AMC 10 · 2021 · #17

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference

similar-trianglespythagorean-theoremisosceles-triangleperpendicular-bisector identify-subproblemsconvert-to-algebra ↑ Prerequisites: similar-triangles

📏 Long solution 💡 3 insights 📊 Diagram

Problem

Trapezoid $ABCD$ has $\overline{AB}\parallel\overline{CD},BC=CD=43$ , and $\overline{AD}\perp\overline{BD}$ . Let $O$ be the intersection of the diagonals $\overline{AC}$ and $\overline{BD}$ , and let $P$ be the midpoint of $\overline{BD}$ . Given that $OP=11$ , the length of $AD$ can be written in the form $m\sqrt{n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. What is $m+n$ ?

$\textbf{(A) }65 \qquad \textbf{(B) }132 \qquad \textbf{(C) }157 \qquad \textbf{(D) }194\qquad \textbf{(E) }215$

Pick an answer.

(A)

(B)

132

(C)

157

(D)

194

(E)

215

View mode:

Toolkit + CCSS Solution

Understand

Restated: In trapezoid $ABCD$, the bases satisfy $AB \parallel CD$, the slanted sides give $BC = CD = 43$, and the diagonal $BD$ is perpendicular to side $AD$. The diagonals $AC$ and $BD$ meet at $O$, and $P$ is the midpoint of $BD$. Given $OP = 11$, write $AD = m\sqrt{n}$ in simplest radical form and find $m + n$.

Givens: Trapezoid $ABCD$ with $AB \parallel CD$; $BC = CD = 43$ (so $\triangle BCD$ is isosceles with base $BD$); $AD \perp BD$ (so $\angle ADB = 90°$); $O = AC \cap BD$ and $P$ = midpoint of $BD$; $OP = 11$; Choices: (A) $65$, (B) $132$, (C) $157$, (D) $194$, (E) $215$

Unknowns: $m + n$, where $AD = m\sqrt{n}$ in simplest radical form

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems, #13 Convert to Algebra

The wording packs four facts (parallel sides, isosceles, perpendicular, midpoint) into one figure — Tool #1 (Diagram) keeps them straight and reveals the key right triangle $\triangle CPD$. Tool #7 (Subproblems) splits the question into three smaller ones: (a) find $AB$ via similar triangles $\triangle BDA \sim \triangle BPC$; (b) find $BD$ via the diagonal ratio $BO:OD = AB:CD = 2:1$ combined with $OP = 11$; (c) find $AD$ via Pythagorean theorem on $\triangle ADB$. Tool #13 (Algebra) handles the linear equation $\tfrac{x}{2} = 11$ that ties step (b) together.

Execute — Answer: D

#1 Draw a Diagram 4.G.A.2 Step 1

Draw the trapezoid with $AB$ on top, $CD$ on the bottom, and mark $BC = CD = 43$.
Since $BC = CD$, triangle $BCD$ is isosceles with base $BD$.
The midpoint $P$ of $BD$ is the foot of the altitude from $C$, so $CP \perp BD$ — that is, $\angle CPB = 90°$.

$$CP \perp BD,\ \angle CPB = 90°$$

💡 In an isosceles triangle, the line from the apex to the midpoint of the base is perpendicular to the base.

#7 Identify Subproblems 8.G.A.4 Step 2

Because $AB \parallel CD$, the diagonal $BD$ is a transversal, so the alternate interior angles satisfy $\angle ABD = \angle BDC$.
In isosceles $\triangle BCD$, the base angles are equal: $\angle DBC = \angle BDC$.
Combining, $\angle ABD = \angle DBC$.
Together with $\angle ADB = \angle CPB = 90°$, triangles $BDA$ and $BPC$ share two equal angles, so they are similar.

$$\triangle BDA \sim \triangle BPC$$

💡 Two right triangles sharing another equal angle must be similar (AA).

#7 Identify Subproblems 7.RP.A.2 Step 3

The similarity gives matching ratios $\tfrac{AB}{BC} = \tfrac{BD}{BP}$.
Since $P$ is the midpoint of $BD$, we have $BD = 2\,BP$, so the ratio equals $2$.
Therefore $AB = 2 \cdot BC = 2 \cdot 43 = 86$.

$$AB = 2 \cdot 43 = 86$$

💡 Doubled hypotenuse means doubled corresponding side.

#1 Draw a Diagram 8.G.A.4 Step 4

Since $AB \parallel CD$, the diagonals of the trapezoid cut each other in the ratio of the parallel sides: $\tfrac{BO}{OD} = \tfrac{AB}{CD} = \tfrac{86}{43} = 2$.
So if $OD = x$, then $BO = 2x$ and $BD = 3x$.
The midpoint $P$ of $BD$ satisfies $DP = \tfrac{3x}{2}$, and the order along the diagonal is $D, O, P, B$, so $OP = DP - DO = \tfrac{3x}{2} - x = \tfrac{x}{2}$.

