AMC 10 · 2010 · #20
Grade 8 geometry-2dTwo circles lie outside regular hexagon ABCDEF. The first is tangent to AB, and the second is tangent to DE. Both are tangent to lines BC and FA. What is the ratio of the area of the second circle to that of the first circle?
Pick an answer.
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.
Toolkit + CCSS Solution
Understand
Restated: A regular hexagon $ABCDEF$ has two circles outside it. Both circles are tangent to the full lines $BC$ and $FA$. One circle is also tangent to side $\overline{AB}$, the other to side $\overline{DE}$. Find the ratio of the area of the second circle (the one touching $\overline{DE}$) to the area of the first (the one touching $\overline{AB}$).
Givens: $ABCDEF$ is a regular hexagon, so all sides are equal and every interior angle is $120^\circ$; The first circle lies outside the hexagon and is tangent to $\overline{AB}$; The second circle lies outside the hexagon and is tangent to $\overline{DE}$; Both circles are tangent to the lines $BC$ and $FA$ (the whole lines, extended as needed); Answer choices: (A) $18$, (B) $27$, (C) $36$, (D) $81$, (E) $108$
Unknowns: The ratio $\dfrac{\text{area of second circle}}{\text{area of first circle}}$
Understand
Restated: A regular hexagon $ABCDEF$ has two circles outside it. Both circles are tangent to the full lines $BC$ and $FA$. One circle is also tangent to side $\overline{AB}$, the other to side $\overline{DE}$. Find the ratio of the area of the second circle (the one touching $\overline{DE}$) to the area of the first (the one touching $\overline{AB}$).
Givens: $ABCDEF$ is a regular hexagon, so all sides are equal and every interior angle is $120^\circ$; The first circle lies outside the hexagon and is tangent to $\overline{AB}$; The second circle lies outside the hexagon and is tangent to $\overline{DE}$; Both circles are tangent to the lines $BC$ and $FA$ (the whole lines, extended as needed); Answer choices: (A) $18$, (B) $27$, (C) $36$, (D) $81$, (E) $108$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #4 Introduce a Variable
This is a pure shapes-and-positions problem, so Tool #1 (Draw a Diagram) is the anchor: extending lines $BC$ and $FA$ until they cross reveals a $60^\circ$ corner that both circles must live in. Tool #7 (Identify Subproblems) then splits the work into one reusable fact — where a circle sits inside that corner — plus two separate radius calculations, one from $\overline{AB}$ and one from $\overline{DE}$. Tool #4 (Introduce a Variable) fixes the hexagon side length at $1$ so every distance becomes a concrete number and the two radii can be compared.
Execute — Answer: D
8.G.A.5 Step 1 Extend the two shared lines to a corner
- Both circles touch the lines $BC$ and $FA$, so draw those two sides as full lines.
- Extend them until they meet at a point $P$.
- Because the hexagon is regular, its interior angles are $120^\circ$, so the sides turn by $60^\circ$ at each corner; the lines $BC$ and $FA$ cross at an angle of $60^\circ$.
- Every circle that is tangent to both lines must sit inside this $60^\circ$ corner, with its center on the bisector that splits the $60^\circ$ into two $30^\circ$ halves.
💡 Two tangent lines meeting form a corner, and any circle touching both lines must nestle into that corner on its bisector.
8.G.B.7 Step 2 A circle in the corner sits twice its radius from the tip
- Take any circle tangent to both sides of the $60^\circ$ corner, with radius $r$.
- Draw from the tip $P$ to the center, then from the center drop a perpendicular of length $r$ to one side.
- This makes a right triangle with a $30^\circ$ angle at $P$.
- Half of an equilateral triangle is exactly this $30^\circ$-$60^\circ$-$90^\circ$ shape, and in it the short leg is half the hypotenuse.
- The short leg is $r$, so the hypotenuse $P$-to-center is $2r$.
- This one fact will be reused for both circles.
💡 Cutting an equilateral triangle in half makes the long side twice the short side, so the center sits two radii from the tip.
8.G.B.7 Step 3 First circle: measure from AB
- Let the hexagon side be $1$.
- The first circle also touches $\overline{AB}$.
- Side $AB$ together with the two corner lines forms a small triangle with all $60^\circ$ angles — an equilateral triangle of side $1$.
- Its height, straight from $P$ down to $\overline{AB}$, is $\tfrac{\sqrt{3}}{2}$.
- The circle's center is $2r_1$ from $P$, and one more radius $r_1$ reaches $\overline{AB}$, so the whole height is $2r_1 + r_1 = 3r_1$.
- Setting $3r_1 = \tfrac{\sqrt{3}}{2}$ gives $r_1 = \tfrac{\sqrt{3}}{6}$.
