AMC 10 · 2009 · #17

Grade 8 geometry-2d
Archetype Similar-Triangle Length / Area Ratio └ Sub-archetype Synthetic Similar-Triangle Length or Area
similar-trianglespythagorean-theorem identify-subproblems ↑ Prerequisites: similar-triangles
📏 Medium solution 💡 3 insights
Problem
Rectangle ABCD has AB = 4 and BC = 3. Segment EF is drawn through B perpendicular to the diagonal DB, and its endpoints are placed so that A lies on segment DE and C lies on segment DF. What is the length of EF?

Pick an answer.

(A)
$\ 9$
(B)
$\ 10$
(C)
$\ \frac {125}{12}$
(D)
$\ \frac {103}{9}$
(E)
$\ 12$

AMC 10 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

EF is hard to measure directly, but B splits it into two pieces EB and BF. Each piece lives in its own right triangle that is similar to a triangle made from the rectangle. So the plan is: draw the figure, find the diagonal, then solve two small similar-triangle subproblems and add the pieces.

1STEP 1

Find the diagonal DB

Diagonal DB cuts off right triangle DAB with legs DA = 3 and AB = 4, so the Pythagorean theorem gives DB = 5.

DB=√(AB²+DA²)=√(4²+3²)=√(25)=5
2STEP 2

Split EF at B

Since angle EDF = angle ADC = 90 degrees, triangle DEF is right-angled at D with altitude DB, and B splits EF into EB + BF.

∠ EDF=∠ ADC=90°, EF=EB+BF
3STEP 3

Similar triangles give EB

Triangles DAB and DBE share angle D and each has a right angle, so they are similar: EBAB=DBDA\frac{EB}{AB} = \frac{DB}{DA} gives EB=203EB = \frac{20}{3}.

△ DAB ∼ △ DBE → EB/AB=DB/DA → EB=(4 · 5)/3=20/3
4STEP 4

Similar triangles give BF

The same pairing on the other side, triangles DCB and DBF, gives BFCB=DBDC\frac{BF}{CB} = \frac{DB}{DC}, so BF=154BF = \frac{15}{4}.

△ DCB ∼ △ DBF → BF/CB=DB/DC → BF=(3 · 5)/4=15/4
5STEP 5

Add the two pieces

Over a common denominator, 203+154=80+4512\frac{20}{3} + \frac{15}{4} = \frac{80+45}{12}, so EF=12512EF = \frac{125}{12}, choice (C).

EF=20/3+15/4=80/12+45/12=125/12
Answer
125/12
EF = 125/12 is about 10.4, which is a bit longer than the diagonal DB = 5 and sensibly larger than either side of the rectangle, since EF stretches beyond both A and C. Choice (C) = 125/12 fits, and the nearby traps 10 and 103/9 are close decoys that the exact fractions rule out.
💡Key takeaway

Cut the mystery segment where the diagonal hits it, match each half to a look-alike right triangle, and add the two halves.

  • Find the diagonal DB
  • Split EF at B
  • Similar triangles give EB
  • Similar triangles give BF
  • Add the two pieces

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