AMC 10 · 2013 · #19
Grade 8 geometry-2dPick an answer.
Three stacked perpendiculars scream 'set up coordinates.' Tool #1 (Diagram): sketch the 13-14-15 triangle, drop the altitude AD, and pin the figure to axes so every 'perpendicular' becomes a slope condition. Tool #7 (Subproblems): first pin down D and A from the altitude, then find the line DE, then locate F on it. Tool #4 (Introduce a Variable): let t=DF and ride the line DE a distance t from D. Tool #13 (Convert to Algebra): the right angle at F (that AF ⊥ BF) turns into one equation in t.
Split the base with the altitude
The altitude splits the base into whole pieces.
The altitude chops the big triangle into two right triangles that share the same height, so Pythagoras pins down every length.
8.G.B.7Identify SubproblemsPut the figure on axes
Axes at that foot make every corner whole.
Anchoring the right angle at the origin lets each 'perpendicular' in the problem show up as a slope you can read off.
5.G.A.2Draw A DiagramFind the direction of line DE
The second perpendicular fixes a direction.
Perpendicular lines have slopes that multiply to -1, so flipping and negating -4/3 gives the direction F must travel from D.
Perpendicular lines have slopes that multiply to minus one, so flipping and negating gives the new direction.
▸ Why?
A quarter turn sends a direction to one at right angles, which shows up exactly as that flip and sign change.
▸ Why?
That right angle also lets the picture's lengths be tied together by the squares on the two legs.
Ride distance t along DE
One distance parametrises the whole line.
Because (4,3) has length 5, scaling it by t/5 moves you exactly t units, so the variable t is literally the distance we want.
8.G.B.8Introduce A VariableTurn the right angle at F into an equation
The right angle becomes one equation.
A right angle at F says the two legs are perpendicular, which is exactly a zero dot product — one clean equation in the single unknown t.
8.EE.C.7Convert To AlgebraSolve and pick the valid point
The nonzero root gives 21, choice (B).
The equation offers two right-angle points; we throw out D (the forbidden duplicate) and keep the real F, then reduce the fraction to read off m+n.
6.NS.B.4Introduce A VariablePin the 13-14-15 triangle to axes so every 'perpendicular' becomes a slope; walk distance t up line DE to reach F, and the right angle at F gives one equation t²-16/5t=0. Toss out the root that is just point D, keep DF=16/5, so m+n=21 — choice (B). It looks like a hard geometry problem, but coordinates turn it into Grade 8 algebra.
- Split the base with the altitude
- Put the figure on axes
- Find the direction of line DE
- Ride distance t along DE
- Turn the right angle at F into an equation
- Solve and pick the valid point
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