A square and a regular pentagon, each of area 1, are coplanar and concentric. Show that the area of the region inside both shapes is greater than 3/4.
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$\begingroup$ Two objects are concentric when they have the same center. $\endgroup$quarague– quarague2025-10-25 07:24:03 +00:00Commented Oct 25 at 7:24
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2 Answers
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The radius of the circle inscribed in the square is ½. Both shapes include this circle*, so the area of the region inside both is greater than π/4 > ¾.
*Why do we only need to consider the circle inscribed in the square?
Intuitively: The pentagon is "more circular" than the square, so its inscribed circle should be larger.
More concretely: A regular n-gon can be divided into n triangles to show that its area is equal to half its perimeter times the radius of its inscribed circle. Given two regular polygons with equal area, the one with more sides has a smaller perimeter and therefore a larger inscribed circle.
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2$\begingroup$ It is non-obvious to me that a pentagon of area 1 contains a circle of radius 1/2. It is true, see Pranay's answer for a proof but without that I don't think this solution is complete. $\endgroup$quarague– quarague2025-10-25 07:26:56 +00:00Commented Oct 25 at 7:26
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3$\begingroup$ I find it intuitive that a circle inscribed in a pentagon covers a greater fraction of its area than one in a square. $\endgroup$xnor– xnor2025-10-25 08:25:12 +00:00Commented Oct 25 at 8:25
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$\begingroup$ @quarague Thanks, I added an explanation. $\endgroup$noedne– noedne2025-10-25 08:54:10 +00:00Commented Oct 25 at 8:54
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$\begingroup$ @xnor That is a clever explanation, nice. $\endgroup$quarague– quarague2025-10-25 12:26:00 +00:00Commented Oct 25 at 12:26
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Let a be the side length of the pentagon and r be the inradius. Then dividing the pentagon into five congruent triangles with the centre as apex and the five sides as bases, the area of each triangle is 1/2 a r = 1/5. On the other hand, the ratio (a/2)/r = tan(36°). Therefore, r2 = cot(36°)/5. Those who are familiar with the (3,4,5) right triangle know that one of the angles is about 36.87°, so cot(36°) > cot(36.87°) = 4/3 > 5/4, which means r > 1/2. Therefore, the incircle of the square is contained inside the pentagon too. And the incircle of the square has area π/4, which is at least 3/4.
