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ABCD is rectangular court with AB = 50m and BC = 30m. Four girls stand at different positions in that court so that the distance between the two girls next to each other is maximised. What is this distance? A. 46m B. 34m C. 26 + (2/3)m D. 20m E. 40m

I posted this question a while back but it got put on hold because no one could solve it, simple as that. This is a question from a PAMO-G qualification exam. It's exactly like this, so I don't know how you want me to "reword" it. You either know the answer or you don't. Now, please, if you do know the answer, help me out with a full solution or at least share your thoughts.

user346756
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2 Answers2

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Firstly, we let two girls be located as far a possible. Therefore, G1 stands at A and G2 stands at corner C.

enter image description here

Next, we put G3 at corner D, which is 30m from A. Of course, this setting is not optimal. We then ask G3 walks x m towards G2 and stop at point E such that AE = EC. We then have $30^2 + x^2 = (30 + 20 – x)^2$.

After solving for x, we get E is 34m from C. The remaining girl is symmetrically placed on the side AB at F.

Added:- Further calculation shows that G3 and G4 is 34.9857 m apart.

Mick
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  • The picture helps to resolve a detail not covered in your answer, how far apart the girls G3 and G4 are in this arrangement. Strictly speaking this shows that the four positions can be at least $34m$ apart, but does not address the possibility that something better could be achieved. – hardmath Jun 29 '16 at 17:14
  • @hardmath Mentioned point has been added. According to the requirement, we need the girls to be placed as far from each other as possible. The natural arrangement is to let all of them stand at the border. – Mick Jun 29 '16 at 17:34
  • I changed my mind (by computing). It isn't possible to increase the distance by repositioning the first two girls away from the corners where you initially placed them. – hardmath Jun 29 '16 at 18:45
  • @hardmath It is just an MC question after all. There shouldn't be too much explanation involved. – Mick Jun 29 '16 at 18:51
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If I'm interpreting this correctly, we're maximizing the distance between the two girls that are the closest to each other. To do this, we need to form a rhombus, specifically the largest possible rhombus inscribed within the rectangle. Let two girls stand at $B$ and $D$, such that one diagonal of the rhombus is maximized, and the other two stand at points $P$ on $AB$ and $Q$ on $CD$ such that $BPDQ$ is a rhombus. A diagram:

enter image description here

Let $x$ equal the distance between points $A$ and $P$ (and between points $C$ and $Q$). Since $PB = PD$ by the definition of a rhombus, we can set up an equation to solve for $x$,

$$ 50-x=\sqrt{30^2+x^2}, $$

where the left side is equal to $PB$ and the right to $PD$ (by the Pythagorean theorem). Solving for $x$, we get $x=16$, thus the girls are spaced $ 50 - 16 = \boxed{\text{(B) }34\text{ m}}$ apart.

DooplissForce
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  • what's the equation that we set up to solve for x? – user346756 Aug 28 '16 at 08:24
  • @GigiChuck It's $50-x = \sqrt{30^2+x^2}$. The left side, which we want to equal $PB$, is $50-x$ because $AB-AP=PB$, and we know that $AB = 50$ and $AP = x$. The right side, which we want to equal $PD$, is $\sqrt{30^2+x^2}$, because $PD^2 = AD^2 + AP^2 \Rightarrow PD^2 = 30^2 + x^2 \Rightarrow PD = \sqrt{30^2 + x^2}$. – DooplissForce Aug 28 '16 at 12:12