2010 AMC 12A Problems/Problem 17

The following problem is from both the 2010 AMC 12A #17 and 2010 AMC 10A #19, so both problems redirect to this page.

Problem

Equiangular hexagon $ABCDEF$ has side lengths $AB=CD=EF=1$ and $BC=DE=FA=r$ . The area of $\triangle ACE$ is $70\%$ of the area of the hexagon. What is the sum of all possible values of $r$ ?

$\textbf{(A)}\ \frac{4\sqrt{3}}{3} \qquad \textbf{(B)} \frac{10}{3} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ \frac{17}{4} \qquad \textbf{(E)}\ 6$

Solution 1

It is clear that $\triangle ACE$ is an equilateral triangle. From the Law of Cosines on $\triangle ABC$ , we get that $AC^2 = r^2+1^2-2r\cos{\frac{2\pi}{3}} = r^2+r+1$ . Therefore, the area of $\triangle ACE$ is $\frac{\sqrt{3}}{4}(r^2+r+1)$ by area of an equilateral triangle.

If we extend $BC$ , $DE$ and $FA$ so that $FA$ and $BC$ meet at $X$ , $BC$ and $DE$ meet at $Y$ , and $DE$ and $FA$ meet at $Z$ , we find that hexagon $ABCDEF$ is formed by taking equilateral triangle $XYZ$ of side length $r+2$ and removing three equilateral triangles, $ABX$ , $CDY$ and $EFZ$ , of side length $1$ . The area of $ABCDEF$ is therefore

$\frac{\sqrt{3}}{4}(r+2)^2-\frac{3\sqrt{3}}{4} = \frac{\sqrt{3}}{4}(r^2+4r+1)$ .

Based on the initial conditions,

$\frac{\sqrt{3}}{4}(r^2+r+1) = \frac{7}{10}\left(\frac{\sqrt{3}}{4}\right)(r^2+4r+1)$

Simplifying this gives us $r^2-6r+1 = 0$ . By Vieta's Formulas we know that the sum of the possible value of $r$ is $\boxed{\textbf{(E)}\ 6}$ .

Solution 2

Step 1: Use Law of Cosines in the same manner as the previous solution to get $AC=\sqrt{r^2+r+1}$ .

Step 2: $\triangle{ABC}$ ~ $\triangle{CDE}$ ~ $\triangle{EFA}$ via SAS congruency. Using the formula $[ABC]=\frac{ab \sin C}{2}= \frac{r \sqrt{3}}{4}$ . The area of the hexagon is equal to $[ACE] + 3[ABC]$ . We are given that $70\%$ of this area is equal to $[ACE]$ ; solving for $AC$ in terms of $r$ gives $AC=\sqrt{7r}$ .

Step 3: $\sqrt{7r}=\sqrt{r^2+r+1} \implies r^2-6r+1=0$ and by Vieta's Formulas , we get $\boxed{\textbf{E}}$ .

Note: To verify that the quadratic $r^2-6r+1$ has two positive roots, we can either solve for the roots directly or note that discriminant is positive, and there are no negative roots (because then $r^2, -6r, 1$ would all be positive).

Solution 3

Find the area of the triangle $ACE$ as how it was done in solution 1. Find the sum of the areas of the congruent triangles $ABC, CDE, EFA$ as how it was done in solution 2. The sum of these areas is the area of the hexagon, hence the areas of the congruent triangles $ABC, CDE, EFA$ is $30\%$ of the area of the hexagon. Hence $\frac{7}{3}$ times the latter is equal to the triangle $ACE$ . Hence $\frac{7}{3}\cdot\frac{3\sqrt{3}}{4}r=\frac{\sqrt{3}}{4}(r^2+r+1)$ . We can simplify this to $7r=r^2+r+1\implies r^2-6r+1=0$ . By Vieta's, we get the sum of all possible values of $r$ is $-\frac{-6}{1}=6\text{ or } \boxed{\textbf{E}}$ . -vsamc

Solution 4 (no trig)

