AMC 10 · 2009 · #16

Grade 8 geometry-2d

equilateral-trianglethirty-sixty-ninety-trianglesymmetry-argument physical-representation ↑ Prerequisites: equilateral-triangle

📏 Medium solution 💡 3 insights

Problem

Points $A$ and $C$ lie on a circle centered at $O$ , each of $\overline{BA}$ and $\overline{BC}$ are tangent to the circle, and $\triangle ABC$ is equilateral. The circle intersects $\overline{BO}$ at $D$ . What is $\frac{BD}{BO}$ ?

Pick an answer.

(A)

$\frac {\sqrt2}{3}$

(B)

$\frac {1}{2}$

(C)

$\frac {\sqrt3}{3}$

(D)

$\frac {\sqrt2}{2}$

(E)

$\frac {\sqrt3}{2}$

AMC 10 2009 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: Two segments drawn from an outside point $B$ just touch a circle (center $O$) at $A$ and $C$, and triangle $ABC$ is equilateral. The segment from $B$ to the center $O$ crosses the circle at $D$. Find the ratio $\frac{BD}{BO}$.

Givens: $A$ and $C$ lie on a circle centered at $O$, so $OA$ and $OC$ are radii.; $\overline{BA}$ and $\overline{BC}$ are tangent to the circle.; Triangle $ABC$ is equilateral, so all three of its angles are $60^\circ$.; $D$ is where segment $\overline{BO}$ crosses the circle.

Unknowns: The ratio $\frac{BD}{BO}$.

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #4 Introduce a Variable, #7 Identify Subproblems

A clear picture turns the words into two right angles and a line of symmetry. Once the radius is named $r$, the whole problem shrinks to one right triangle whose angles are known, and the ratio drops out of that triangle.

Execute — Answer: B

#1 Draw a Diagram 8.G.A.5 Step 1

Draw it and mark the right angles

Sketch the circle with center $O$, the two tangent segments $BA$ and $BC$, and triangle $ABC$.
A tangent meets the radius at the touching point at a right angle, so $\angle OAB = 90^\circ$ and $\angle OCB = 90^\circ$.
Let the radius be $r$, so $OA = OC = r$.

$$\angle OAB = 90^\circ,\quad \angle OCB = 90^\circ,\quad OA = OC = r$$

💡 Where a tangent kisses a circle, it stands square to the radius.

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

Use symmetry to find the angle at B

The two tangent lengths from $B$ are equal, so reflecting the figure across line $BO$ swaps $A$ and $C$.
That means $BO$ splits the equilateral triangle's $60^\circ$ angle at $B$ into two equal halves, so $\angle ABO = 30^\circ$.

$$\angle ABC = 60^\circ \;\Rightarrow\; \angle ABO = \tfrac{1}{2}\cdot 60^\circ = 30^\circ$$

💡 A mirror line through the tip cuts the tip angle in half.

#4 Introduce a Variable 8.G.B.7 Step 3

Solve the right triangle for BO

Look at right triangle $OAB$: the angle at $A$ is $90^\circ$ and the angle at $B$ is $30^\circ$, so it is a 30-60-90 triangle.
In such a triangle the side opposite the $30^\circ$ angle is exactly half the hypotenuse.
Here $OA = r$ is opposite the $30^\circ$ angle and $BO$ is the hypotenuse, so $BO = 2r$.

$$OA = r \text{ (opposite } 30^\circ),\quad BO = 2\,OA = 2r$$

💡 In a 30-60-90 triangle the shortest side is half the longest.

#7 Identify Subproblems 6.RP.A.1 Step 4

Locate D and take the ratio

$D$ is where $\overline{BO}$ crosses the circle, so $OD$ is a radius: $OD = r$.
Since $D$ sits between $B$ and $O$, we get $BD = BO - OD = 2r - r = r$.
Therefore $\frac{BD}{BO} = \frac{r}{2r} = \frac12$, which is choice $\textbf{(B)}$.

$$OD = r,\quad BD = BO - OD = 2r - r = r,\quad \frac{BD}{BO} = \frac{r}{2r} = \frac12$$

💡 The whole distance $BO$ is two radii, and $BD$ is exactly one of them.

[1] #1 8.G.A.5 Sketch the circle with center $O$, the two tangent segments $BA$ and $BC$, and t

[2] #1 8.G.A.1 The two tangent lengths from $B$ are equal, so reflecting the figure across line

[3] #4 8.G.B.7 Look at right triangle $OAB$: the angle at $A$ is $90^\circ$ and the angle at $B

[4] #7 6.RP.A.1 $D$ is where $\overline{BO}$ crosses the circle, so $OD$ is a radius: $OD = r$.

Review

Reasonableness: Set $r = 1$. Then $BO = 2$ and $OD = 1$, so $D$ is the midpoint of $BO$ and $BD = 1$. The ratio $\frac{BD}{BO} = \frac12$, a clean fraction between $0$ and $1$, which is exactly what a point strictly between $B$ and $O$ should give. It matches choice (B).

Alternative: Place $O$ at the origin with $B$ on the positive $x$-axis. The tangent length is $AB = r\sqrt3$ (from $BO^2 = OA^2 + AB^2$), and using $\angle ABO = 30^\circ$ gives $BO = OA/\sin 30^\circ = r/\tfrac12 = 2r$. Then $BD = BO - r = r$ and the ratio is again $\frac12$.

CCSS standards used (min grade 8)

8.G.A.5 Use informal arguments to establish facts about angle sum and exterior angles (Recognizing the tangent-radius right angles and working with the triangle's angles.)
8.G.A.1 Verify experimentally the properties of rotations, reflections, and translations (Using the mirror symmetry across line BO to halve the 60 degree angle at B.)
8.G.B.7 Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Using the 30-60-90 right triangle side relationship to get BO = 2r.)
6.RP.A.1 Understand the concept of a ratio and use ratio language (Forming and simplifying the ratio BD/BO.)

⭐ A tangent stands square to the radius; that right angle plus the triangle's symmetry makes BO two radii long, and BD is just one of them.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Draw it and mark the right angles

Use symmetry to find the angle at B

Solve the right triangle for BO

Locate D and take the ratio

Review

CCSS standards used (min grade 8)

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