Fiona wants to draw a 2-dimensional shape whose perimeter passes through all of the points (P, Q, R, S) on the grid of squares shown. Which of the following can she draw?

(i) A circle (ii) An equilateral triangle (iii) A square

Kindly provided by UKMT
Fiona wants to draw a 2-dimensional shape whose perimeter passes through all of the points (P, Q, R, S) on the grid of squares shown. Which of the following can she draw? (i) A circle (ii) An equilateral triangle (iii) A square
20 Comments
s k
5/12/2014 07:49:16 pm
a square
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Paul
5/12/2014 11:32:54 pm
Thanks SK. Can you elaborate at all? And can any of the others be drawn?
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Simon Geard
6/12/2014 03:06:01 am
The short answer is none of them if she has to keep them entierly on the grid shown (which IMHO makes this a trick question). More interesting is if the paper is sufficiently large in which case I think the answer is (ii) and (iii).
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6/12/2014 05:27:21 am
Quick and easy partial answer: Not a circle since the centre must lie on the perpendicular bisector of PS and QR. These lines are parallel and not coincident.
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Andrey Dyukmedzhiev
6/12/2014 05:38:42 am
The answer is a square - follow the link below to see the picture:
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Andrez Dyukmedyhiev
6/12/2014 07:50:43 am
(ADDENDUM and correction to my post above) If all 4 points are on the contour (no one is a vertex) then the lines with the following equations: y = 3 (line QR on the diagram), y = x√3 (through the origin P) and y = (x - 6)√3 (through S on x-axis) are sides of an equilateral triangle.
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6/12/2014 08:11:38 am
It is a Square.
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Williamson
6/12/2014 09:37:30 am
Another thing about proving that an equilateral triangle can not be constructed is by proving that if the line passing through PQ, L1 intersect the line pass through R, L2 at a different point, cos A = Vector U. Vector V/ Norm(U)*Norm (V) (where Vector U and Vector V are both directional vectors of L1 and L2 respectively) is not equal to 1/3 since cos 60* = 1/3. Let me know if it is OK to proceed by proving it this way.
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Andrey Dyukmedzhiev
6/12/2014 10:58:16 am
Williamson, the equilateral triangle is possible with sides
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Williamson
6/12/2014 11:42:00 am
Yes you are right. You need to have a good time to think of the problem. Let us know if you have any different ideas. I would like to know if there is any other possibilities.
Andrey Dyukmedzhiev
7/12/2014 11:00:06 am
Again I must apologize, there is one minus sign ( '-' ) missing:
Williamson
7/12/2014 01:35:40 pm
lol Maybe you might need a rest :). Thank you for your help :). It was a good problem.
Simon Geard
6/12/2014 10:05:39 am
I have constructed the equilateral triangle I described in my reply above, a scaled image of which can be seen here:
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Williamson
6/12/2014 10:22:38 am
Very nice. I would not be able to see an equilateral triangle can fit in this. But how one is about to prove that will be really important. The length of each side my look equal in the eyes, but we need to prove that it can be done. Probably your input will give me a away to prove that. :)
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Williamson
6/12/2014 11:25:20 am
Simon, here is the proof that you make me reconsider after seeing your picture. I proved one side of this by doing the converse proof. But I will need to do the direct proof as well.
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Simon Geard
6/12/2014 12:40:31 pm
Your proof looks fine to me although I don't think it is necessary. By construction angle T and angle U are 60 degrees therefore the triangle is equilateral. 6/12/2014 11:00:01 am
In the broadest definition of the word "shape", she can draw a complex shape combining circle, square and eq-triangle. The only constraint implied by the problem statement is "a contiguous shape", so she has to choose component shapes that intersect somewhere on the overall shape.
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Paul
6/12/2014 10:26:18 pm
Thanks to everyone for a wonderful discussion. I always get more than I bargained for when I post a new puzzle. Excellent responses.
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ANN MODICA
13/7/2015 09:14:25 pm
I taught high school and I guess I taught had to give two many tests.
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Paul
13/7/2015 10:53:19 pm
True, Ann, but a good teacher would would surely ask the pupil "why?" and try to develop the problem, and thus the learning, further.
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