Question 15

The inner quadrilateral formed by joining the midpoints of the sides of a square is also a square.

Step 1 — Proving UVWX is a rhombus

Let the outer square be CASE. Let the side length of square CASE be $2a$. The points U, V, W, and X are the midpoints of the sides CA, CE, ES, and SA respectively. This means: $CU = UA = a$ $CV = VE = a$ $EW = WS = a$ $SX = XA = a$

Consider the four corner triangles: $\triangle VCU$, $\triangle UAX$, $\triangle XSW$, and $\triangle WEV$. Let's look at $\triangle VCU$. It is a right-angled triangle because $\angle C$ is an angle of the square CASE, so $\angle C = 90^\circ$. We have $VC = a$ and $CU = a$. By the Pythagorean theorem, the length of the hypotenuse $VU$ is: $VU^2 = VC^2 + CU^2$ $VU^2 = a^2 + a^2$ $VU^2 = 2a^2$ $VU = \sqrt{2a^2}$

$\boxed{VU = a\sqrt{2}}$

Similarly, for $\triangle UAX$: $UA = a$ and $AX = a$. $UX^2 = UA^2 + AX^2$ $UX^2 = a^2 + a^2$ $UX^2 = 2a^2$ $UX = a\sqrt{2}$

For $\triangle XSW$: $SX = a$ and $SW = a$. $XW^2 = SX^2 + SW^2$ $XW^2 = a^2 + a^2$ $XW^2 = 2a^2$ $XW = a\sqrt{2}$

For $\triangle WEV$: $WE = a$ and $EV = a$. $WV^2 = WE^2 + EV^2$ $WV^2 = a^2 + a^2$ $WV^2 = 2a^2$ $WV = a\sqrt{2}$

Since $VU = UX = XW = WV = a\sqrt{2}$, all four sides of the quadrilateral UVWX are equal. A quadrilateral with all four sides equal is a rhombus.

Step 2 — Proving UVWX has right angles

Now we need to show that the angles of UVWX are $90^\circ$. Consider $\triangle VCU$. We know $VC = CU = a$ and $\angle C = 90^\circ$. Since two sides are equal, it is an isosceles right-angled triangle. The angles opposite to the equal sides must be equal. So, $\angle CVU = \angle CUV$. The sum of angles in a triangle is $180^\circ$. $\angle C + \angle CVU + \angle CUV = 180^\circ$ $90^\circ + \angle CUV + \angle CUV = 180^\circ$ $90^\circ + 2 \times \angle CUV = 180^\circ$ $2 \times \angle CUV = 180^\circ - 90^\circ$ $2 \times \angle CUV = 90^\circ$ $\angle CUV = 45^\circ$ Similarly, for $\triangle UAX$, $\angle AUX = 45^\circ$. For $\triangle XSW$, $\angle SXW = 45^\circ$. For $\triangle WEV$, $\angle EWV = 45^\circ$.

Now let's look at the angle $\angle VUX$ inside the quadrilateral UVWX. The points C, U, A lie on a straight line (the side of the square CASE). The sum of angles on a straight line is $180^\circ$. So, the angles around point U on the line CA add up to $180^\circ$. $\angle CUV + \angle VUX + \angle AUX = 180^\circ$ We found $\angle CUV = 45^\circ$ and $\angle AUX = 45^\circ$. $45^\circ + \angle VUX + 45^\circ = 180^\circ$ $90^\circ + \angle VUX = 180^\circ$ $\angle VUX = 180^\circ - 90^\circ$

$\boxed{\angle VUX = 90^\circ}$

We can apply the same logic for the other angles of UVWX: $\angle UXW = 90^\circ$ (from $\angle AUX + \angle UXW + \angle SXW = 180^\circ$) $\angle XWV = 90^\circ$ (from $\angle SXW + \angle XWV + \angle EWV = 180^\circ$) $\angle WVU = 90^\circ$ (from $\angle EWV + \angle WVU + \angle CVU = 180^\circ$)

Since all sides of UVWX are equal and all its angles are $90^\circ$, the quadrilateral UVWX is a square.

