Difference between revisions of "2010 AMC 10A Problems/Problem 19"
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<math>\textbf{(A)}\ \frac{4\sqrt{3}}{3} \qquad \textbf{(B)} \frac{10}{3} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ \frac{17}{4} \qquad \textbf{(E)}\ 6</math> | <math>\textbf{(A)}\ \frac{4\sqrt{3}}{3} \qquad \textbf{(B)} \frac{10}{3} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ \frac{17}{4} \qquad \textbf{(E)}\ 6</math> | ||
− | == | + | == Solutions == |
− | + | This diagram only shows one of the possible values of r. | |
<asy> | <asy> | ||
unitsize(5cm); | unitsize(5cm); |
Revision as of 22:59, 15 March 2020
Problem
Equiangular hexagon has side lengths and . The area of is of the area of the hexagon. What is the sum of all possible values of ?
Solutions
This diagram only shows one of the possible values of r. ~Brian
Solution 1
It is clear that is an equilateral triangle. From the Law of Cosines on triangle ABC, we get that . Therefore, the area of is .
If we extend , and so that and meet at , and meet at , and and meet at , we find that hexagon is formed by taking equilateral triangle of side length and removing three equilateral triangles, , and , of side length . The area of is therefore
.
Based on the initial conditions,
Simplifying this gives us . By Vieta's Formulas (or Girard identities, or Newton-Girard identities) we know that the sum of the possible value of is .
Solution 2
As above, we find that the area of is .
We also find by the sine triangle area formula that , and thus This simplifies to .
Solution 3 (no trig)
Extend to point M so that it creates right triangle where . It is given that the hexagon is equiangular, therefore . (exterior angles of a polygon add up to 360 )
We can use either Pythagorean theorem or the properties of a triangle to find the length of and . The legs of are and .
Using Pythagorean theorem, we get . We can then follow to solve for . .
Alternatively, we can find the area of . We know that the three smaller triangles: , , and are congruent because of . Therefore one of the smaller triangles accounts for of the total area. The height of the smaller triangle is just so the area is . We can then find the area of the hexagon using .
We can even find the area of and and solve for because the ratio of the areas is to .
~Zeric Hang
See also
2010 AMC 10A (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.