Difference between revisions of "2017 AMC 12B Problems/Problem 15"
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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 | + | <math>\textbf{(A)}\ 9\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 36\qquad\textbf{(E)}\ 37</math> |
− | [[2017 | + | ==Diagram== |
+ | <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}</math> | ||
+ | |||
+ | ===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}</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 {\left(\frac{3x}{2}-x\right)}^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\left(9x^2+\frac{x^2}{4}\right)</math>. | ||
+ | Therefore the ratio of <math>[A'B'C']</math> to <math>[ABC]</math> is <math>\frac{\sqrt{3}\cdot\left(9x^2+\frac{x^2}{4}\right)}{\frac{x^2\sqrt{3}}{4}}=\boxed{\textbf{(E) } 37}</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}</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}</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}</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*4*3+1*1=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}</math> | ||
+ | |||
+ | ~Solution by EricShi1685 | ||
+ | |||
+ | ===Solution 9: Law of Cosines=== | ||
+ | Solution by HydroQuantum | ||
+ | |||
+ | |||
+ | Let <math>AB=BC=CA=x</math>. | ||
+ | |||
+ | |||
+ | Recall The Law of Cosines. Letting <math>A'B'=B'C'=C'A'=y</math>, <cmath>y^2=(3x)^2+(x+3x)^2-2(3x)(x+3x)(\cos 120) = </cmath> <cmath>(3x)^2+(4x)^2-2(3x)(4x)(\cos 120)=9x^2+16x^2-24x(\cos 120)=25x^2+12x^2=37x^2.</cmath> Since both <math>\triangle ABC</math> and <math>\triangle A'B'C'</math> are both equilateral triangles, they must be similar due to <math>AA</math> similarity. This means that <math>\frac{A'B'}{AB}</math> <math>=</math> <math>\frac{B'C'}{BC}</math> <math>=</math> <math>\frac{C'A'}{CA}</math> <math>=</math> <math>\frac{[\triangle A'B'C']}{[\triangle ABC]}</math> <math>=</math> <math>\frac{37}{1}</math>. | ||
+ | |||
+ | |||
+ | Therefore, our answer is <math>\boxed{\textbf{(E) }37}</math>. | ||
+ | |||
+ | ===Solution 10: Inspection(easiest solution)=== | ||
+ | Note that the height and base of <math>\triangle A'CC'</math> are respectively 4 times and 3 times that of <math>\triangle ABC</math>. Therefore the area of <math>\triangle A'CC'</math> is 12 times that of <math>\triangle ABC</math>. | ||
+ | |||
+ | By symmetry, <math>\triangle A'CC' \cong \triangle B'AA' \cong \triangle C'BB'</math>. Adding the areas of these three triangles and <math>\triangle ABC</math> for the total area of <math>\triangle A'B'C'</math> gives a ratio of <math>(12 + 12 + 12 + 1) : 1</math>, or <math>\boxed{\textbf{(E) } 37}</math>. | ||
+ | |||
+ | ===Solution 11: Coordinates=== | ||
+ | |||
+ | First we note that <math>A'B'C'\sim ABC</math> due to symmetry. WLOG, let <math>B = (0, 0)</math> and <math>AB = 1</math> Therefore, <math>C = (1, 0), A = \frac{1}{2}, \left(\frac{\sqrt{3}}{2}\right)</math>. Using the condition that <math>CC' = 3</math>, we get <math>C' = (4, 0)</math> and <math>B' = \left(\frac{-3}{2}, \frac{-3\sqrt{3}}{2}\right)</math>. It is easy to check that <math>B'C' = \sqrt{37}</math>. Since the area ratios of two similar figures is the square of the ratio of their lengths, the ratio is <math>\boxed{\textbf{(E) } 37}</math> | ||
+ | |||
+ | Solution by <i>mathwiz0803</i> | ||
+ | |||
+ | |||
+ | ===Solution 12: Computing the Areas=== | ||
+ | |||
+ | Note that angle <math>C'BB'</math> is <math>120</math>°, as it is supplementary to the equilateral triangle. Then, using area <math>= \frac{1}{2}ab\sin\theta</math> and letting side <math>AB = 1</math> for ease, we get: <math>4\cdot3\cdot\frac{\sin120}{2} = 3\sqrt{3}</math> as the area of <math>C'BB'</math>. Then, the area of <math>ABC</math> is <math>\frac{\sqrt{3}}{4}</math>, so the ratio is <math>\frac{3(3\sqrt{3})+\frac{\sqrt{3}}{4}}{\frac{\sqrt{3}}{4}} = \boxed{\textbf{(E) } 37}</math> | ||
+ | |||
+ | Solution by Aadileo | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{AMC12 box|year=2017|ab=B|num-b=14|num-a=16}} | ||
+ | {{AMC10 box|year=2017|ab=B|num-b=18|num-a=20}} | ||
+ | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] |
Latest revision as of 19:16, 16 October 2021
Contents
- 1 Problem
- 2 Diagram
- 2.1 Solution 1 (Uses Trig)
- 2.2 Solution 2
- 2.3 Solution 3
- 2.4 Solution 4 (Elimination)
- 2.5 Solution 5 (Barycentric Coordinates)
- 2.6 Solution 6 (Area Comparison)
- 2.7 Solution 7 (Quick Proportionality)
- 2.8 Solution 8 (Sin area formula)
- 2.9 Solution 9: Law of Cosines
- 2.10 Solution 10: Inspection(easiest solution)
- 2.11 Solution 11: Coordinates
- 2.12 Solution 12: Computing the Areas
- 3 See Also
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 ?
Diagram
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 30-60-90 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: Law of Cosines
Solution by HydroQuantum
Let .
Recall The Law of Cosines. Letting , Since both and are both equilateral triangles, they must be similar due to similarity. This means that .
Therefore, our answer is .
Solution 10: Inspection(easiest solution)
Note that the height and base of are respectively 4 times and 3 times that of . Therefore the area of is 12 times that of .
By symmetry, . Adding the areas of these three triangles and for the total area of gives a ratio of , or .
Solution 11: Coordinates
First we note that due to symmetry. WLOG, let and Therefore, . Using the condition that , we get and . It is easy to check that . Since the area ratios of two similar figures is the square of the ratio of their lengths, the ratio is
Solution by mathwiz0803
Solution 12: Computing the Areas
Note that angle is °, as it is supplementary to the equilateral triangle. Then, using area and letting side for ease, we get: as the area of . Then, the area of is , so the ratio is
Solution by Aadileo
See Also
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 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.