1846 Sangaku Maximum

Let us define the tilting angle as $x$ and let us mark vertices $A$ , $B$ , $C$ , $D$ and $E$ as in figure below. Let us also call the purple side as $y$ .

We will have:

$\large \displaystyle y = 1 + \overline{DE}$

Notice also that:

$\large \displaystyle \overline{BC} = \cos(x) + \sin(x) - 1$

$\large \displaystyle \overline{AB} = 1 + \cos(x)$

$\large \displaystyle \overline{AD} = \cos(x) + \sin(x) + 1$

Triangles $\Delta ABC$ and $\Delta ADE$ are similar, so:

$\large \displaystyle \frac{\overline{DE}}{\overline{BC}} = \frac{\overline{AD}}{\overline{AB}}$

$\large \displaystyle \overline{DE} = \frac{ (\cos(x) + \sin(x) -1)(\cos(x) + \sin(x) + 1)}{1 + \cos(x)}$

$\large \displaystyle \overline{DE} = \frac{ (\cos(x) + \sin(x))^2 - 1}{1 + \cos(x)}$

$\large \displaystyle \overline{DE} = \frac{ \sin(2x)}{1 + \cos(x)}$

$\color{#20A900} \boxed { \large \displaystyle y = 1 + \frac{ \sin(2x)}{1 + \cos(x)} }$

Differentiating $y$ with respect to $x$ and making it equal to $0$ :

$\large \displaystyle \frac{dy}{dx} = \frac{2\cos(2x) (1 + \cos(x)) - \sin(2x) ( -\sin(x) ) }{ (1+\cos(x))^2 } = 0$

$\large \displaystyle \frac{dy}{dx} = \frac{ 2 (2\cos^2(x)-1) (1 + \cos(x)) + 2\sin^2(x) \cos(x) }{ (1+\cos(x))^2 } = 0$

$\large \displaystyle \frac{dy}{dx} = \frac{ 2 (2\cos^2(x)-1) (1 + \cos(x)) + 2( 1 - \cos^2(x) ) \cos(x) }{ (1+\cos(x))^2 } = 0$

Since $0 \leq x \leq \frac{\pi}{2}$ , $1 + \cos(x) \neq 0$ . So:

$\large \displaystyle \cos^3(x) +2\cos^2(x) -1 = 0$

$\large \displaystyle (\cos(x) + 1)(\cos^2(x) + \cos(x) -1) = 0$

$\cos(x) = -1$ is impossible since $0 \leq x \leq \frac{\pi}{2}$ . The other two solutions are $\cos(x) = - \frac{1 + \sqrt{5}}{2} \approx -1.618$ , which is also impossible, and:

$\color{#20A900} \boxed { \large \displaystyle \cos(x) = \frac{ \sqrt{5} - 1}{2} }$

Which leads to:

$\color{#20A900} \boxed { \large \displaystyle \sin(x) = \frac{ \sqrt{2\sqrt{5} - 2 }}{2} }$

So:

$\large \displaystyle y = 1 + \frac{ \sin(2x)}{1 + \cos(x)}$

$\large \displaystyle y = 1 + \frac{ 2\sin(x)\cos(x)}{1 + \cos(x)}$

$\large \displaystyle y = 1 + \frac{ \left ( \sqrt{2\sqrt{5} - 2} \right )(\sqrt{5} - 1)}{\sqrt{5} + 1}$

Multiplying above and below by $\sqrt{5} - 1$ :

$\large \displaystyle y = 1 + \frac{ \left ( \sqrt{2\sqrt{5} - 2} \right )(6 - 2\sqrt{5})}{4}$

$\large \displaystyle y = 1 + \frac{ \sqrt{( 2\sqrt{5} - 2)(6 - 2\sqrt{5})^2}}{4}$

$\large \displaystyle y = 1 + \frac{ \sqrt{160\sqrt{5} - 352}}{\sqrt{16} }$

$\color{#20A900} \boxed{ \large \displaystyle y = 1 + \sqrt{10\sqrt{5} - 22} }$

So:

$\color{#3D99F6} \large \displaystyle a= 10, b = 5, c = 22 \rightarrow \boxed{\large \displaystyle a+b+c = 37}$

Label the trapezoid as $ABCD$ and altitude of the top vertex of the blue unit square as $EF$ ; and let the angle a side of the blue square makes with the vertical be $\theta$ as shown.

