## Problem 1

All small squares are of equal size and the big rectangle is $11 \times 13$.Find the area of the green region in the figure below.

### Solution (Credit: Atul Gupta)

The key insight is that the cross comprising of $5$ squares is within the $11 \times 11$ square $QRST$.

So we now have $UV=2$.From the similar triangles $AUV$ and $AWZ$ it is easy to see that $WZ=4$. Hence, $ZX=WX-WZ=13-4=9$. From the similar triangles $ZCD$ and $ZBX$, we have $ZD=3$.

Triangles $AUV$ and $ZDC$ are congruent, so the side of the small square $AV=ZC=\sqrt{2^2+3^2} = \sqrt{13}$.

Therefore, the area of the green region is $13\cdot11 - 13\cdot6 = 65$.

## Problem 2

Find angle $x$ in the figure below.

### Solution

Reflecting $ΔADB$ about segment $DB$ we get the equilateral triangle $ΔABF$ because $AB=AF$ and $∠ ABF = 60^∘$.

As $AB = BF$, we also have $CE = \frac{1}{2} BF = AC$ as $C$ and $E$ are the midpoints of $AF$ and $AB$.

Triangle $AEC$ is equilateral because we have $AC=CE$ and $\angle FAB = ∠EAC = 60^∘$. We now have $AE=EC=AC$.

We also have $\angle CDB = 180^{∘} - ∠CBD - ∠BCD = 180^{∘} - 30^{∘} - (180^{∘} -45^{∘}) = 15^{∘}$, $\angle ECD = \angle ECA - \angle ACD = \angle 60^{∘} -45^{∘} = 15^{∘}$. Therefore, $DE = EC = EA$.

As $DE=EA$, $\angle ADE = \angle DAE = 45^{∘}$ because $\angle AED = 90^{∘}$.

Therefore $x = \angle ADE - \angle CDB = 45^{∘} - 15^{∘} = 30^{∘}$.

## Problem 3

The distance between the circumcentre and the incentre of a right angled triangle is $6$ units. If both the circumradius and the inradius are positive integers, what is the area of the triangle?

### Solution

In the figure below,

$\Delta ABC$ is a right angled triangle with $\angle B = 90^{\circ}$. Let $r$ be the inradius and $R$ be the circumradius of the triangle. $D$ is the mid point of $AC$. Let $DF=u$.

As $D$ is the circumcentre, we have $AD=DC=BD=R$. We have $AF=AG=R+u$, $CF=CH=R-u$, $GB=BH=r$, hence $AB=AG+GB=R+u+r$ and $BC=CH+HB=R-u+r$.

Using the Apollonius theorem in triangle $\Delta ABC$, we have

\begin{align*} AB^2 + BC^2 &= 2(BD^2 + CD^2) \ \implies (R+u+r)^2 + (R-u+r)^2 &= 4R^2 \ \implies u^2 + r^2 + 2rR &= R^2 \end{align*}

From the right angled triangle $\Delta DEF$, we also have $u^2+ r^2 = d^2$.

Eliminating $u$ from the above two equations, we have $d^2 + 2rR = R^2$.

As $d=8$, $r$ and $R$ are integers, from the equation $R(R-2r) = 64$, we get $R=16$ and $r=6$.

Area of the triangle

\begin{align*} \Delta ABC &= r\cdot s = r\left(\frac{AB+BC+CA}{2}\right) \ &= r(2R+r) = 6\cdot38 = 228. \end{align*}

## Problem 4

Take a regular hexagon and connect each vertex to the midpoint of the opposite side, choosing the closest counter clockwise of the two opposite sides. This defines a central regular hexagon with the same center as the starting hexagon. What proportion of the total area does the central hexagon occupy?

Source. Rick Mabry, Mathematics Magazine 91(3) June 2018 , pp. 184-185.

### Solution

The inner hexagon is 1/13 of the starting hexagon. This can be derived in various ways, using geometry or trigonometry and algebra, but the discoverer, Rick Mabry, had the cute dissection solution shown in the diagram below.

## Problem 5

An octagon with the given side lengths is inscribed in a circle. Find the area of the octagon?

### Solution (Credit: Atul Gupta)

The octagon can be transformed as follows while preserving the area

The side of the square is $3+2\sqrt{2}$.

The area of the octagon is $(3+2\sqrt{2})^2-4\cdot\frac{1}{2}\cdot(\sqrt{2})^2 = 13+12\sqrt{2}$.

## Problem 6

Consider a unit square. Four quarter circles of radius $1$ are drawn with each of the vertices as center. What is the area of the region that is common to all the four quarter circles.

### Solution

The common area to all the four quarter circles is the area bounded by the circular arcs $LN, NO, OP$ and $PL$ in the diagram below. From symmetry it is easy to see that the area is $8$ times the area of the region $LEM$. The area of the region $LEM$ = Area of the sector $ALM$ - Area of the $\Delta AEL$. $\Delta LAB$ is an equilateral triangle, therefore $\angle LAE = \angle LAB -\angle EAB = 60^\circ - 45^\circ = 15^\circ$. The area of the sector $ALM = \frac{\pi}{24}$. The area of $\Delta AEL = \frac{1}{2}\cdot 1 \cdot \frac{1}{\sqrt{2}} \cdot \sin 15^\circ$. Hence the required area is given by $8 \left(\frac{\pi}{24} - \frac{\sin 15^\circ}{2\sqrt{2}}\right)=\frac{\pi}{3} - \sqrt{3} + 1$.

## Problem 7

Consider a sphere with the same centre as the centre of a cube. Starting with a very small radius for the sphere, we increase it slowly just until the cube becomes $8$ separate corner pieces. So something like the diagram below. What is the ratio of the volume carved out from the cube by this sphere to the volume of the cube?

### Solution

Let the side of the cube have a length of $1$. The radius of the sphere when it just breaks the cube into $8$ corner pieces is $\sqrt{\frac{1}{2}^2+\frac{1}{2}^2} = \sqrt{\frac{1}{2}}$ (the length of the line joining the centre of the cube to a centre of an edge of the cube).

The volume of the part of the sphere inside the cube can be calculated by subtracting the volume of the $6$ spherical caps that protrude from the $6$ sides of the cube.

The volume of a spherical cap as shown in the diagram below is given by $\frac{1}{6}\pi h(3a^2+h^2)$.

In the context of the problem, $h = \frac{1}{\sqrt{2}} - \frac{1}{2}$ and $a= \frac{1}{2}$.

Therefore, the volume of the part of the sphere inside the cube is given by

$$\frac{4}{3} \frac{\pi}{2\sqrt{2}} - 6 \cdot \frac{1}{6} \pi \left(\frac{1}{\sqrt{2}} - \frac{1}{2}\right) \left(3\frac{1}{2}^2 + \frac{1}{2} + \frac{1}{4} -\frac{1}{\sqrt{2}} \right) \ = \pi\frac{2}{3\sqrt{2}} - \pi\left(\frac{1}{\sqrt{2}} - \frac{1}{2}\right)\left(\frac{3}{2} - \frac{1}{\sqrt{2}} \right)\ = \pi\frac{2}{3\sqrt{2}} - \pi\left( \frac{3}{2\sqrt{2}} -\frac{1}{2} -\frac{3}{4} + \frac{1}{2\sqrt{2}}\right )\ = \pi \left( \frac{5}{4} - \frac{4}{3\sqrt{2}}\right) = \frac{\pi}{12} \left(15 - 8\sqrt{2}\right)$$

### References

Spherical Cap volume on Wikipedia