Christmas Streak 55/88: 2018 CSAT (Korean SAT) #29 of Natural Sciences

If we introduce the rotated coordinate frame $Ouvw$ where $u \; = \; \tfrac{1}{\sqrt{5}}(x+2z) \hspace{2cm} v \; = \; y \hspace{2cm} w \; = \; \tfrac{1}{\sqrt{5}}(2x - z)$ then the sphere has equation $u^2 + v^2 + w^2 = 6$ , while $\mathcal{C}$ is the intersection of this sphere with the plane $u=\sqrt{5}$ , so has equation $u=\sqrt{5}$ , $v^2 + w^2 = 1$ . Thus the point $P$ has $(u,v,w)$ -coordinates $(\sqrt{5},-1,0)$ , or $(x,y,z)$ -coordinates $(1,-1,2)$ . Thus $Q$ has $(x,y,z)$ -coordinates $(1,-1,0)$ , and hence its $(u,v,w)$ -coordinates are $(\tfrac{1}{\sqrt{5}},-1,\tfrac{2}{\sqrt{5}})$ . A general point $X$ on $\mathcal{C}$ has $(u,v,w)$ -coordinates $(\sqrt{5},\cos\theta,\sin\theta)$ . Thus. in terms of the $Ouvw$ -coordinate system, $\overrightarrow{PX} + \overrightarrow{QX} \; = \; \left(\begin{array}{c} 0 \\ \cos\theta + 1 \\ \sin\theta \end{array}\right) + \left(\begin{array}{c} \tfrac{4}{\sqrt{5}} \\ \cos\theta + 1 \\ \sin\theta - \tfrac{2}{\sqrt{5}}\end{array}\right) \; = \; \left(\begin{array}{c} \tfrac{4}{\sqrt{5}} \\ 2\cos\theta + 2 \\ 2\sin\theta - \tfrac{2}{\sqrt{5}} \end{array} \right)$ and hence $\big| \overrightarrow{PX} + \overrightarrow{QX}\big|^2 \; = \; 12 + \tfrac{8}{\sqrt{5}}(\sqrt{5}\cos\theta - \sin\theta) \; \le \; 12 + \tfrac{8}{\sqrt{5}} \times \sqrt{6} \; = \; 12 + \tfrac85\sqrt{30}$ making the answer $10(12 + \tfrac{8}{5}) = \boxed{136}$ .

The distance between the center of the sphere $(0,~0,~0)$ and the intersecting plane $x+2z-5=0$ is $\dfrac{5}{\sqrt{5}}=\sqrt{5}.$

The radius of the circle is $\sqrt{6},$ and so the radius of $C$ is $\sqrt{\sqrt{6}^2-\sqrt{5}^2}=1.$

Note that $C$ is an intersection of $x^2+y^2+z^2=6$ and $x=-2z+5.$

$(2z-5)^2+y^2+z^2=6~\Leftrightarrow~5(z-2)^2-1+y^2=0$

Therefore $1-y^2\ge 0~\Leftrightarrow~-1\le y \le 1.$

If $y=-1,$ then $z=2$ and $x=1.$ So $\mathrm{P}(1,~-1,~2)$ and $\mathrm{Q}(1,~-1,~0).$

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Note that $\overrightarrow{\mathrm{PX}}+\overrightarrow{\mathrm{QX}}=2\overrightarrow{\mathrm{MX}}$ where $\mathrm{M}(1,~-1,~1)~(\text{midpoint of }\mathrm{P}\text{ and }\mathrm{Q})$

Then this problem is about mazimizing $\overline{\mathrm{MX}}.$

Shoot a foot of perpendicular from $\mathrm{P}$ to $x+2z-5=0$ and call it $\mathrm{H}.$ Then $\overline{\mathrm{MH}}=\dfrac{2}{\sqrt{5}}.$

And since $\overline{\mathrm{MX}}=\sqrt{\overline{\mathrm{MH}}^2+\overline{\mathrm{HX}}^2},$ so this problem is about maximizing $\overline{\mathrm{HX}}.$

It's obvious that the maximum occurs when $\mathrm{H},~\mathrm{C},~\mathrm{X}$ are on a line in this sequence.

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$\mathrm{C}$ is the foot of perpendicular from the origin to the plane $x+2z-5=0$ which is parallel to the $y$ -axis.

Therefore, $\mathrm{C}(?,~0,~?).$

Thinking on a $xz$ -plane, we can figure out that $\mathrm{C}(1,~0,~2).$

Then we get $\overline{\mathrm{MC}}=\sqrt{2}.$ So we get $\overline{\mathrm{CH}}=\sqrt{\overline{\mathrm{MC}}^2-\overline{\mathrm{MH}}^2}=\sqrt{\sqrt{2}^2-\left(\dfrac{2}{\sqrt{5}}\right)^2}=\dfrac{\sqrt{30}}{5}.$

Obviously we have $\overline{\mathrm{CX}}=1.$

Since we're looking for the maximum,

$\overline{\mathrm{MX}}=\sqrt{\overline{\mathrm{MH}}^2+\overline{\mathrm{HX}}^2} \\ =\sqrt{\left(\dfrac{2}{\sqrt{5}}\right)^2+\left(\overline{\mathrm{HC}}+\overline{\mathrm{CX}}\right)^2} \\ =\sqrt{\dfrac{4}{5}+\left(\dfrac{\sqrt{30}}{5}+1\right)^2} \\ = \sqrt{\dfrac{4}{5}+\dfrac{6}{5}+1+\dfrac{2\sqrt{30}}{5}} \\ =\sqrt{3+\dfrac{2\sqrt{30}}{5}}.$

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$\therefore~\left|\overrightarrow{\mathrm{PX}}+\overrightarrow{\mathrm{QX}}\right|^2=4\left|\overrightarrow{\mathrm{MX}}\right|^2=4\left(3+\dfrac{2\sqrt{30}}{5}\right)=12+\dfrac{8\sqrt{30}}{5}.$

$10(a+b)=10\left(12+\dfrac{8}{5}\right)=120+16=\boxed{136}.$