$OP = \tfrac{x}{2}$, where $OD = x$

💡 Parallel bases make the two triangles meeting at $O$ similar; the side ratio sets the diagonal split.

#13 Convert to Algebra 6.EE.B.7 Step 5

Plug in $OP = 11$: $\tfrac{x}{2} = 11$, so $x = 22$.
Then $BD = 3x = 66$.

$$x = 22,\quad BD = 66$$

💡 A single linear equation in one unknown — solve and read off $BD$.

#7 Identify Subproblems 8.G.B.7 Step 6

Triangle $ADB$ has a right angle at $D$, so apply the Pythagorean theorem with legs $AD,\ BD$ and hypotenuse $AB$.
$AD^2 = 86^2 - 66^2 = (86-66)(86+66) = 20 \cdot 152 = 3040$.

$$AD^2 = 86^2 - 66^2 = 20 \cdot 152 = 3040$$

💡 Difference of squares avoids squaring big numbers.

#13 Convert to Algebra 8.EE.A.2 Step 7

Simplify $\sqrt{3040}$.
Factor: $3040 = 16 \cdot 190$, and $190 = 2 \cdot 5 \cdot 19$ is square-free.
So $AD = 4\sqrt{190}$, giving $m = 4$ and $n = 190$.
Therefore $m + n = 4 + 190 = 194$ — choice (D).

$$AD = 4\sqrt{190},\ m + n = 194 \Rightarrow \textbf{(D)}$$

💡 Pull the largest perfect square out of the radicand and add.

[1] #1 4.G.A.2 Draw the trapezoid with $AB$ on top, $CD$ on the bottom, and mark $BC = CD = 43$

[2] #7 8.G.A.4 Because $AB \parallel CD$, the diagonal $BD$ is a transversal, so the alternate

[3] #7 7.RP.A.2 The similarity gives matching ratios $\tfrac{AB}{BC} = \tfrac{BD}{BP}$. Since $P

[4] #1 8.G.A.4 Since $AB \parallel CD$, the diagonals of the trapezoid cut each other in the ra

[5] #13 6.EE.B.7 Plug in $OP = 11$: $\tfrac{x}{2} = 11$, so $x = 22$. Then $BD = 3x = 66$.

[6] #7 8.G.B.7 Triangle $ADB$ has a right angle at $D$, so apply the Pythagorean theorem with l

[7] #13 8.EE.A.2 Simplify $\sqrt{3040}$. Factor: $3040 = 16 \cdot 190$, and $190 = 2 \cdot 5 \cdo

Review

Reasonableness: Quick sanity: $AB = 86$ and $BD = 66$, so $AD = \sqrt{86^2 - 66^2} \approx \sqrt{3040} \approx 55.1$ — a positive length shorter than the hypotenuse $86$, exactly as a leg should be. And $4\sqrt{190} \approx 4 \cdot 13.78 \approx 55.1$ matches. The square-free check on $190 = 2 \cdot 5 \cdot 19$ confirms the form is minimal, so $m + n = 194$ is the intended choice (D).

Alternative: Tool #13 (Convert to Algebra) with coordinates: place $D$ at the origin with $BD$ along the positive $x$-axis. Then $B = (BD, 0)$, $A = (0, h)$ since $AD \perp BD$, and $C$ lies on the perpendicular bisector of $BD$ with $CD = 43$. Use $AB \parallel CD$ to relate slopes, plug in $OP = 11$, and solve for $h = AD$. The Pythagorean step is then automatic from the coordinates.

CCSS standards used (min grade 8)

4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines (Recognizing the isosceles triangle $\triangle BCD$ and the perpendicular altitude $CP$ from its apex to its base.)
8.G.A.4 Understand that a two-dimensional figure is similar to another using transformations (Establishing $\triangle BDA \sim \triangle BPC$ from the right angle plus the shared/alternate angle, and the diagonal-split similarity $\triangle ABO \sim \triangle CDO$.)
7.RP.A.2 Recognize and represent proportional relationships between quantities (Reading off $AB = 2 \cdot BC = 86$ from the similarity ratio $\tfrac{BD}{BP} = 2$.)
6.EE.B.7 Solve real-world problems by writing and solving equations of the form px = q (Solving the linear equation $\tfrac{x}{2} = 11$ to get $x = 22$ and then $BD = 66$.)
8.G.B.7 Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Computing $AD^2 = AB^2 - BD^2 = 86^2 - 66^2 = 3040$ in the right triangle $\triangle ADB$.)
8.EE.A.2 Use square root and cube root symbols to represent solutions (Simplifying $\sqrt{3040} = 4\sqrt{190}$ to extract $m = 4$ and $n = 190$.)

⭐ Once you spot the right angle inside the isosceles triangle $\triangle BCD$, the rest is plug-and-chug: a similar-triangle ratio gives $AB = 86$, the diagonal split $BO:OD = 2:1$ plus $OP = 11$ gives $BD = 66$, and Pythagoras on $\triangle ADB$ gives $AD = 4\sqrt{190}$, so $m + n = \textbf{(D) }194$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 8)

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