### How Big?

If the sides of the triangle in the diagram are 3, 4 and 5, what is the area of the shaded square?

### Do Unto Caesar

At the beginning of the night three poker players; Alan, Bernie and Craig had money in the ratios 7 : 6 : 5. At the end of the night the ratio was 6 : 5 : 4. One of them won $1 200. What were the assets of the players at the beginning of the evening? ### Oh for the Mathematics of Yesteryear A garrison of 600 men has just enough bread ... but, with the news that the enemy was planning an attack... How many ounces of bread a day must each man in the garrison be allowed, to hold out 45 days against the siege of the enemy? # Hexagon Slices ##### Stage: 3 Short Challenge Level: Each exterior angle of a regular hexagon is$60^{\circ} (360^{\circ}/6)$, so when sides HB and IC are extended to meet at A , an equilateral triangle, ABC is created. Let the sides of this triangle be of length x. As BC, DE and FG are all parallel, triangles ABC, ADE and AFG are all equilateral. So, DE = DA =$p$+$x$; and FG = FA =$q$+$p$+$x$. The perimeter of trapezium BCED =$x + p + x + 2p = 2x + 3p$; The perimeter of trapezium DEGF =$(p + x) + (q + p + x ) + 2q = 2x + 2p + 3q$; The perimeter of hexagon FGIKJH =$2((q + p + x) + 2r ) = 2x + 2p + 2q + 4r$. So,$2x$+$3p$=$2x$+$2p$+$3q$; hence$p$=$3q$. Also$2x$+$2p$+$3q$=$2x$+$2p$+$2q$+$4r$; hence$q$=$4r$. So,$p$:$q$:$r$=$12r$:$4r$:$r$=$12$:$4$:$1\$

This problem is taken from the UKMT Mathematical Challenges.
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