Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1953 AHSME Problems/Problem 32"

(Created page with solution.)
 
(→‎Solution)
 
Line 18: Line 18:
 
draw(A--B--C--D--cycle);
 
draw(A--B--C--D--cycle);
  
−
pair Aa, Ab, Ba, Bb;
+
pair Ab, Ba, Bc, Cb, Cd, Dc, Da, Ad;
−
Aa = A + (D-A)*dir(30);
+
Ab = A + 0.5*(D-A)*dir(60);
−
Ab = A + (D-A)*dir(60);
+
Ba = B + 0.5*(D-A)*dir(-60);
−
Ba = B + (C-B)*dir(-30);
+
Bc = B + 0.7*(D-A)*dir(-30);
−
Bb = B + (C-B)*dir(-60);
+
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);
  
−
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--Ba), blue);
−
dot(intersectionpoint(A--Ab, B--Bb), blue);
+
dot(intersectionpoint(B--Bc, C--Cb), blue);
 +
dot(intersectionpoint(C--Cd, D--Dc), blue);
 +
dot(intersectionpoint(D--Da, A--Ad), blue);
 
</asy>
 
</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 <math>\boxed{\text{C}}</math>, and we are done.
+
By looking at these intersections (marked here in blue), we see that these four points form a rhombus. Therefore, the correct answer is <math>\boxed{\text{D}}</math>, and we are done.

Latest revision as of 17:32, 21 August 2025

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.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1953_AHSME_Problems/Problem_32&oldid=258523"