Then $EF = \sin \theta + \cos \theta$ , $DF = 1 + \cos \theta$ , and $DC=1+\sin \theta + \cos \theta$ , then

$\begin{aligned} \frac {BC-1}{EF-1} & = \frac {DC}{DF} \\ \implies BC & = 1 + \frac {DC(EF-1)}{DF} & \small \blue{\text{Let }BC = h} \\ h & = 1 + \frac {(\sin \theta + \cos \theta +1 )(\sin \theta + \cos \theta -1)}{1+\cos \theta} \\ & = 1 + \frac {2\sin \theta \cos \theta}{1+\cos \theta} \\ & = 1 + \frac {2 \left(2 \sin \frac \theta 2 \cos \frac \theta 2\right)\left( 2\cos^2 \frac \theta 2-1\right)}{1+2\cos^2 \frac \theta 2-1} \\ & = 1 + 2 \sin \theta - 2\tan \frac \theta 2 & \small \blue{\text{Let }t = \tan \frac \theta 2} \\ & = 1 + \frac {4t}{1+t^2} - 2t \\ & = 1 + \frac {2t(1-t^2}{1+t^2} & \small \blue{\text{See note: }h \text{ is maximum,}} \\ \implies \max(h) & = \frac {2\sqrt{\sqrt 5-2}(3-\sqrt 5)}{\sqrt 5 -1} & \small \blue{\text{when }t = \sqrt{\sqrt 5-2}} \\ & = 1 + \frac {2\sqrt{\sqrt 5-2}(3-\sqrt 5)(\sqrt 5+1)}{(\sqrt 5 -1)(\sqrt 5 +1)} \\ & = 1 + \sqrt{\sqrt 5-2}(\sqrt 5-1) \\ & = 1 +\sqrt{(\sqrt 5-2)(6-2\sqrt 5)} \\ & = 1 + \sqrt{10 \sqrt 5-22} \end{aligned}$

Therefore $a+b+c = 10 + 5 + 22 = \boxed {37}$ .

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Note: To find $\max(h)$ , we first find $t$ , when $\dfrac {dh}{dt} = 0$ .

$\begin{aligned} \frac {dh}{dt} & = \frac {4(1+t^2) - 4t(2t)}{(1+t^2)^2} - 2 & \small \blue{\text{Equating to zero}} \\ \implies \frac {4-4t^2}{(1+t^2)^2} & = 2 \\ 2-2t^2 & = t^4 + 2t^2 + 1 \\ t^4 + 4t^2 - 1 & = 0 \\ \implies t & = \sqrt{\sqrt 5 - 2} \end{aligned}$

Draw in these lines and label the diagram as follows:

From right $\triangle HCG$ and $GH = 1$ , $CG = \sin \theta$ and $CH = \cos \theta$ . By symmetry, $CG = HI = EJ = FK = \sin \theta$ and $CH = EI = FJ = GK = \cos \theta$ .

$\triangle AME \sim \triangle ANL$ by AA symmetry, so $\frac{LN}{AN} = \frac{EM}{AM}$ . Letting $y = LK$ ,

• $LN = LK - NK = y - 1$

• $AN = CD + CG + KG = \sin \theta + \cos \theta + 1$

• $EM = BI = CH + HI - BC = \sin \theta + \cos \theta - 1$

• $AM = AB + BM = AB + EI = \sin \theta + 1$

Substituting these into the ratio gives $\frac{y - 1}{\sin \theta + \cos \theta + 1} = \frac{\sin \theta + \cos \theta - 1}{\sin \theta + 1}$ which simplifies to $y = \frac{2 \sin \theta \cos \theta}{\sin \theta + 1} + 1$ .

The maximum of $y = LK$ occurs when the derivative $y'$ is zero, therefore, $y' = \frac{2(1 + \sin \theta)(\cos^2 \theta - \sin^2 \theta) - 2 \sin \theta \cos^2 \theta}{(\sin \theta + 1)^2} = 0$ . Using the identity $\cos^2 \theta = 1 - \sin^2 \theta$ , this simplifies to $(\sin \theta + 1)(\sin^2 \theta + \sin \theta - 1) = 0$ , which has one positive solution of $\sin \theta = \frac{\sqrt{5} - 1}{2}$ .

This makes $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{\frac{\sqrt{5} - 1}{2}}$ , which makes the maximum of $LK$ to be $y = \frac{2 \sin \theta \cos \theta}{\sin \theta + 1} + 1 = 1 + \sqrt{10\sqrt{5} - 22}$ , so that $a = 10$ , $b = 5$ , $c = 22$ , and $a + b + c = \boxed{37}$ .

Let the angle of bottom left side of right square as x, determine the slope M(x) of inclined line of trapezoid: M=[{(2^0.5)sin(x+pi/4)-1}/{1+cos(x)-(2^0.5)cos(x+pi/4)}], and the vertical right side H of trapezoid: H=[1+{1+sin(x)+cos(x)}M]. Maximum of H(x) gives H=1+[10{(5)^0.5}-22]^0.5. Therefore, a+b+c=10+5+22 Answer=37