💡 From the tip to $\overline{AB}$ you pass the center (two radii) and then one more radius, so the height is three radii.
8.G.B.7 Step 4 Second circle: measure from DE, farther out
- The second circle touches $\overline{DE}$, the side opposite $\overline{AB}$, and lies outside the hexagon beyond $\overline{DE}$.
- So its center is past $\overline{DE}$: the distance $2r_2$ from $P$ to the center equals the distance from $P$ to $\overline{DE}$ plus one more radius $r_2$.
- That means the distance from $P$ to $\overline{DE}$ is $2r_2 - r_2 = r_2$.
- Now measure that distance: it is the height down to $\overline{AB}$, $\tfrac{\sqrt{3}}{2}$, plus the width across the hexagon from $\overline{AB}$ to the opposite side $\overline{DE}$, which is $\sqrt{3}$.
- So $r_2 = \tfrac{\sqrt{3}}{2} + \sqrt{3} = \tfrac{3\sqrt{3}}{2}$.
💡 This circle pokes past its side instead of tucking before it, so the tip-to-side distance equals just one radius, not three.
7.G.B.4 Step 5 Square the radius ratio
- A circle's area is $\pi r^2$, so the ratio of the two areas is the ratio of the radii squared.
- The radii ratio is $\dfrac{r_2}{r_1} = \dfrac{3\sqrt{3}/2}{\sqrt{3}/6} = \dfrac{3\sqrt{3}}{2}\cdot\dfrac{6}{\sqrt{3}} = 9$.
- Squaring gives $9^2 = 81$.
- The answer is $\textbf{(D)}\ 81$.
💡 Area grows with the square of the radius, so a $9$-times-wider circle has $81$ times the area.
8.G.A.5 Both circles touch the lines $BC$ and $FA$, so draw those two sides as full line 8.G.B.7 Take any circle tangent to both sides of the $60^\circ$ corner, with radius $r$. 8.G.B.7 Let the hexagon side be $1$. The first circle also touches $\overline{AB}$. Side 8.G.B.7 The second circle touches $\overline{DE}$, the side opposite $\overline{AB}$, an 7.G.B.4 A circle's area is $\pi r^2$, so the ratio of the two areas is the ratio of the Review
Reasonableness: The radii are $r_1 = \tfrac{\sqrt{3}}{6} \approx 0.29$ and $r_2 = \tfrac{3\sqrt{3}}{2} \approx 2.60$, a ratio of about $9$ — the second circle really is much larger, which matches a picture where it wraps the far side of the hexagon. Squaring $9$ gives $81$, one of the choices. The subtle point that keeps the answer from being $9$ (the linear factor) is that areas scale as the square, and the answer being $81 = 9^2$ rather than, say, $3^2 = 9$ comes from the near circle tucking in before $\overline{AB}$ (three radii from the tip) while the far circle pokes out past $\overline{DE}$ (one radius from the tip). Answer $\textbf{(D)}$ holds.
Alternative: Tool #1 with coordinates: place the hexagon's center at the origin with circumradius $1$, write the equations of lines $FA$, $BC$, and $DE$, and require each circle's center (on the $60^\circ$ bisector) to be equidistant from the relevant lines. The first circle is the incircle of an equilateral triangle of side $1$ and the second is the escribed circle of an equilateral triangle of side $3$; their radii $\tfrac{\sqrt{3}}{6}$ and $\tfrac{3\sqrt{3}}{2}$ give the same ratio $9$, hence area ratio $81$.
CCSS standards used (min grade 8)
8.G.A.5Use informal arguments to establish facts about angles of triangles and the angle-angle criterion for similar triangles (Using the hexagon's $120^\circ$ interior angles to show lines $BC$ and $FA$ cross at $60^\circ$, and that each tangent side plus the two lines makes an equilateral triangle.)8.G.B.7Apply the Pythagorean Theorem to determine unknown side lengths in right triangles (Getting the $30$-$60$-$90$ fact that the center sits $2r$ from the corner tip, the equilateral height $\tfrac{\sqrt{3}}{2}$, and the hexagon width $\sqrt{3}$, which give $r_1 = \tfrac{\sqrt{3}}{6}$ and $r_2 = \tfrac{3\sqrt{3}}{2}$.)7.G.B.4Know and use the formulas for the area and circumference of a circle (Turning the radius ratio $9$ into the area ratio by using area $=\pi r^2$, giving $9^2 = 81$.)
⭐ Extend the two shared lines to a corner, remember a circle's center sits two radii from the tip, then square the radius ratio to get the area ratio.
⭐ Extend the two shared lines to a corner, remember a circle's center sits two radii from the tip, then square the radius ratio to get the area ratio.
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