AMC 10 · 2010 · #19

Grade 8 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity

law-of-cosinesequilateral-trianglearea-trianglesvieta-formulasquadratic-equations identify-subproblems ↑ Prerequisites: law-of-cosines

📏 Long solution 💡 4 insights

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 $r$ ?

Pick an answer.

(A)

$\frac{4\sqrt{3}}{3}$

(B)

$\frac{10}{3}$

(C)

(D)

$\frac{17}{4}$

(E)

AMC 10 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: An equiangular hexagon $ABCDEF$ (every interior angle $120^\circ$) has sides alternating $AB=CD=EF=1$ and $BC=DE=FA=r$. The triangle $\triangle ACE$ joining every other vertex has area equal to $70\%$ of the hexagon's area. Find the sum of all values of $r$ that make this true.

Givens: Hexagon is equiangular: each interior angle is $120^\circ$; Short sides $AB=CD=EF=1$; Long sides $BC=DE=FA=r$; $[\triangle ACE] = 0.70 \cdot [\text{hexagon } ABCDEF]$; Answer choices: (A) $\tfrac{4\sqrt{3}}{3}$, (B) $\tfrac{10}{3}$, (C) $4$, (D) $\tfrac{17}{4}$, (E) $6$

Unknowns: All valid values of $r$; Their sum

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #1 Draw a Diagram, #4 Introduce a Variable

The area equation hides two separate area computations, so split the work (Tool #7 Identify Subproblems): first find the area of $\triangle ACE$, then find the area of the whole hexagon, both in terms of $r$. Drawing the figure (Tool #1) reveals two structural gifts — the three corner triangles $ABC, CDE, EFA$ are congruent, which forces $\triangle ACE$ to be equilateral, and the hexagon is a big equilateral triangle with three unit corners sliced off. Since everything is already labelled with the variable $r$ (Tool #4), setting $[\triangle ACE] = 0.7 \cdot [\text{hexagon}]$ collapses to a single quadratic in $r$, and the question only asks for the sum of its roots.

Execute — Answer: E

#1 Draw a Diagram 8.G.A.2 Step 1

Triangle ACE is equilateral

Look at the three corner triangles $ABC$, $CDE$, $EFA$.
Each has a short side $1$, a long side $r$, and the included vertex angle is the hexagon's $120^\circ$.
By SAS they are congruent, so their third sides $AC$, $CE$, $EA$ are all equal.
Three equal sides means $\triangle ACE$ is equilateral.

$$\triangle ABC \cong \triangle CDE \cong \triangle EFA \;(\text{SAS}) \;\Rightarrow\; AC = CE = EA$$

💡 Identical corner triangles must hand back identical long edges, so the inner triangle can only be equilateral.

#7 Identify Subproblems 8.G.B.7 Step 2

Find side AC

Work inside corner triangle $ABC$ with $AB=1$, $BC=r$, and $\angle B = 120^\circ$.
Drop a perpendicular from $A$ to line $BC$; because $\angle ABC = 120^\circ$, the foot lands outside the segment and the little right triangle at $B$ is a $30\text{-}60\text{-}90$ triangle with hypotenuse $AB=1$.
Its legs are $\tfrac{1}{2}$ (horizontal) and $\tfrac{\sqrt{3}}{2}$ (vertical).
The horizontal distance from the foot to $C$ is $r + \tfrac{1}{2}$, so the Pythagorean theorem gives $AC^2$.

$$AC^2 = \left(\tfrac{\sqrt{3}}{2}\right)^2 + \left(r + \tfrac{1}{2}\right)^2 = \tfrac{3}{4} + r^2 + r + \tfrac{1}{4} = r^2 + r + 1$$

💡 A $120^\circ$ corner splits into a clean $30\text{-}60\text{-}90$ triangle, so Pythagoras finishes the length with no trig needed.

#7 Identify Subproblems 6.G.A.1 Step 3

Area of triangle ACE

An equilateral triangle with side $s$ has height $\tfrac{\sqrt{3}}{2}s$ (another $30\text{-}60\text{-}90$ split), so its area is $\tfrac{1}{2}\cdot s \cdot \tfrac{\sqrt{3}}{2}s = \tfrac{\sqrt{3}}{4}s^2$.
Here $s^2 = AC^2 = r^2 + r + 1$.

$$[\triangle ACE] = \tfrac{\sqrt{3}}{4}\,(r^2 + r + 1)$$

💡 For an equilateral triangle you only need $s^2$, and we already have it as $r^2+r+1$.

#7 Identify Subproblems 7.G.B.6 Step 4

Area of the hexagon

Extend the three short sides ($AB$, $CD$, $EF$) until they meet.
The equiangular hexagon becomes one big equilateral triangle with three small equilateral triangles cut off the corners.
Each cut corner has side $1$ (the short sides), and the big triangle's side is $r + 2$ (a long side $r$ flanked by two unit corner pieces).
Subtract: big minus three unit triangles.

$$[\text{hex}] = \tfrac{\sqrt{3}}{4}(r+2)^2 - 3\cdot\tfrac{\sqrt{3}}{4}(1)^2 = \tfrac{\sqrt{3}}{4}\big[(r+2)^2 - 3\big] = \tfrac{\sqrt{3}}{4}\,(r^2 + 4r + 1)$$

💡 A tidy equiangular hexagon is just a big equilateral triangle with its three tips snipped off.

