AMC 10 · 2021 · #24
Grade 11 geometry-2d
Pick an answer.
Tool #4 (Introduce a Variable) — the whole figure is pinned down by just two numbers: the sizes of the two diagonals and the angle θ between them. Name them and every given becomes an equation. Tool #1 (Draw a Diagram) — the picture shows the diagonals crossing at a center point E; that center is the hinge of the whole solution, because Q is the mirror image of P through E and S is the mirror image of R. Tool #13 (Convert to Algebra) — the area condition, the PQ condition, and the RS condition become three algebra statements in two unknowns. Tool #15 (Organize Information in More Ways) — dividing the PQ equation by the RS equation makes the shared cosθ cancel and instantly hands over the ratio AC : BD. Tool #16 (Change Focus) — instead of hunting for θ, use sin²θ + cos²θ = 1 to erase θ entirely and solve a plain quadratic in the diagonal length.
Use the centre
The centre halves both distances.
A parallelogram looks identical after a half-turn about its center, so every point and its partner sit at equal distances from that center.
A parallelogram looks identical after a half turn about its centre, so partners sit at equal distances from it.
▸ Why?
A rotation moves the figure onto itself without stretching, so matched points keep matched distances.
▸ Why?
Half of a full turn is a straight angle, so each point ends up directly opposite where it began.
Turn projections into cosines
Each is a diagonal times cosine.
Dropping a perpendicular onto a line is exactly the "multiply by cos of the angle" move — the shadow of a segment on a line is its length times the cosine of the angle it makes with that line.
10.G-SRT.C.6Introduce A VariableGet the diagonal ratio
Dividing cancels the cosine.
Two quantities scaled by the same unknown factor have a ratio that forgets the factor — so divide instead of solve.
9.A-SSE.A.2Organize Information In More WaysGet the sine from the area
The area supplies the sine.
The area formula 1/2absin C applied to the four corner triangles rebuilds the whole parallelogram out of its diagonals and their crossing angle.
11.G-SRT.D.9Convert To AlgebraErase the angle
The identity erases the angle.
When an unwanted angle shows up once as a sine and once as a cosine, sin²θ + cos²θ = 1 is the eraser.
11.F-TF.C.8Change Focus Count The ComplementSolve the quadratic
Solve the leftover quadratic.
A quartic that only uses even powers is a quadratic in disguise — substitute v = k² and the standard formula finishes it.
9.A-REI.B.4Introduce A VariableRead the three numbers and add
Adding gives 81.
The last step is bookkeeping: force the expression into the exact requested form and add up the pieces.
9.A-SSE.A.1Organize Information In More WaysEvery projection onto a line is just "length times cosine of the angle," so PQ = ACcosθ and RS = BDcosθ — and dividing those two makes the angle vanish, handing you AC/BD = 6/8 = 3/4 for free. With AC = 3k and BD = 4k, the area condition and sin²θ + cos²θ = 1 collapse into one quadratic, giving d² = 32 + 8√(41) and the answer 32 + 8 + 41 = 81, choice (A).
- Use the center of the parallelogram
- Turn each projection into cosine
- Divide to get the diagonal ratio
- Write the area with the diagonals
- Erase the angle
- Solve the quadratic in k²
- Read off m, n, and p