1953 AHSME Problems/Problem 32

Each angle of a rectangle is trisected. The intersections of the pairs of trisectors adjacent to the same side always form:

$\textbf{(A)}\ \text{a square} \qquad \textbf{(B)}\ \text{a rectangle} \qquad \textbf{(C)}\ \text{a parallelogram with unequal sides}\\ \textbf{(D)}\ \text{a rhombus}\qquad \textbf{(E)}\ \text{a quadrilateral with no special properties}$

Solution

Draw a diagram:

$[asy] unitsize(1cm); pair A, B, C, D; A = (0,0); B = (0,3); C = (4.5,3); D = (4.5,0); draw(A--B--C--D--cycle); pair Aa, Ab, Ba, Bb; Aa = A + (D-A)*dir(30); Ab = A + (D-A)*dir(60); Ba = B + (C-B)*dir(-30); Bb = B + (C-B)*dir(-60); draw(Aa--A--Ab, red); draw(Ba--B--Bb, red); dot(intersectionpoint(A--Aa, B--Ba), blue); dot(intersectionpoint(A--Aa, B--Bb), blue); dot(intersectionpoint(A--Ab, B--Ba), blue); dot(intersectionpoint(A--Ab, B--Bb), blue); [/asy]$

By looking at these intersections (marked here in blue), we see that these four points form a parallelogram with no special properties. Therefore, the correct answer is $\boxed{\text{C}}$ , and we are done.

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1953 AHSME Problems/Problem 32

Solution