2021 CMC 12A Problems/Problem 6

The following problem is from both the 2021 CMC 12A #6 and 2021 CMC 10A #7, so both problems redirect to this page.

Problem

How many of the following statements are true for every parallelogram $\mathcal{P}$ ?

      i. The perpendicular bisectors of the sides of  all share at least one common point.
      ii. The perpendicular bisectors of the sides of  are all distinct.
      iii. If the perpendicular bisectors of the sides of  all share at least one common point then  is a square.
      iv. If the perpendicular bisectors of the sides of  are all distinct then these bisectors form a parallelogram.

$\textbf{(A) } 0\qquad\textbf{(B) } 1\qquad\textbf{(C) } 2\qquad\textbf{(D) } 3\qquad\textbf{(E) } 4\qquad$

Solution

Statement (i) is obviously false; suppose $\mathcal{P}$ is a non-rectangular parallelogram, then the statement does not hold.

Statement (ii) is also false; suppose $\mathcal{P}$ is a rectangle, then the perpendicular bisectors of opposite sides are the same.

Statement (iii) is also false; if we let $\mathcal{P}$ be a non-square rectangle, the bisectors still share at least one common point at the center of the rectangle.

Statement (iv) is true. A parallelogram has two pairs of parallel opposite sides. Thus the perpendicular bisectors of those opposite sides will also be parallel, and thus if they are distinct, they will form a parallelogram.

In conclusion, only one statement is true, namely statement (iv), so the answer is $\boxed{\textbf{(B) } 1}$ .

2021 CMC 12 (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All CMC 12 Problems and Solutions

2021 CMC 10 (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All CMC 10 Problems and Solutions

The problems on this page are copyrighted by the MAC's Christmas Mathematics Competitions.

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2021 CMC 12A Problems/Problem 6

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