AMC 10 · 2012 · #17
Grade 10 geometry-2dPick an answer.
The entire configuration is controlled by a single number, the tilt of one pair of sides, so name it (tool #4). A sketch first (tool #1) sorts the four given points into the two parallel families: (3,0) and (5,0) sit on one pair of opposite side-lines, a gap of 2 apart on the x-axis, while (7,0) and (13,0) sit on the other pair, a gap of 6 apart. Because the two families are perpendicular, the same tilt controls both gaps, and each gap converts into the same side length s — one equation in one unknown (tool #13). Then, instead of hunting the four vertices, chase the center directly (tool #16): it must lie on the midline of each pair of opposite sides, and those midlines pass through the gap midpoints (4,0) and (10,0). The tilt equation has two roots, so finish by eliminating the wrong one with the first-quadrant condition and exhibiting the surviving square explicitly (tool #3).
Sort the four points into pairs
The four points sort into two pairs.
The letters P, Q, R, S quietly tell you which two of the given points sit on parallel walls of the square.
10.G-CO.A.1Draw A DiagramRule out axis-parallel sides
A square with sides on the axis is impossible.
If one pair of sides were axis-parallel, the other pair would have to be the x-axis twice over, which no square permits.
10.G-CO.A.1Eliminate PossibilitiesTurn each gap into the side length
Each gap gives the side through the tilt.
A gap measured flat along the x-axis shrinks to the true distance between two tilted lines by the sine of the tilt.
A gap measured flat along one axis shrinks to the true distance between two tilted lines by the sine of the tilt.
▸ Why?
The flat gap is the hypotenuse of a right triangle whose far leg is the perpendicular distance.
▸ Why?
Parallel lines keep a constant perpendicular gap, so one measurement stands for the whole pair.
Solve for the tilt
Matching them solves for that tilt.
Two perpendicular pairs of parallel lines always box in a rectangle, so "equal spacing" is precisely the whole square condition — nothing more to check.
9.A-REI.B.3Convert To AlgebraChase the center, not the corners
The centre sits on two middle lines.
The midpoints of the two x-axis gaps sit on the square's own center-lines, so the center is simply where those center-lines cross.
10.G-GPE.B.4Change Focus Count The ComplementIntersect the two midlines
Intersecting them gives the centre.
Two straight lines cross exactly once, so one small system pins the center down completely.
8.EE.C.8Convert To AlgebraKill the mirror case and verify
Its coordinates add to 32/5, choice (D).
Two mirror squares fit the four points equally well; the first-quadrant clause is the tiebreaker, and writing the vertices out proves the winner truly exists.
8.G.B.8Eliminate PossibilitiesWhen four lines pin a shape down, don't chase the corners: find the one tilt that makes both gaps shrink to the same side length, then meet the two center-lines.
- Sort the four points into pairs
- Rule out axis-parallel sides
- Turn each gap into the side length
- Solve for the tilt
- Chase the center, not the corners
- Intersect the two midlines
- Kill the mirror case and verify