Problem 1

Area of Squares

Answer Format: Proof

As shown in the illustration below, two squares are contained within a semicircle. Determine the total area of the squares and prove it's consistency or inconsistency among all cases.

Detailed Description: The upper-left point of square A and the upper-right point of square B touch the semicircle arc. The bottom-right point of A and the bottom-left point of B lie on the same location on the straight line base of the semicircle.

Note: This proof is not unique. Your proof might be different but equally valid. If your proof is different, I would love to check it out! Send me an email and I'll get back to you as fast as I can!

Claim:

A+B=r2A + B = r^2 where rr is the radius of the semicircle.

Proof:

Let rr be the radius of the semicircle and PAP_A, PBP_B be the points of contact with the semicircle as shown below.

The y-positions of PAP_A and PBP_B are equal to the side lengths of squares AA and BB. If θA\theta_A and θB\theta_B represent the angles of vectors PAP_A and PBP_B respectively, we can write AA and BB as,

A=(PAy)2=(rsinθA)2B=(PBy)2=(rsinθB)2 \begin{aligned} A &= (P_{Ay})^2 = (rsin\theta_{A})^2\\ B &= (P_{By})^2 = (rsin\theta_{B})^2\\ \end{aligned}

Finding the sum,

A+B=(rsinθA)2+(rsinθB)2=r2sin2θA+r2sin2θB=r2(sin2θA+sin2θB) \begin{aligned} A+B&=(rsin\theta_{A})^2 + (rsin\theta_{B})^2\\ &=r^2sin^2\theta_{A} + r^2sin^2\theta_{B}\\ &=r^2(sin^2\theta_{A} + sin^2\theta_{B}) \end{aligned}

The total area is equal to r2r^2 multiplied by sin2θA+sin2θBsin^2\theta_{A} + sin^2\theta_{B}.


To evaluate sin2θA+sin2θBsin^2\theta_{A} + sin^2\theta_{B}, notice that the total base of the squares is equal to the sum of the square heights.

This is represented in equation form like so,

∣PAx∣+∣PBx∣=∣PAy∣+∣PBy∣ \begin{aligned} |P_{Ax}| + |P_{Bx}| &= |P_{Ay}| + |P_{By}| \end{aligned}

We can evaluate this to determine a relationship between θA\theta_A and θB\theta_B. Remember sum-to-product identities: sinθ+sinφ=2sin(θ+φ2)cos(θ−φ2)sin\theta+sin\varphi=2sin(\frac{\theta+\varphi}{2})cos(\frac{\theta-\varphi}{2}) and cosθ−cosφ=−2sin(θ+φ2)sin(θ−φ2)cos\theta-cos\varphi=-2sin(\frac{\theta+\varphi}{2})sin(\frac{\theta-\varphi}{2}).

∣PAx∣+∣PBx∣=∣PAy∣+∣PBy∣−PAx+PBx=PAy+PBy−rcosθA+rcosθB=rsinθA+rsinθBcosθB−cosθA=sinθA+sinθB−2sin(θB+θA2)sin(θB−θA2)=2sin(θA+θB2)cos(θA−θB2)−sin(θB−θA2)=cos(θA−θB2)sin(−θB−θA2)=cos(θA−θB2)cos(π2+θB−θA2)=cos(θA−θB2)π2+θB−θA2=θA−θB2π+θB−θA=θA−θBπ=2θA−2θBπ2=θA−θBθB=θA−π2 \begin{aligned} |P_{Ax}| + |P_{Bx}| &= |P_{Ay}| + |P_{By}|\\ -P_{Ax} + P_{Bx} &= P_{Ay} + P_{By}\\ -rcos\theta_{A} + rcos\theta_{B} &= rsin\theta_{A} + rsin\theta_{B}\\ cos\theta_{B} - cos\theta_{A} &= sin\theta_{A} + sin\theta_{B}\\ -2sin(\frac{\theta_B+\theta_A}{2})sin(\frac{\theta_B-\theta_A}{2}) &= 2sin(\frac{\theta_A+\theta_B}{2})cos(\frac{\theta_A-\theta_B}{2})\\ -sin(\frac{\theta_B-\theta_A}{2}) &= cos(\frac{\theta_A-\theta_B}{2})\\ sin(-\frac{\theta_B-\theta_A}{2}) &= cos(\frac{\theta_A-\theta_B}{2})\\ cos(\frac{\pi}{2}+\frac{\theta_B-\theta_A}{2}) &= cos(\frac{\theta_A-\theta_B}{2})\\ \frac{\pi}{2}+\frac{\theta_B-\theta_A}{2} &= \frac{\theta_A-\theta_B}{2}\\ \pi+\theta_B-\theta_A &= \theta_A-\theta_B\\ \pi &= 2\theta_A-2\theta_B\\ \frac{\pi}{2} &= \theta_A-\theta_B\\ \theta_B &= \theta_A-\frac{\pi}{2}\\ \end{aligned}

Plugging into our original equation,

A+B=r2(sin2θA+sin2θB)=r2(sin2θA+sin2(θA−π2))=r2(sin2θA+cos2θA)=r2 \begin{aligned} A+B &= r^2(sin^2\theta_A + sin^2\theta_B)\\ &= r^2(sin^2\theta_A + sin^2(\theta_A-\frac{\pi}{2}))\\ &= r^2(sin^2\theta_A + cos^2\theta_A)\\ &= r^2 \end{aligned}

QEDQED