Appearances are deceptive

Geometry Level 2

In the regular star polygon above, $\text{\_\_\_\_\_\_\_\_\_\_\_}.$

The black and blue areas are equal The black area is larger The blue area is larger

4 solutions

Ali Qureshi
Jul 3, 2016

the picture explains

How do we know the 2 blue areas are equal

Jerry McKenzie - 4 years ago

It may help to observe that the blue triangle in the interior is the same size as the red triangle adjacent to it. There is one of each colored triangles in the areas that make the blue and black areas.

Pun It - 2 years, 7 months ago

I don't agree to the answer of this problem. The area of the blue area is larger than the area of the black area.

A Former Brilliant Member - 4 years, 9 months ago

I know right

Christopher Draper - 3 years, 4 months ago

Scott Caffrey
Jul 2, 2016

S = the area of an isosceles tip of the star. Imagine folding one of these tips into the centre of the shape such that it cuts the internal pentagon into three shapes. One of these shapes would have area S. The remaining area would be two congruent isosceles triangles, T.

The total area of the whole shape will then equal 6S + 2T.

The area of the trapezium is 3S + T.

0.5(6S + 2T) = 3S + T therefore the area of the trapezium is half the area of the whole star. The two areas are equal!

First time I've tried to comment on a solution, hope you guys approve!

Good explanation!! Cheers..

Santhosh DS - 4 years, 11 months ago

Nice explanation yes, but dosen't the assumption that "folding one of the tips into the center will result in the vertex of the folded tip to fall exactly on the opposite corner of the pentagon" need proof?

Michael McCafferty - 3 years ago

OK, I see the proof. Since the tips are isosceles triangles whose sides are parallel to the two sides of the pentagon adjacent to the isosceles base, and since the interior angles of the pentagon are 108 degrees and the exterior angles are 72 degrees, by alternate interior angles one can show that a triangle drawn from the base of the pentagon to the opposite corner will be congruent to an isosceles triangle drawn outward using the pentagon side as a base. Therefore the altitude of the tip is exactly the same distance as that from the side of the pentagon to the opposite corner, hence the folded tip will fall exactly on the opposite corner.

Michael McCafferty - 3 years ago

Alkis Piskas
Oct 23, 2018

Let's call "limbs" each of the 5 isosceles triangles of the star. The black and blue areas contain 2 limbs each. So, let's take them out of the scene. What remains in the blue part is area ABDE and in the black part area BDF . Let's compare them.

Blue ABDE = ABE + BDE. But BDE is symetrical to the limb DEG. So ABDE = ABE + one limb .

Black BDF = BCD + CDF = BCD + one limb .

Now, triangles BCD and ABE are equal since they are formed by one point of the interior regular pentagon and 2 adjacent points. So the areas ABDE and BDF are equal .

Sudoku Subbu
Jul 2, 2016

I didn't even work it out on a paper. as the nme suggests i kept the right option

Appearances are deceptive

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