Step 3 — Construction and Measurement

To verify this by construction and measurement:

1. Draw a square CASE with a ruler. For example, make each side $10 \text{ cm}$ long.
2. Find the midpoint of each side. For a $10 \text{ cm}$ side, the midpoint will be at $5 \text{ cm}$. Mark these midpoints as U, V, W, X.
3. Connect the midpoints to form the quadrilateral UVWX.
4. Measure the sides of UVWX. You will find that each side is approximately $7.07 \text{ cm}$ (which is $5\sqrt{2} \text{ cm}$). All sides are equal.
5. Measure the angles of UVWX using a protractor. You will find that each angle is $90^\circ$. This confirms that UVWX is a square.

Step 4 — Other ways to construct a square within a square

Figure (b) shows another way to construct a square inside a square. Let the outer square be ABCD. We can take points P, Q, R, S on the sides AB, BC, CD, DA respectively, such that the segments cut off from the corners are equal. For example, let $AP = BQ = CR = DS = x$. Let the side length of the outer square be $L$. Then, $PB = QC = RD = SA = L - x$.

Consider the four corner triangles: $\triangle SAP$, $\triangle PBQ$, $\triangle QCR$, $\triangle RDS$. Let's look at $\triangle SAP$. It is a right-angled triangle because $\angle A = 90^\circ$. We have $SA = L - x$ and $AP = x$. By the Pythagorean theorem, the length of the hypotenuse $SP$ is: $SP^2 = SA^2 + AP^2$ $SP^2 = (L-x)^2 + x^2$

Similarly, for $\triangle PBQ$: $PB = L - x$ and $BQ = x$. $PQ^2 = PB^2 + BQ^2$ $PQ^2 = (L-x)^2 + x^2$

For $\triangle QCR$: $QC = L - x$ and $CR = x$. $QR^2 = QC^2 + CR^2$ $QR^2 = (L-x)^2 + x^2$

For $\triangle RDS$: $RD = L - x$ and $DS = x$. $RS^2 = RD^2 + DS^2$ $RS^2 = (L-x)^2 + x^2$

Since $SP^2 = PQ^2 = QR^2 = RS^2$, it means $SP = PQ = QR = RS$. So, the quadrilateral PQRS has all four sides equal, making it a rhombus.

Now let's check the angles of PQRS. Since $AP = BQ = CR = DS = x$ and $SA = PB = QC = RD = L-x$, and all corner angles of the outer square are $90^\circ$, all four corner triangles ($\triangle SAP$, $\triangle PBQ$, $\triangle QCR$, $\triangle RDS$) are congruent by the SAS (Side-Angle-Side) congruence rule. This means their corresponding angles are equal. Let $\angle ASP = \alpha$ and $\angle APS = \beta$. In $\triangle SAP$, $\alpha + \beta + 90^\circ = 180^\circ$, so $\alpha + \beta = 90^\circ$. Due to congruence: $\angle ASP = \angle BPQ = \angle CQR = \angle DRS = \alpha$ $\angle APS = \angle BQP = \angle CRQ = \angle DSR = \beta$

Now consider the angle $\angle SPQ$ of the inner quadrilateral PQRS. The points A, P, B lie on a straight line. So, the angles around point P on the line AB add up to $180^\circ$. $\angle APS + \angle SPQ + \angle BPQ = 180^\circ$ $\beta + \angle SPQ + \alpha = 180^\circ$ Since $\alpha + \beta = 90^\circ$: $90^\circ + \angle SPQ = 180^\circ$ $\angle SPQ = 180^\circ - 90^\circ$ $\angle SPQ = 90^\circ$

Similarly, we can show that $\angle PQR = 90^\circ$, $\angle QRS = 90^\circ$, and $\angle RSP = 90^\circ$. Since all sides of PQRS are equal and all its angles are $90^\circ$, the quadrilateral PQRS is a square.

This method works for any value of $x$ (as long as $0 < x < L$). If $x = L/2$, then $x = L-x$, which means the points are midpoints, and this becomes the same as part (a). If $x$ is very small, the inner square is almost as big as the outer square. If $x$ is close to $L$, the inner square is small and rotated.

Answer

(a) The quadrilateral UVWX is a square. (b) Other ways to construct a square within a square are by taking points on the sides such that the segments cut off from the corners are equal (e.g., $AP = BQ = CR = DS$). This creates congruent right-angled triangles at each corner, leading to an inner square.