AMC 10 · 2019 · #6
Grade 8 geometry-2dPick an answer.
Plant AB on a coordinate axis (Tool #1 + #9 makes the picture concrete). Split the requirements into (a) area → height of C above AB, and (b) perimeter → sum of slanted sides AC + BC (Tool #7 sub-questions). Then test whether ANY C at the required height can give the required slant sum — try the best (closest) candidate first (Tool #6); if even that fails, no C works.
Set up coordinates
Place the two points symmetrically.
Setting AB on the x-axis turns geometry into easy distance calculations.
8.G.B.8Draw A DiagramArea fixes the height
The area pins the height.
Area = 1/2 · base · height pins down how far C is from AB.
With the base fixed, the area pins down exactly how far the third corner sits from that base.
▸ Why?
An area is half the base times the height, so with the base known the height is forced.
▸ Why?
Every corner at that height gives the same area, so the corner is free to slide along one line.
The other two sides
The perimeter fixes their sum.
The two slanted sides must add to 40.
6.EE.B.7Identify SubproblemsFind the smallest possible sum
Find the minimum at that height.
The symmetric point above the base is usually the trickiest — easiest place to test the perimeter.
8.G.B.8Guess And CheckCompare the two
The minimum is already larger than required.
Even the smallest possible |AC|+|BC| overshoots the budget.
8.NS.A.2Guess And CheckCheck every position
No position can satisfy both.
Going further left or right only stretches the slanted distances — symmetric point is the tightest.
8.G.B.7Solve An Easier Related ProblemCount them
So the answer is 0.
If even the best candidate fails, no point works — answer is 0.
K.MD.B.3Eliminate PossibilitiesThis AMC 12 problem only needs Grade 8 Pythagorean thinking you already know: area =100 forces C to sit 20 above (or below) AB. Even the closest such C gives slants summing to 10√(17)≈ 41.2 ＞ 40, so the perimeter 50 is impossible — 0 points.