2017 AMC 10B Problems/Problem 19
Contents
- 1 Problem
- 2 Diagram
- 3 Solution 1 (Uses Trig)
- 4 Solution 2
- 5 Solution 3
- 6 Solution 4 (Elimination)
- 7 Solution 5 (Barycentric Coordinates)
- 8 Solution 6 (Area Comparison)
- 9 Solution 7 (Quick Proportionality)
- 10 Solution 8 (Sin area formula)
- 11 Solution 9 (Same as Solution 8 but faster)
- 12 Solution 10 (Area Ratios)
- 13 Solution 11 (Quick Guess If You Have No Time)
- 14 Solution 12 (Mass Points and Routh)
- 15 Video Solution by OmegaLearn (Law of Cosines)
- 16 Video Solution by OmegaLearn (Meta-Solving Technique)
- 17 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 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 .
Solution 10 (Area Ratios)
Connect and . Let . . So the area of is equal to . We can do the same thing for . . Thus, the area of is equal to . We will now find the area of in terms of . . The area of is equal to the sum of the areas of and , which is . So the area of is equal to and the area of is equal to so the ratio of the area of to the area of is equal to -Heavytoothpaste
Solution 11 (Quick Guess If You Have No Time)
, , , and are all perfect squares, which makes them seem unlikely, so we can guess that the answer probably is as it is the only one with a side length that is implied to not be an integer.
Solution 12 (Mass Points and Routh)
This looks like an easy application of Routh's Theorem, except we are only given information about the ratios of the cevians, not the side lengths of .
Let's figure those out. Extend and to meet and at and , respectively (Note we only need to look at 2 of 3 cevians to figure everything out). Call the unknown lengths so our new diagram showing the cevian ratios is as follows.
with the balanced mass points on the right. Now by the symmetry in the original diagram, so:
Finally we can apply Routh's Theorem:
Hence
~ proloto
Video Solution by OmegaLearn (Law of Cosines)
https://youtu.be/4_x1sgcQCp4?t=5373
~ pi_is_3.14
Video Solution by OmegaLearn (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 |
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