1960 AMC 12 Problem 15

Attempt Problem 15 of the 1960 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1960 AMC 12 solutions, or check the answer key.

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15.

Triangle I is equilateral with side

A,

perimeter

P,

area

K,

and circumradius

R

(radius of the circumscribed circle). Triangle II is equilateral with side

a,

perimeter

p,

area

k,

and circumradius

r.

A

is different from

a,

then:

P:p=R:r

only sometimes

P:p=R:r

always

P:p=K:k

only sometimes

R:r=K:k

always

R:r=K:k

only sometimes

All equilateral triangles are similar

Perimeter and circumradius both scale linearly with side length

For equilateral triangles,

P=3A

and

R=\frac{A}{\sqrt3},

with analogous formulas for

p

and

r.

Therefore

P:p=A:a=R:r

for every such pair of triangles.

Thus, the correct answer is B.

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