Competition · AMC preparation · step 4 of 4

AMC 10 · 2013B · #23

Grade 8 geometry-2d

coordinate-geometrypythagorean-theoreminteger-pythagorean-triples convert-to-algebraidentify-subproblems ↑ Prerequisites: pythagorean-theoremcoordinate-geometry

📏 Long solution 💡 3 insights

Problem

In triangle

ABC

AB=13

BC=14

CA=15

. Distinct points

D

E

F

lie on segments

BC

CA

DE

respectively, with

AD \perp BC

DE \perp AC

, and

AF \perp BF

. The length

DF

can be written as

\frac{m}{n}

, where

m

and

n

are relatively prime positive integers. What is

m+n

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

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.

1STEP 1

Split the base with the altitude

Drop the altitude: $x^2+h^2=13^2$ and $(14-x)^2+h^2=15^2$ give $BD=5$ , $DC=9$ , $AD=12$ .

BD=5, DC=9, AD=12

2STEP 2

Put the figure on axes

Set $D=(0,0)$ with $BC$ on the $x$ -axis: $B=(-5,0)$ , $C=(9,0)$ , $A=(0,12)$ — the altitude is now the $y$ -axis.

D=(0,0), B=(-5,0), C=(9,0), A=(0,12)

3STEP 3

Find the direction of line DE

$AC$ has slope $-\frac{4}{3}$ , so $DE \perp AC$ has the negative reciprocal: $DE$ is $y=\frac{3}{4}x$ through $D$ .

slope(AC)=-4/3 → slope(DE)=3/4, DE: y=3/4x

4STEP 4

Ride distance t along DE

Let $t=DF$ ; stepping $t$ from $D$ along the unit vector $(\frac{4}{5},\frac{3}{5})$ lands at $F=(\frac{4t}{5},\frac{3t}{5})$ .

F=(4t/5, 3t/5), DF=t

5STEP 5

Turn the right angle at F into an equation

$AF \perp BF$ means $FA \cdot FB=0$ : $\frac{4t}{5}(\frac{4t}{5}+5)+(\frac{3t}{5}-12)\frac{3t}{5}=0$ , which collapses to $t^2-\frac{16t}{5}=0$ .

FA·FB=0 → t²-16t/5=0

6STEP 6

Solve and pick the valid point

$t(t-\frac{16}{5})=0$ gives $t=0$ (point $D$ , rejected) or $DF=\frac{16}{5}$ , which fits inside $DE=\frac{36}{5}$ ; so $m+n=21$ .

t(t-16/5)=0 → DF=16/5, m+n=16+5=21 → (B)

Answer

The two roots t=0 and t=16/5 have a clean meaning: both D and F lie on the circle with diameter AB (every point on that circle sees AB at 90°), and line DE cuts that circle in exactly those two points. Rejecting D leaves F at distance 16/5=3.2 from D, comfortably inside DE=36/5=7.2, so F sits on the segment as required. The fraction 16/5 is already reduced, giving m+n=21 — answer (B), which is on the list.

💡Key takeaway

Pin 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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