1965 AMC 12 Problem 25

Attempt Problem 25 of the 1965 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 1965 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

25.

Let ABCDABCD be a quadrilateral with ABAB extended to EE so that AB=BE.AB=BE. Lines ACAC and CECE are drawn to form angle ACE.ACE. For this angle to be a right angle it is necessary that quadrilateral ABCDABCD have:
all angles equal
all sides equal
two pairs of equal sides
one pair of equal sides
one pair of equal angles
Answer: D
Concepts:right trianglemidpointcircumcircle, circumcenter, and circumradius
Small hint:
Because AB=BE,AB=BE, point BB is the midpoint of AEAE
Big hint:
In a right triangle, the midpoint of the hypotenuse is equidistant from all three vertices
Solution:
If ∠ACE=90∘,\angle ACE=90^\circ, then AEAE is the hypotenuse of right triangle ACE.ACE. Its midpoint BB is equidistant from A,C,E.A,C,E. Thus AB=BC,AB=BC, so quadrilateral ABCDABCD necessarily has at least one pair of equal sides. No condition involving DD follows.
Therefore, the correct answer is D.
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