perplexus.info :: Shapes : Hexagon Hinder

Hexagon Hinder (Posted on 2016-02-12)

PQRSTU is a regular hexagon
A,B,C,D,E and F are the respective midpoints of sides PQ, QR, RS, ST, TU and UP.
PB, QC, RD, SE, TF and UA bound a smaller regular hexagon.
The ratio of the area of the smaller hexagon to the area of PQRSTU is equal to x/y, where x and y are relatively prime positive integers.

Determine x and y

See The Solution Submitted by K Sengupta

Rating: 5.0000 (1 votes)

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Simple solution

| Comment 1 of 2

Call the center of the hexagons O

Call the point where AF and OP cross X

Let QP=1 so that OP=1 and AP=1/2

Triangle OAP is 1/12 of the large hexagon. It has hypotenuse 1

Triangle AXP is the missing portion of this triangle from the smaller hexagon. It has hypotenuse 1/2.

So the areas of these triangles are in the ratio 1/4.

So the ratio of the hexagons is 3/4.

Posted by Jer on 2016-02-12 09:02:01

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