fixed point for planes of triangle, max area, sphere and perpendiculars related

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#1 h May 17, 2019, 2:52 PM • 2 Y

Y by Adventure10, Mango247

On a sphere with radius $1$ , a point $P$ is given. Three mutually perpendicular the rays emanating from the point $P$ intersect the sphere at the points $A$ , $B$ and $C$ . Prove that all such possible $ABC$ planes pass through fixed point, and find the maximum possible area of the triangle $ABC$

This post has been edited 1 time. Last edited by parmenides51, May 17, 2019, 3:04 PM
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#2 h May 17, 2019, 5:06 PM • 2 Y

Y by Adventure10, Mango247

Denote $S$ as the center of the sphere. Let $X$ be a point such that $\vec{SX}=\frac{1}{3}\cdot \vec{SP}$ . We will prove that $X$ is the required point. If plane $ABC$ contains line $PS$ then it's right. Now assume that $PS$ and the plane have only one common point. Take points $D,E,F,G$ such that quadrilaterals $PAEC, PBGC, PADB,CEFG$ are rectangles. Then solid $PADBCEFG$ is a cuboid. So center of this cuboid is the center $S$ of our sphere. Let $M$ be a midpoint of segment $AB$ . Because $CP\perp PBA\perp SM$ we know that points $P,C,S,M$ lie in one plane. Because $CP\perp PM\perp SM\wedge |CP|=2|SM|$ we know that $X$ is the point of intersection of segments $CM$ and $PS$ using similarities (you have to make a foot of this hypothetic point $X$ onto segment $PM$ ). So $X$ which is a fixed point lies on a segment $CM$ which belongs to plane $ABC$ .

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#3 h May 17, 2019, 5:11 PM • 2 Y

Y by Adventure10, Mango247

Maximal area
Denote $|PA|=a\wedge| PB|=b\wedge |PC|=c$ . We know that diameter is equal to diagonal of cuboid so $a^2+b^2+c^2=4$ . By Heron's and AM-GM, then AM-QM
$\frac{16}{3}[ABC]^2=\frac{a+b+c}{3}\cdot (a+b-c)(a-b+c)(-a+b+c)\le\left(\frac{a+b+c}{3}\right)^4\le \left(\frac{a^2+b^2+c^2}{3}\right)^2=\frac{16}{9}$ . Hence $[ABC]\le \frac{1}{\sqrt3}$ attainable iff $a=b=c$

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#4 h May 17, 2019, 5:31 PM • 2 Y

Y by Adventure10, Mango247

We follow a more indirect approach. Set $P$ as the origin of the system (which is variable). Let the center of the sphere be $(a, b, c)$ . The points where the sphere hits the $x, y, z$ axes respectively are $A=(2a, 0, 0), B=(0, 2b, 0), C=(0, 0, 2c)$ respectively (because the perpendicular from the center to a chord bisects it). So the plane is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c} = 2$ . The point $\left(\frac{2a}{3}, \frac{2b}{3}, \frac{2c}{3}\right)$ is on the plane, which is the point dividing the segment joining $P$ and the center in the ratio $1:2$ , and is thus fixed for varying $P$ . As for the area, we need to maximize (due to properties of the area vector) $\frac{1}{2}|A \times B + B \times C + C \times A| = \sqrt{a^2 b^2 + b^2 c^2 + c^2 a^2}$ under the constraint $a^2+b^2+c^2 = 1$ . But then we have $3(a^2 b^2 + b^2 c^2 + c^2 a^2) \le a^4 + b^4 + c^4 + 2(a^2 b^2 + b^2 c^2 + c^2 a^2) = (a^2+b^2+c^2)^2 = 1$ , so the expression we need to maximise is at most $\frac{1}{\sqrt{3}}$ . This is indeed attainable at $a=b=c$ , so we are done.

This post has been edited 1 time. Last edited by WizardMath, May 17, 2019, 5:32 PM
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#5 h May 18, 2019, 2:41 PM • 2 Y

Y by Adventure10, Mango247

Another analytic approach to a fixed point
Take this sphere as $x^2+y^2+z^2=1$ . Let $P=(a,b,c)$ . We can rotate the sphere so that $PX$ is parallel to axis $OX$ and plane $PXY$ is parallel to plane $z=0$ . So $X=(-a,b,c)$ and $Y=(a,-b,c)$ . Using dot product it's easy to verify that $Z=(a,b,-c)$ . Now we use coplanarity criterion on plane $XYZ$ and point $\left(\frac x3,\frac y3,\frac z3\right)$ .Well,
$\left|\begin{array}{ccc}\frac{4x}{3}&\frac{-2y}{3}&\frac{-2z}{3}\\2x&0&-2z\\2x&-2y&0\end{array}\right|=0$ so that's it.

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