Question 54

Challenge question #54 (03/07/2017)

Junior Question

A calisson is a French sweet that looks like two equilateral triangles glued together.

Calissons are packed into regular hexagonal boxes, arranged in a way such that there are no gaps between them like in the diagram alongside.

So, there are $3$ distinct orientations for the calissons in a box: Maths Image

For any arrangement of the calissons of side length $1$, find the proportion of each orientation of calisson in a box with side length $6$.

Answer

Senior Question

A circle with radius $x$ cm is inscribed within a square as shown. A line [MN] is drawn where M is the midpoint of [AD] and N is the point on [BC] such that BN $= \tfrac{1}{3}$ NC.

Find the shaded area in terms of $x$.

Answer

shaded area $= x^{2}\left(\tfrac{25}{34} - \tfrac{\pi}{4} + \arctan(\tfrac{1}{4})\right)$ cm$^{2}$

Worked Solution

Let O be the centre of the circle and P be a point such that [MP] is the diameter of the circle.

$\therefore\quad$AM $= x$ cm and NP $=\tfrac{1}{2}x$ cm.

Now$\quad a+b=$

area of square $-$ area of quarter circle

$=$

$x^{2}-\tfrac{1}{4}\pi x^{2}$

$=$

$\left(1-\tfrac{\pi}{4}\right)x^{2}$

and$\quad b+c+d=$

area of triangle NPM

$=$

$\tfrac{1}{2}\left(2x\right)\left(\tfrac{1}{2}x\right) = \tfrac{1}{2} x^{2}$

Let$\quad \text{N}\hat{\text{M}}\text{P} =$

$\alpha$ radians.

$\therefore\quad\text{M}\hat{\text{O}}\text{Q} =$

$\pi - 2\alpha$

$\{$angles in a triangle$\}$

$\text{and}\quad \text{Q}\hat{\text{O}}\text{P} =$

$2\alpha$

$\{$exterior angle of a triangle$\}$

$\therefore\quad c=$ area of sector OPQ

$=\tfrac{1}{2}x^{2}\left(2\alpha\right)$

$=x^{2}\alpha$

and$\quad d =$

area of triangle MOQ

$=$

$\tfrac{1}{2}x^{2}\sin\left(\pi - 2\alpha\right)$

$=$

$\tfrac{1}{2}x^{2}\sin2\alpha$

$\therefore\quad b =$

$\tfrac{1}{2}x^{2}-c-d$

$=$

$\tfrac{1}{2}x^{2}-x^{2}\alpha-\tfrac{1}{2}x^{2}\sin2\alpha$

$=$

$x^{2}\left(\tfrac{1}{2}-\alpha-\tfrac{1}{2}\sin2\alpha\right)$

$\therefore\quad a =$

$\left(1 - \tfrac{\pi}{4}\right)x^{2} - b$

$=$

$\left(1 - \tfrac{\pi}{4}\right)x^{2} - x^{2}\left(\tfrac{1}{2}-\alpha-\tfrac{1}{2}\sin2\alpha\right)$

$=$

$x^{2}\left(\tfrac{1}{2}-\tfrac{\pi}{4}+\alpha+\tfrac{1}{2}\sin2\alpha\right)$

$=$

$x^{2}\left(\tfrac{1}{2}-\tfrac{\pi}{4}+\alpha+\sin\alpha\cos\alpha\right)$

Now in triangle MPN,

$\tan\alpha =$

$\dfrac{\tfrac{1}{2}x}{2x} = \tfrac{1}{4}\text{,}\quad\sin\alpha = \tfrac{1}{\sqrt{17}}\text{,}\quad $and$\quad\cos\alpha = \tfrac{4}{\sqrt{17}}$

$\therefore\quad\alpha=$

$\arctan\left(\tfrac{1}{4}\right)$

So$\quad a =$

$x^{2}\left( \tfrac{1}{2} - \tfrac{\pi}{4} + \arctan\left(\tfrac{1}{4}\right) + \left(\tfrac{1}{\sqrt 17}\right)\left(\tfrac{4}{\sqrt 17}\right) \right)$

$=$

$x^{2}\left( \tfrac{1}{2} + \tfrac{4}{17} - \tfrac{\pi}{4} + \arctan\left(\tfrac{1}{4} \right)\right)$

$\therefore\quad$shaded area $=$

$x^{2}\left(\tfrac{25}{34} - \tfrac{\pi}{4} + \arctan(\tfrac{1}{4})\right)$ cm$^{2}$

Question 54

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