Suppose five circles, each 4 inches in diameter, are cut from a rectangular strip of paper 12 inches
long. If the least amount of paper is to be wasted, what is the width of the paper strip?
(A) 5
(B) 4 +2 Sqr 3
(C) 8
(D) 4(1+ Sqr 3)
(E) not enough information
Since this is a hard problem, we can eliminate (E), “not enough information.” And because it is too easily
derived, we can eliminate (C), (8 = 4 + 4). Further, we can eliminate (A), 5, because answer-choices (B)
and (D) form a more complicated set. At this stage we cannot apply any more elimination rules; so if we
could not solve the problem, we would guess either (B) or (D).
Let’s solve the problem directly. The drawing below shows the position of the circles so that the
paper width is a minimum.
Now, take three of the circles in isolation, and connect the centers of these circles to form a triangle:
Since the triangle connects the centers of circles of diameter 4, the triangle is equilateral with sides of
length 4. Drawing an altitude gives
Applying the Pythagorean Theorem to either right triangle gives Squaring yields
Subtracting 4 from both sides of this equation yields
Taking the square root of both sides yields
Removing the perfect square 4 from the radical yields
Summarizing gives
Adding to the height, h =2 sqr 3, the distance above the triangle and the distance below the triangle to the
edges of the paper strip gives
width = (2 + 2) + 2 sqr 3 = 4 + 2 sqr3
The answer is (B).
long. If the least amount of paper is to be wasted, what is the width of the paper strip?
(A) 5
(B) 4 +2 Sqr 3
(C) 8
(D) 4(1+ Sqr 3)
(E) not enough information
Since this is a hard problem, we can eliminate (E), “not enough information.” And because it is too easily
derived, we can eliminate (C), (8 = 4 + 4). Further, we can eliminate (A), 5, because answer-choices (B)
and (D) form a more complicated set. At this stage we cannot apply any more elimination rules; so if we
could not solve the problem, we would guess either (B) or (D).
Let’s solve the problem directly. The drawing below shows the position of the circles so that the
paper width is a minimum.
Now, take three of the circles in isolation, and connect the centers of these circles to form a triangle:
Since the triangle connects the centers of circles of diameter 4, the triangle is equilateral with sides of
length 4. Drawing an altitude gives
Applying the Pythagorean Theorem to either right triangle gives Squaring yields
Subtracting 4 from both sides of this equation yields
Taking the square root of both sides yields
Removing the perfect square 4 from the radical yields
Summarizing gives
Adding to the height, h =2 sqr 3, the distance above the triangle and the distance below the triangle to the
edges of the paper strip gives
width = (2 + 2) + 2 sqr 3 = 4 + 2 sqr3
The answer is (B).
where is the figure?
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