | |||
+ | ===Solution 8 (Sin area formula) === | ||
+ | Drawing the diagram, we see that the large triangle, <math>A'B'C'</math>, is composed of three congruent triangles with the triangle <math>ABC</math> at the center. Let each of the sides of triangle <math>ABC</math> be <math>x</math>. Therefore, using the equilateral triangle area formula, the <math>[ABC] = \frac{x^2\sqrt{3}}{4}</math>. We also know now that the sides of the triangles are <math>3x</math> and <math>3x + x</math>, or <math>4x</math>. We also know that since <math>BB'</math> are collinear, as are the others, angle <math>C'BB'</math> is <math>180 - 60</math>, which is <math>120</math> degrees. Because that angle is an included angle, we get the area of all three congruent triangle's are <math>\frac{12x^2\sin120}{2} \cdot 3</math>. Simplifying that yields <math>\frac{36x^2\sqrt{3}}{4}</math>. Adding that to the <math>[ABC]</math> yields <math>\frac{37x^2\sqrt{3}}{4}</math>. From this, we can compare the ratios by canceling everything out except for the <math>37</math>, so the answer is <math>\boxed{\textbf{(E) }37:1}</math> | ||
+ | |||
+ | ~Solution by EricShi1685 | ||
+ | |||
+ | |||
+ | ===Solution 9 (Same as Solution 8 but faster)=== | ||
+ | WLOG, let the side length of the smaller triangle be 1. The area of the big portion (A'B'A) is then <math>\frac{1}{2}\cdot3\cdot4\cdot\sin\left(120\right)=\frac{1}{2}\cdot3\cdot4\cdot\sin\left(60\right)=\frac{1}{2}\cdot3\cdot4\cdot\frac{\sqrt{3}}{2}=3\sqrt{3}</math>. Now simply multiply by three and add <math>\frac{\sqrt{3}}{4}</math> (the area of the small triangle) we get <math>\frac{37\sqrt{3}}{4}</math> and so the ratio is <math>37:1</math>. <math>\boxed{\textbf{(E) }37:1}</math> | ||
+ | |||
+ | == Video Solution (Law of Cosines)== | ||
+ | https://youtu.be/4_x1sgcQCp4?t=5373 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | == Video Solution (Meta-Solving Technique) == | ||
+ | https://youtu.be/GmUWIXXf_uk?t=710 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2017|ab=B|num-b=18|num-a=20}} | {{AMC10 box|year=2017|ab=B|num-b=18|num-a=20}} | ||
{{AMC12 box|year=2017|ab=B|num-b=14|num-a=16}} | {{AMC12 box|year=2017|ab=B|num-b=14|num-a=16}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 07:06, 4 November 2021
Contents
Problem
Let be an equilateral triangle. Extend side beyond to a point so that . Similarly, extend side beyond to a point so that , and extend side beyond to a point so that . What is the ratio of the area of to the area of ?
Solution
Solution 1 (Uses Trig)
Note that by symmetry, is also equilateral. Therefore, we only need to find one of the sides of to determine the area ratio. WLOG, let . Therefore, and . Also, , so by the Law of Cosines, . Therefore, the answer is
Solution 2
As mentioned in the first solution, is equilateral. WLOG, let . Let be on the line passing through such that is perpendicular to . Note that is a with right angle at . Since , and . So we know that . Note that is a right triangle with right angle at . So by the Pythagorean theorem, we find Therefore, the answer is .
Solution 3
Let . We start by noting that we can just write as just . Similarly , and . We can evaluate the area of triangle by simply using Heron's formula, . Next in order to evaluate we need to evaluate the area of the larger triangles . In this solution we shall just compute of these as the others are trivially equivalent. In order to compute the area of we can use the formula . Since is equilateral and , , are collinear, we already know Similarly from above we know and to be , and respectively. Thus the area of is . Likewise we can find to also be . . Therefore the ratio of to is
Solution 4 (Elimination)
Looking at the answer choices, we see that all but has a perfect square in the ratio. With some intuition, we can guess that the sidelength of the new triangle formed is not an integer, thus we pick .
Solution by sp1729
Solution 5 (Barycentric Coordinates)
We use barycentric coordinates wrt , to which we can easily obtain that , , and . Now, since the coordinates are homogenized (), we can directly apply the area formula to obtain that so the answer is
Solution 6 (Area Comparison)
First, comparing bases yields that . By congruent triangles, so
Solution 7 (Quick Proportionality)
Scale down the figure so that the area formulas for the and equilateral triangles become proportional with proportionality constant equivalent to the product of the corresponding sides. By the proportionality, it becomes clear that the answer is .
~ solution by mathchampion1
Solution 8 (Sin area formula)
Drawing the diagram, we see that the large triangle, , is composed of three congruent triangles with the triangle at the center. Let each of the sides of triangle be . Therefore, using the equilateral triangle area formula, the . We also know now that the sides of the triangles are and , or . We also know that since are collinear, as are the others, angle is , which is degrees. Because that angle is an included angle, we get the area of all three congruent triangle's are . Simplifying that yields . Adding that to the yields . From this, we can compare the ratios by canceling everything out except for the , so the answer is
~Solution by EricShi1685
Solution 9 (Same as Solution 8 but faster)
WLOG, let the side length of the smaller triangle be 1. The area of the big portion (A'B'A) is then . Now simply multiply by three and add (the area of the small triangle) we get and so the ratio is .
Video Solution (Law of Cosines)
https://youtu.be/4_x1sgcQCp4?t=5373
~ pi_is_3.14
Video Solution (Meta-Solving Technique)
https://youtu.be/GmUWIXXf_uk?t=710
~ pi_is_3.14
See Also
2017 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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 |
2017 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.