Circles and Geometric Shapes | EOT

Question 26

How would you use the following figure to justify the statement that the sum of the opposite angles of a cyclic quadrilateral is $180^\circ$ ?

Solution

Understand the Question

Let the rectangle have width $w$ , height $h$ , and total area $R = wh$ .
Place the rectangle on a coordinate plane to represent the vertices and calculate the areas of triangles $A$ , $B$ , and $C$ in terms of coordinates.
Express the coordinates in terms of the areas $A$ and $B$ , and substitute them into the formula for area $C$ to derive a relationship involving $R$ , $A$ , $B$ , and $C$ .
Check whether the proposed formula $R = \dfrac{2(A+C)(B+C)}{C}$ satisfies this geometric relationship.

Step 1 · Define Rectangle and Coordinates

Question diagram

Let the rectangle have width $w$ and height $h$ , so its area is: $R = wh$

Set up coordinates with the bottom-left corner at the origin:

Bottom-left: $(0,0)$
Bottom-right: $(w,0)$
Top-left: $(0,h)$
Top-right: $(w,h)$
Top-edge point: $P_T = (x_T, h)$
Internal point: $P_I = (x_I, y_I)$

Step 2 · Express Triangle Areas

For Triangle $A$ with vertices $(0,0)$ , $(0,h)$ , and $(x_T,h)$ :

Base on the left edge $= h$
Height $= x_T$ $A = \dfrac{1}{2} \times h \times x_T$

For Triangle $B$ with vertices $(0,0)$ , $(w,0)$ , and $(x_I,y_I)$ :

Base on the bottom edge $= w$
Height $= y_I$ $B = \dfrac{1}{2} \times w \times y_I$

For Triangle $C$ with vertices $(x_T,h)$ , $(w,h)$ , and $(x_I,y_I)$ :

Base on the top edge $= w - x_T$
Height $= h - y_I$ $C = \dfrac{1}{2} \times (w-x_T) \times (h-y_I)$

Step 3 · Derive Relationship for Rectangle Area $R$

From the area equations of $A$ and $B$ , express $x_T$ and $y_I$ : $x_T = \dfrac{2A}{h}, \quad y_I = \dfrac{2B}{w}$

Substitute these into the expression for $C$ : $C = \dfrac{1}{2} \left(w - \dfrac{2A}{h}\right)\left(h - \dfrac{2B}{w}\right)$

$2C = \left(w - \dfrac{2A}{h}\right)\left(h - \dfrac{2B}{w}\right)$

Expanding the right-hand side:

\begin{aligned} 2C &= wh - w\left(\dfrac{2B}{w}\right) - h\left(\dfrac{2A}{h}\right) + \left(\dfrac{2A}{h}\right)\left(\dfrac{2B}{w}\right) \\[0.6em] &= wh - 2B - 2A + \dfrac{4AB}{wh} \end{aligned}

Since $R = wh$ : $2C = R - 2A - 2B + \dfrac{4AB}{R}$

Multiplying by $R$ : $2CR = R^2 - 2AR - 2BR + 4AB$

Rearranging into standard quadratic form:

\begin{aligned} R^2 - 2AR - 2BR - 2CR + 4AB &= 0 \\[0.6em] R^2 - (2A + 2B + 2C)R + 4AB &= 0 \end{aligned}

Step 4 · Evaluate the Given Formula

Let the given expression for the area be $R_{\text{formula}}$ : $R_{\text{formula}} = \dfrac{2(A+C)(B+C)}{C}$

Expanding $R_{\text{formula}}$ :

\begin{aligned} R_{\text{formula}} &= \dfrac{2(AB + AC + BC + C^2)}{C} \\[0.6em] &= \dfrac{2AB}{C} + \dfrac{2AC}{C} + \dfrac{2BC}{C} + \dfrac{2C^2}{C} \\[0.6em] &= \dfrac{2AB}{C} + 2A + 2B + 2C \end{aligned}

Let $S = 2A + 2B + 2C$ , so $R_{\text{formula}} = \dfrac{2AB}{C} + S$ .

Substitute $R_{\text{formula}}$ into the quadratic equation $R^2 - SR + 4AB = 0$ :

\begin{aligned} \left(\dfrac{2AB}{C} + S\right)^2 - S\left(\dfrac{2AB}{C} + S\right) + 4AB &= 0 \\[0.6em] \left(\dfrac{2AB}{C}\right)^2 + 2S\left(\dfrac{2AB}{C}\right) + S^2 - S\left(\dfrac{2AB}{C}\right) - S^2 + 4AB &= 0 \\[0.6em] \left(\dfrac{2AB}{C}\right)^2 + S\left(\dfrac{2AB}{C}\right) + 4AB &= 0 \end{aligned}

Substitute $S = 2A + 2B + 2C$ back:

\begin{aligned} \dfrac{4A^2B^2}{C^2} + (2A + 2B + 2C)\left(\dfrac{2AB}{C}\right) + 4AB &= 0 \\[0.6em] \dfrac{4A^2B^2}{C^2} + \dfrac{4A^2B}{C} + \dfrac{4AB^2}{C} + \dfrac{4ABC}{C} + 4AB &= 0 \end{aligned}

Dividing through by $4AB$ (since $A, B > 0$ ):

\begin{aligned} \dfrac{AB}{C^2} + \dfrac{A}{C} + \dfrac{B}{C} + 1 + 1 &= 0 \\[0.6em] \dfrac{AB}{C^2} + \dfrac{A}{C} + \dfrac{B}{C} + 2 &= 0 \end{aligned}

Multiplying by $C^2$ : $AB + AC + BC + 2C^2 = 0$

Since $A, B, C > 0$ are positive geometric areas, the sum $AB + AC + BC + 2C^2$ must be strictly positive, leading to a contradiction.

Diagram 1

Answer

The geometric relationship leads to the quadratic equation $R^2 - (2A+2B+2C)R + 4AB = 0$ . Substituting $R = \dfrac{2(A+C)(B+C)}{C}$ yields $AB + AC + BC + 2C^2 = 0$ , which is impossible for positive areas $A, B, C$ .

Common Mistakes

Assuming Validity Without Verification: Directly assuming the given formula is an identity without checking if it satisfies the underlying quadratic equation derived from the coordinates.
Sign and Expansion Errors: Miscalculating cross-terms when expanding $\left(w - \dfrac{2A}{h}\right)\left(h - \dfrac{2B}{w}\right)$ , leading to incorrect signs for the linear coefficients in the quadratic equation.

Question 26

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