#4 Introduce a Variable 8.EE.C.7 Step 5

Set up and simplify the equation

Impose the $70\%$ condition.
The common factor $\tfrac{\sqrt{3}}{4}$ cancels from both sides, leaving a plain equation in $r$.
Distribute the $0.7$, move everything to one side, and divide by $0.3$.

$$r^2 + r + 1 = 0.7\,(r^2 + 4r + 1) \;\Rightarrow\; 0.3r^2 - 1.8r + 0.3 = 0 \;\Rightarrow\; r^2 - 6r + 1 = 0$$

💡 The messy $\tfrac{\sqrt{3}}{4}$ factor is identical on both sides, so the geometry boils down to one clean quadratic.

#4 Introduce a Variable 8.EE.C.7 Step 6

Sum of all valid r

Solve $r^2 - 6r + 1 = 0$.
The discriminant is $36 - 4 = 32 > 0$, so there are two real roots; since their sum $6$ and product $1$ are both positive, both roots are positive and valid.
The two roots are $3 \pm 2\sqrt{2}$, and their sum equals $6$ (equivalently, the sum of roots of $r^2 - 6r + 1$ is $-\tfrac{-6}{1} = 6$).

$$r = \dfrac{6 \pm \sqrt{32}}{2} = 3 \pm 2\sqrt{2}; \quad (3 + 2\sqrt{2}) + (3 - 2\sqrt{2}) = 6 \;\Rightarrow\; \textbf{(E)}\ 6$$

💡 You never need the individual roots — the sum of the solutions of $r^2-6r+1$ is just the middle coefficient flipped in sign.

[1] #1 8.G.A.2 Look at the three corner triangles $ABC$, $CDE$, $EFA$. Each has a short side $1

[2] #7 8.G.B.7 Work inside corner triangle $ABC$ with $AB=1$, $BC=r$, and $\angle B = 120^\circ

[3] #7 6.G.A.1 An equilateral triangle with side $s$ has height $\tfrac{\sqrt{3}}{2}s$ (another

[4] #7 7.G.B.6 Extend the three short sides ($AB$, $CD$, $EF$) until they meet. The equiangular

[5] #4 8.EE.C.7 Impose the $70\%$ condition. The common factor $\tfrac{\sqrt{3}}{4}$ cancels fro

[6] #4 8.EE.C.7 Solve $r^2 - 6r + 1 = 0$. The discriminant is $36 - 4 = 32 > 0$, so there are tw

Review

Reasonableness: Sanity-check one root. Take $r = 3 + 2\sqrt{2} \approx 5.828$. Then $r^2 + r + 1 \approx 33.97 + 5.83 + 1 = 40.80$ and $r^2 + 4r + 1 \approx 33.97 + 23.31 + 1 = 58.28$; the ratio $40.80/58.28 \approx 0.700$, exactly the required $70\%$. The other root $r = 3 - 2\sqrt{2} \approx 0.172$ is a genuinely different but still positive hexagon, so two values of $r$ really do exist and both must be counted. Their sum is $6$, matching (E). The answer is a plain integer, consistent with a competition asking for a 'sum of all possible values'.

Alternative: Skip the big-triangle picture and instead build the hexagon from $\triangle ACE$ plus the three corner triangles. Each corner triangle $ABC$ has area $\tfrac{1}{2}\cdot AB \cdot BC \cdot \sin 120^\circ = \tfrac{\sqrt{3}}{4}r$, so the three of them total $\tfrac{3\sqrt{3}}{4}r$. Then $[\text{hex}] = [\triangle ACE] + \tfrac{3\sqrt{3}}{4}r = \tfrac{\sqrt{3}}{4}(r^2 + r + 1) + \tfrac{3\sqrt{3}}{4}r = \tfrac{\sqrt{3}}{4}(r^2 + 4r + 1)$ — the same hexagon area, so the same quadratic $r^2 - 6r + 1 = 0$ and the same sum $6$.

CCSS standards used (min grade 8)

8.G.A.2 Understand that a two-dimensional figure is congruent to another using transformations (Showing the three corner triangles are congruent (SAS), which forces $\triangle ACE$ to be equilateral.)
8.G.B.7 Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Finding $AC^2 = r^2 + r + 1$ from the $30\text{-}60\text{-}90$ split of the $120^\circ$ corner.)
6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing (Computing the equilateral-triangle area $\tfrac{\sqrt{3}}{4}s^2$ for $\triangle ACE$.)
7.G.B.6 Solve real-world problems involving area, surface area, and volume (Getting the hexagon area as a big equilateral triangle minus three unit corner triangles.)
8.EE.C.7 Solve linear equations in one variable (Reducing the $70\%$ area condition to $r^2 - 6r + 1 = 0$ and reading off the sum of its roots.)

⭐ Write both areas in terms of $r$: the inner triangle is equilateral with side$^2 = r^2+r+1$, and the hexagon is a big triangle minus three unit corners with area factor $r^2+4r+1$. The $70\%$ rule turns into $r^2 - 6r + 1 = 0$, and the sum of its two roots is $\textbf{(E)}\ 6$.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Triangle ACE is equilateral

Find side AC

Area of triangle ACE

Area of the hexagon

Set up and simplify the equation

Sum of all valid r

Review

CCSS standards used (min grade 8)

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