Quadrilaterals | FIO

Question 15

CASE is a square. The points U, V, W and X are the midpoints of the sides of the square. What type of quadrilateral is UVWX? Find this by using geometric reasoning, as well as by construction and measurement. Find other ways of constructing a square within a square such that the vertices of the inner square lie on the sides of the outer square, as shown in Figure (b).

Solution

Understand the Question

A square has four equal sides and four right angles ( $90^\circ$ ).
Joining the midpoints of the sides of a square divides each corner into an isosceles right-angled triangle.
Using the Pythagoras theorem, we can find the lengths of the inner quadrilateral's sides to prove they are all equal (a rhombus).
Using angle sums in triangles and linear pairs on a straight line, we can prove each corner angle of the inner quadrilateral is $90^\circ$ , confirming it is a square.
This geometric property can also be generalised by dividing each side of the outer square into segments of lengths $x$ and $L-x$ .

Step 1 · Prove UVWX is a Rhombus

Let the side length of square $\text{CASE}$ be $2a$ .

Since $\text{U, V, W, X}$ are midpoints of $\text{CA, CE, ES, SA}$ respectively: $\text{CU} = \text{UA} = a$ $\text{CV} = \text{VE} = a$ $\text{EW} = \text{WS} = a$ $\text{SX} = \text{XA} = a$ Diagram 1

In right-angled $\Delta \text{VCU}$ (since $\angle \text{C} = 90^\circ$ ):

\begin{aligned} \text{VU}^2 &= \text{VC}^2 + \text{CU}^2 \\ &= a^2 + a^2 \\ &= 2a^2 \\ \text{VU} &= \sqrt{2a^2} = a\sqrt{2} \end{aligned}

Similarly, for $\Delta \text{UAX}$ , $\Delta \text{XSW}$ , and $\Delta \text{WEV}$ :

\begin{aligned} \text{UX}^2 &= \text{UA}^2 + \text{AX}^2 = a^2 + a^2 = 2a^2 \implies \text{UX} = a\sqrt{2} \\ \text{XW}^2 &= \text{SX}^2 + \text{SW}^2 = a^2 + a^2 = 2a^2 \implies \text{XW} = a\sqrt{2} \\ \text{WV}^2 &= \text{WE}^2 + \text{EV}^2 = a^2 + a^2 = 2a^2 \implies \text{WV} = a\sqrt{2} \end{aligned}

Since $\text{VU} = \text{UX} = \text{XW} = \text{WV} = a\sqrt{2}$ , all four sides are equal, so $\text{UVWX}$ is a rhombus.

Step 2 · Prove All Interior Angles of UVWX are 90°

In isosceles right-angled $\Delta \text{VCU}$ , $\text{VC} = \text{CU} = a$ and $\angle \text{C} = 90^\circ$ :

\begin{aligned} \angle \text{C} + \angle \text{CVU} + \angle \text{CUV} &= 180^\circ \\ 90^\circ + 2\angle \text{CUV} &= 180^\circ \\ 2\angle \text{CUV} &= 90^\circ \\ \angle \text{CUV} &= 45^\circ \end{aligned}

Similarly: $\angle \text{AUX} = 45^\circ, \quad \angle \text{SXW} = 45^\circ, \quad \angle \text{EWV} = 45^\circ$

Since $\text{C, U, A}$ lie on a straight line:

\begin{aligned} \angle \text{CUV} + \angle \text{VUX} + \angle \text{AUX} &= 180^\circ \\ 45^\circ + \angle \text{VUX} + 45^\circ &= 180^\circ \\ 90^\circ + \angle \text{VUX} &= 180^\circ \\ \angle \text{VUX} &= 90^\circ \end{aligned}

By symmetry, the remaining angles are also right angles: $\angle \text{UXW} = 90^\circ, \quad \angle \text{XWV} = 90^\circ, \quad \angle \text{WVU} = 90^\circ$

Since all four sides are equal and all four interior angles are $90^\circ$ , $\text{UVWX}$ is a square.

Step 3 · Verification by Construction and Measurement

Draw a square $\text{CASE}$ of side length $10\text{ cm}$ .
Mark midpoints $\text{U, V, W, X}$ at $5\text{ cm}$ along each side.
Join the points to form quadrilateral $\text{UVWX}$ .
Measuring each side gives $\text{UV} = \text{VW} = \text{WX} = \text{XU} \approx 7.07\text{ cm}$ (which equals $5\sqrt{2}\text{ cm}$ ).
Measuring each angle with a protractor gives $\angle \text{U} = \angle \text{V} = \angle \text{W} = \angle \text{X} = 90^\circ$ , confirming $\text{UVWX}$ is a square.

Step 4 · General Method to Inscribe a Square Inside a Square

Let outer square be $\text{ABCD}$ with side length $L$ .

Choose points $\text{P, Q, R, S}$ on sides $\text{AB, BC, CD, DA}$ respectively such that: $\text{AP} = \text{BQ} = \text{CR} = \text{DS} = x$ $\text{PB} = \text{QC} = \text{RD} = \text{SA} = L - x$

By the Pythagoras theorem in the corner right triangles:

\begin{aligned} \text{SP}^2 &= \text{SA}^2 + \text{AP}^2 = (L - x)^2 + x^2 \\ \text{PQ}^2 &= \text{PB}^2 + \text{BQ}^2 = (L - x)^2 + x^2 \\ \text{QR}^2 &= \text{QC}^2 + \text{CR}^2 = (L - x)^2 + x^2 \\ \text{RS}^2 &= \text{RD}^2 + \text{DS}^2 = (L - x)^2 + x^2 \end{aligned}

Therefore, $\text{SP} = \text{PQ} = \text{QR} = \text{RS}$ .

By $\text{SAS}$ congruence, $\Delta \text{SAP} \cong \Delta \text{PBQ} \cong \Delta \text{QCR} \cong \Delta \text{RDS}$ .

Let $\angle \text{ASP} = \alpha$ and $\angle \text{APS} = \beta$ . Since $\alpha + \beta = 90^\circ$ :

\begin{aligned} \angle \text{APS} + \angle \text{SPQ} + \angle \text{BPQ} &= 180^\circ \\ \beta + \angle \text{SPQ} + \alpha &= 180^\circ \\ 90^\circ + \angle \text{SPQ} &= 180^\circ \\ \angle \text{SPQ} &= 90^\circ \end{aligned}

Hence, $\text{PQRS}$ is a square for any $0 < x < L$ .

Answer

$\text{UVWX}$ is a square. Other squares can be constructed by choosing points that divide all four sides in the same cyclic ratio ( $x : L-x$ ).

Common Mistakes

Incomplete Proof: Stopping after showing that all four sides are equal ( $a\sqrt{2}$ ). Proving equal sides only guarantees a rhombus; you must also prove at least one angle is $90^\circ$ to confirm it is a square.
Linear Pair Angle Error: Forgetting that $\angle \text{CUV} + \angle \text{VUX} + \angle \text{AUX} = 180^\circ$ forms a straight line angle along the side of the square.
Cyclic Order in Construction: When choosing points for the general square, lengths must cycle uniformly ( $x$ followed by $L-x$ in the same clockwise or anticlockwise order around all sides).

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