1956 AMC 12 Problem 39

Attempt Problem 39 of the 1956 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 1956 AMC 12 solutions, or check the answer key.

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39.

The hypotenuse cc and one arm aa of a right triangle are consecutive integers. The square of the second arm is:

caca

ca\dfrac ca

c+ac+a

cac-a

none of these

Answer: C
Concepts:Pythagorean theoremdifference of squaresconsecutive integers
Difficulty rating: 1470
Small Hint:

If the second arm is b,b, then b2=c2a2b^2=c^2-a^2

Big Hint:

Factor the difference of squares and use ca=1c-a=1

Solution:

By the Pythagorean theorem, b2=c2a2=(ca)(c+a). b^2=c^2-a^2=(c-a)(c+a). Since cc and aa are consecutive and c>a,c\gt a, we have ca=1.c-a=1. Hence b2=c+a.b^2=c+a.

Thus, the correct answer is C.

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Problem 39 in Other Years

1950 AMC 12 · 1951 AMC 12 · 1952 AMC 12 · 1953 AMC 12 · 1954 AMC 12 · 1955 AMC 12 · 1957 AMC 12 · 1958 AMC 12 · 1959 AMC 12 · 1960 AMC 12 · 1961 AMC 12 · 1962 AMC 12 · 1963 AMC 12 · 1964 AMC 12 · 1965 AMC 12 · 1966 AMC 12 · 1967 AMC 12