AMC 10 · 2007 · #25
Grade 10 geometry-3dPick an answer.
A bent 3-D loop is very hard to picture, and picturing it is also the dangerous part: if you sketch one shape that happens to fit and measure it, you have only shown that shape is possible — not that it is the only one. If a second, different shape also fit, the problem would have no single answer. So the safe move is Tool #13 (Convert to Algebra): drop in coordinates and turn every single condition into an equation, then solve the system instead of guessing at a picture. The square corner at B hands you a ready-made frame with B at the origin and BA, BC along two axes (Tool #17 to see why that frame is legal), the parallel condition becomes "D and E have the same height" (Tool #4 names that height h), and each right angle becomes a Pythagorean equation (Tool #7 splits the loop into those separate small equations). Solving the system leaves only a couple of possibilities, and they all give the same triangle.
Set the frame at the square corner
Setting the frame at a right angle gives clean coordinates.
A right angle with two known legs is a built-in pair of axes, so let it be the axes.
10.G-GPE.B.4Visualize Spatial RelationshipsParallel means one shared height
Parallel means the two points share one height.
Parallel to the floor means level with the floor: same height at both ends.
10.G-CO.A.1Introduce A VariableTurn the right angles into lengths
Each right angle becomes a length through the Pythagorean theorem.
A right angle is the same information as one extra length, and lengths are easier to write down.
8.G.B.6Identify SubproblemsWrite the equations for E and D
Writing the equations ties each point to one relation.
Subtracting two squared-distance equations kills all the squares and leaves a straight line's worth of information.
Subtracting two squared-distance equations kills all the squares and leaves a straight line's worth of information.
▸ Why?
Both equations carry the identical squared block, so the subtraction removes it entirely.
▸ Why?
Each squared distance is the sum of the squared coordinate gaps, which is where that shared block comes from.
DE has only two places to go
The shared segment has only two possible directions.
Once every other rule is used up, the last length equation has only two solutions — DE must lie right on top of a side you already drew.
9.A-REI.B.4Convert To AlgebraFinish the first case exactly
Following the first case gives every coordinate.
With the direction of DE known, the remaining equations pin the points to one exact spot.
8.G.B.8Introduce A VariableRecognize the triangle and finish
The lengths reveal another right angle, giving area 2, choice (C).
Sides 2,2,2√(2) can only be half of a square, and half of a 2 × 2 square has area 2.
6.G.A.1Identify SubproblemsWhen a shape in space is too tangled to draw, put it on coordinate axes and turn every length and right angle into an equation — here the equations leave DE only two possible directions, and both give the same right triangle with legs 2 and 2.
- Set the frame at the square corner
- Parallel means one shared height
- Turn the right angles into lengths
- Write the equations for E and D
- DE has only two places to go
- Finish the first case exactly
- Recognize the triangle and finish