$[asy] import graph; size(8cm); pen dps = fontsize(10); defaultpen(dps); real r = 0.7; // Define hexagon vertices clockwise with AB on top pair A = (0, 0); pair B = (1, 0); pair C = B + r * dir(-60); pair D = C + dir(-120); pair E = D + r * dir(180); pair F = E + dir(120); // Draw the hexagon draw(A--B--C--D--E--F--cycle); // Draw triangle ACE draw(A--C--E--cycle); // Draw line CF draw(C--F); // Drop perpendiculars from B and E to CF pair foot_B = foot(B, C, F); pair foot_E = foot(E, C, F); // Draw perpendiculars (dashed) draw(B--foot_B, dashed); draw(E--foot_E, dashed); // Right angle markers (smaller) draw(rightanglemark(B, foot_B, C, 4)); draw(rightanglemark(E, foot_E, C, 4)); // Vertex labels label("$A$", A, NW); label("$B$", B, NE); label("$C$", C, SE); label("$D$", D, S); label("$E$", E, SW); label("$F$", F, NW); // Feet labels label("$P$", foot_B, S); label("$Q$", foot_E, N); // Side length labels label("$1$", midpoint(A--B), N); label("$r$", midpoint(B--C), dir(45)); label("$1$", midpoint(C--D), E); label("$r$", midpoint(D--E), S); label("$1$", midpoint(E--F), W); label("$r$", midpoint(F--A), W); [/asy]$ Drawing not to scale.* To find r, we'll form an equation by finding the area of hexagon $ABCDEF$ and $\triangle{ACE}$ separately. First, connect a directly opposite diagonal such as $CF$ . Because the hexagon is equiangular, edges $AF$ and $BC$ protrude from points $A$ and $B$ at the same angle but in opposite directions and equal length. This places points $C$ and $F$ at the same "height" relative to segment $AB$ , so segment $CF$ is parallel to $AB$ and also $DE$ (opposite sides are parallel since you rotated through three $120^\circ$ angles between each other).

Next, we'll calculate the area of isosceles trapezoids $ABCF$ and $CDEF$ . Drop a perpendicular down from $B$ to $CF$ , and call the intersection to $CF$ , $P$ . Because adjacent angles between parallel sides in a trapezoid sum to $180^\circ$ and $\angle ABC = 120^\circ$ , $\angle BCP = 60^\circ$ meaning $\triangle{BPC}$ is a 30-60-90 right triangle. Now we get $BP = \frac{\sqrt{3}}{2}r$ and $CP = \frac{r}{2}$ . $CF$ is equal to $AB + 2CP = r+1$ . Hence, the area of trapezoid $ABCF$ is $\frac{1}{2} \cdot \frac{\sqrt{3}}{2}r \cdot (r+1+1) = \frac{\sqrt{3}}{4} r(r+2)$ . Similarly, on trapezoid $CDEF$ we find $\triangle EQF$ is 30-60-90, and $EQ = \frac{\sqrt{3}}{2}, FQ = \frac{1}{2}$ . Then, area of $CDEF$ is $\frac{\sqrt{3}}{4} (2r + 1)$ . Thus, the area of the hexagon is the sum of the two areas, $\frac{\sqrt{3}}{4} (r^2 + 4r + 1)$ .

Next, notice that $\triangle ACE$ is equilateral. And by Pythagorean theorem in $\triangle CQE$ , side $CE = \sqrt{r^2 + r + 1}$ . Applying area of equilateral triangle formula, area $\triangle ACE = \frac{\sqrt{3}}{4} (r^2 + r + 1)$ . Using this area and the area given from 70% of the area of the hexagon, we get the equation $\frac{7}{10} \cdot \frac{\sqrt{3}}{4} (r^2 + 4r + 1) = \frac{\sqrt{3}}{4} (r^2 + r + 1)$ . Simplifying, we get the quadratic $r^2-6r+1=0$ , which by Vieta's yields the answer $6, \boxed{\textbf{E}}$ .

~henry

Proof Triangle ACE is Equilateral.

We know $\triangle{ABC}$ , $\triangle{CDE}$ , and $\triangle{EFA}$ are congruent by SAS, so the side opposite the 120 degree angle is also the same (since the triangles are congruent). Thus $\triangle{ACE}$ is equilateral. Q.E.D. ~mathboy282

Video Solution by Pi Academy

https://youtu.be/x8b1D35qxME?si=8jTbTBi-LPOUcP3t

~ Pi Academy

Video Solution 2

https://youtu.be/rsURe5Xh-j0?t=961

2010 AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

2010 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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All AMC 12 Problems and Solutions

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