AMC 10 · 2017 · #10
Grade 8 arithmeticPick an answer.
AMC 10 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Tool #4 (Introduce a Variable): a, b, c are all unknown, so I keep them as letters and let the conditions pin them down. Tool #7 (Identify Subproblems): the two facts give two separate handles — 'perpendicular' controls the slopes, and 'passes through (1,-5)' controls the constants — so I solve each piece and then combine. Tool #13 (Convert to Algebra): I turn each geometric statement into an equation and finish with ordinary algebra.
Find each slope
Solve each equation for y: ax-2y=c gives slope a/2, and 2x+by=-c gives slope -2/b.
Solving each equation for y exposes its slope as the coefficient of x.
8.EE.B.6Use Matrix LogicUse perpendicular
Perpendicular means the slopes multiply to -1: a/2·(-2/b)=-1, so -a/b=-1, which simplifies to a=b.
The perpendicular condition turns into one clean equation linking a and b.
8.EE.C.7Convert To AlgebraPlug in the point
Substitute (1,-5) into both lines: line 1 gives a+10=c, and line 2 gives c=5b-2.
A point on a line makes its equation true, so substitution gives real equations in the unknowns.
6.EE.B.5Identify SubproblemsSolve the system
With a=b, the two expressions for c meet: a+10=5a-2, so 12=4a and a=3.
Two different expressions for the same c must be equal, which solves for a.
8.EE.C.8Identify SubproblemsRead off c
Put a=3 into c=a+10: c=3+10=13, which is choice (E).
With a known, either equation for c delivers the answer.
8.EE.C.8Use Matrix LogicRead each line's slope, use 'perpendicular means slopes multiply to -1' to link a and b, then plug the shared point into both equations and solve — that gives c=13, choice (E).
- Find each slope
- Use perpendicular
- Plug in the point
- Solve the system
- Read off c