AMC 10 · 2021 · #17
Grade 8 geometry-2dProblem
Trapezoid ABCD has AB∥CD,BC=CD=43, and AD⊥BD. Let O be the intersection of the diagonals AC and BD, and let P be the midpoint of BD. Given that OP=11, the length of AD can be written in the form mn, where m and n are positive integers and n is not divisible by the square of any prime. What is m+n?
Pick an answer.
AMC 10 2021 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 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
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) $215$
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
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, $\angle CPB = 90°$.
💡 In an isosceles triangle, the line from the apex to the midpoint of the base is perpendicular to the base.
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.
💡 Two right triangles sharing another equal angle must be similar (AA).
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 $AB = 2 \cdot BC = 2 \cdot 43 = 86$.
💡 Doubled hypotenuse means doubled corresponding side.
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 $OP = DP - DO = \tfrac{3x}{2} - x = \tfrac{x}{2}$.
💡 Parallel bases make the two triangles meeting at $O$ similar; the side ratio sets the diagonal split.
6.EE.B.7 Step 5 - Plug in $OP = 11$: $\tfrac{x}{2} = 11$, so $x = 22$.
- Then $BD = 3x = 66$.
💡 A single linear equation in one unknown — solve and read off $BD$.
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 $AB$.
- $AD^2 = 86^2 - 66^2 = (86-66)(86+66) = 20 \cdot 152 = 3040$.
💡 Difference of squares avoids squaring big numbers.
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).
💡 Pull the largest perfect square out of the radicand and add.
4.G.A.2 Draw the trapezoid with $AB$ on top, $CD$ on the bottom, and mark $BC = CD = 43$ 8.G.A.4 Because $AB \parallel CD$, the diagonal $BD$ is a transversal, so the alternate 7.RP.A.2 The similarity gives matching ratios $\tfrac{AB}{BC} = \tfrac{BD}{BP}$. Since $P 8.G.A.4 Since $AB \parallel CD$, the diagonals of the trapezoid cut each other in the ra 6.EE.B.7 Plug in $OP = 11$: $\tfrac{x}{2} = 11$, so $x = 22$. Then $BD = 3x = 66$. 8.G.B.7 Triangle $ADB$ has a right angle at $D$, so apply the Pythagorean theorem with l 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.2Classify 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.4Understand 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.2Recognize and represent proportional relationships between quantities (Reading off $AB = 2 \cdot BC = 86$ from the similarity ratio $\tfrac{BD}{BP} = 2$.)6.EE.B.7Solve 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.7Apply 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.2Use 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 $m + n = \textbf{(D) }194$.
⭐ 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 $m + n = \textbf{(D) }194$.
More like this
Same archetype — closest grade level first.
