AMC 10 · 2009 · #20
Grade 8 geometry-2dTriangle ABC has a right angle at B, AB=1, and BC=2. The bisector of ∠BAC meets BC at D. What is BD?
Pick an answer.
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.
Toolkit + CCSS Solution
Understand
Restated: In right triangle $ABC$ the right angle is at $B$, with $AB=1$ and $BC=2$. The ray from $A$ that cuts $\angle BAC$ into two equal angles hits side $\overline{BC}$ at a point $D$. Find the length $BD$.
Givens: Triangle $ABC$ has a right angle at $B$.; The two legs are $AB=1$ and $BC=2$.; $AD$ bisects $\angle BAC$, meaning it splits that corner angle into two equal halves.; $D$ is the point where this bisector meets the side $\overline{BC}$.; Answer choices: (A) $\frac{\sqrt3-1}{2}$, (B) $\frac{\sqrt5-1}{2}$, (C) $\frac{\sqrt5+1}{2}$, (D) $\frac{\sqrt6+\sqrt2}{2}$, (E) $2\sqrt3-1$.
Unknowns: The length $BD$, the part of $\overline{BC}$ from $B$ up to $D$.
Understand
Restated: In right triangle $ABC$ the right angle is at $B$, with $AB=1$ and $BC=2$. The ray from $A$ that cuts $\angle BAC$ into two equal angles hits side $\overline{BC}$ at a point $D$. Find the length $BD$.
Givens: Triangle $ABC$ has a right angle at $B$.; The two legs are $AB=1$ and $BC=2$.; $AD$ bisects $\angle BAC$, meaning it splits that corner angle into two equal halves.; $D$ is the point where this bisector meets the side $\overline{BC}$.; Answer choices: (A) $\frac{\sqrt3-1}{2}$, (B) $\frac{\sqrt5-1}{2}$, (C) $\frac{\sqrt5+1}{2}$, (D) $\frac{\sqrt6+\sqrt2}{2}$, (E) $2\sqrt3-1$.
Plan
Primary tool: #4 Introduce a Variable
Secondary: #7 Identify Subproblems, #1 Draw a Diagram
The Angle Bisector Theorem says the bisector from $A$ divides the opposite side $BC$ in the same ratio as the two sides that make the angle: $BD/DC=AB/AC$. To use it I first need the hypotenuse $AC$, so Tool #7 (Identify Subproblems) peels off a quick Pythagorean sub-task. The figure (Tool #1) shows the right angle and the bisector, confirming which sides form the ratio. Then Tool #4 (Introduce a Variable) names $BD=x$ so the ratio becomes one clean equation I can solve.
Execute — Answer: B
8.G.B.7 Step 1 Find the hypotenuse AC
- Before using the bisector rule I need all three sides of the triangle.
- The right angle is at $B$, so $AC$ is the hypotenuse and the two legs are $AB=1$ and $BC=2$.
- The Pythagorean theorem gives $AC^2=AB^2+BC^2=1^2+2^2=5$, so $AC=\sqrt5$.
💡 The right angle hands you $a^2+b^2=c^2$, which is the fastest way to get the missing hypotenuse.
7.RP.A.2 Step 2 Set up the bisector ratio
- The Angle Bisector Theorem says the bisector from $A$ cuts the opposite side $BC$ into two pieces whose lengths keep the ratio of the sides next to angle $A$: $\frac{BD}{DC}=\frac{AB}{AC}=\frac{1}{\sqrt5}$.
- Let $BD=x$.
- Since $BD+DC=2$, the other piece is $DC=2-x$.
- This turns the ratio into a single equation in $x$.
💡 A bisector shares the far side in the exact proportion of the two near sides, so a ratio is all you need.
8.EE.C.7 Step 3 Solve for x
- Cross-multiply the proportion: $\sqrt5\,x=2-x$.
- Bring the $x$ terms together: $\sqrt5\,x+x=2$, so $x(\sqrt5+1)=2$.
- Dividing gives $x=\frac{2}{\sqrt5+1}$.
💡 Clearing the fraction leaves a plain linear equation, even with a $\sqrt5$ riding along as a constant.
8.NS.A.1 Step 4 Rationalize to match a choice
- The answer choices have no radical in the denominator, so multiply top and bottom by the conjugate $\sqrt5-1$.
- The denominator becomes $(\sqrt5+1)(\sqrt5-1)=5-1=4$, giving $x=\frac{2(\sqrt5-1)}{4}=\frac{\sqrt5-1}{2}$.
- So $BD=\frac{\sqrt5-1}{2}$, which is choice (B).
💡 Multiplying by the conjugate uses $(a+b)(a-b)=a^2-b^2$ to wipe the root out of the denominator.
8.G.B.7 Before using the bisector rule I need all three sides of the triangle. The right 7.RP.A.2 The Angle Bisector Theorem says the bisector from $A$ cuts the opposite side $BC 8.EE.C.7 Cross-multiply the proportion: $\sqrt5\,x=2-x$. Bring the $x$ terms together: $\ 8.NS.A.1 The answer choices have no radical in the denominator, so multiply top and botto Review
Reasonableness: Numerically $BD=\frac{\sqrt5-1}{2}\approx\frac{2.236-1}{2}\approx0.618$, so $DC\approx1.382$. Both are positive and add to $2$, and $BD<DC$, which fits the prediction that the bisector lands nearer the shorter side $AB$. Checking the ratio, $BD/DC\approx0.618/1.382\approx0.447=1/\sqrt5$, exactly the Angle Bisector Theorem ratio.
Alternative: Use the half-angle idea instead. In right triangle $ABC$, $\tan(\angle BAC)=BC/AB=2$. The bisector makes $\angle BAD=\tfrac12\angle BAC$, and in right triangle $ABD$ (right angle at $B$), $BD=AB\tan(\angle BAD)=\tan(\tfrac12\angle BAC)$. Writing $t=\tan(\tfrac12\angle BAC)$, the double-angle rule $\frac{2t}{1-t^2}=2$ gives $t^2+t-1=0$, so $t=\frac{-1+\sqrt5}{2}$, matching $BD=\frac{\sqrt5-1}{2}$.
CCSS standards used (min grade 8)
8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Computing the hypotenuse $AC=\sqrt{1^2+2^2}=\sqrt5$ from the two legs.)7.RP.A.2Recognize and represent proportional relationships between quantities (Turning the Angle Bisector Theorem into the proportion $\frac{x}{2-x}=\frac{1}{\sqrt5}$ for $BD=x$.)8.EE.C.7Solve linear equations in one variable (Cross-multiplying and solving $x(\sqrt5+1)=2$ to get $x=\frac{2}{\sqrt5+1}$.)8.NS.A.1Know that numbers that are not rational are called irrational numbers (Rationalizing $\frac{2}{\sqrt5+1}$ with the conjugate to reach the simplified form $\frac{\sqrt5-1}{2}$.)
⭐ An angle bisector splits the far side in the same ratio as the two sides that make the angle, so find that ratio, set the piece you want equal to $x$, and solve.
⭐ An angle bisector splits the far side in the same ratio as the two sides that make the angle, so find that ratio, set the piece you want equal to $x$, and solve.
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