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:

$[asy] unitsize(1cm); pair A, B, C, D; A = (0,0); B = (0,3); C = (4.5,3); D = (4.5,0); draw(A--B--C--D--cycle); pair Ab, Ba, Bc, Cb, Cd, Dc, Da, Ad; Ab = A + 0.5*(D-A)*dir(60); Ba = B + 0.5*(D-A)*dir(-60); Bc = B + 0.7*(D-A)*dir(-30); Cb = C + 0.7*(D-A)*dir(210); Cd = C + 0.5*(D-A)*dir(240); Dc = D + 0.5*(D-A)*dir(120); Da = D + 0.7*(D-A)*dir(150); Ad = A + 0.7*(D-A)*dir(30); draw(A--Ab, red); draw(A--Ad, red); draw(B--Ba, red); draw(B--Bc, red); draw(C--Cb, red); draw(C--Cd, red); draw(D--Da, red); draw(D--Dc, red); dot(intersectionpoint(A--Ab, B--Ba), blue); dot(intersectionpoint(B--Bc, C--Cb), blue); dot(intersectionpoint(C--Cd, D--Dc), blue); dot(intersectionpoint(D--Da, A--Ad), blue); [/asy]$

By looking at these intersections (marked here in blue), we see that these four points form a rhombus. Therefore, the correct answer is $\boxed{\text{D}}$ , and we are done.

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1953 AHSME Problems/Problem 32

Solution