Trirectangular Corner Locus

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STRATEGY: For any two such A and B select a coordinate frame as shown. Then repeatedly use " angle in a semicircle = 90° " to get the locus of P in XZ plane. Finally, by symmetry get the surface as surface of revolution of that locus about Z axis.

From the figure $A = (\cos\alpha,\sin\alpha,0)~~~~B= (\cos\alpha,-\sin\alpha,0)$ Take the symmetric case where $C = (-1,0,0)$

Consider the $\color{#D61F06}{red}$ semicircle shown with AB as diameter. Center M, Radius = $\sin \alpha$ and a point Q vertically above M giving $\angle AQB = 90°$

Now consider another semicircle in XZ plane with CM as diameter shown in $\color{#20A900}{green}$ . All angles inscribed here will be 90° too. So if we can get our point Q on this semicircle, the problem is solved!

So rotate MQ about M till it meets this circle at P as shown in the 'Elevation' below. THAT point P is a trirectangular corner!

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Let $\beta = \angle PMC$ and since $MA = MQ = MP = \sin \alpha$ we get $Px = \cos \alpha - \sin \alpha . \cos \beta; Py = 0; Pz = \sin \alpha . \sin \beta$

From $\triangle CPM, \cos \beta = \frac{\sin \alpha}{1+\cos \alpha}$ and $\sin \beta = \sqrt{\frac{2 \cos \alpha}{1 + \cos \alpha}}$

$\Note:$ I had given a long winded way of showing that locus of P in XZ plane to be the ellipse $P_{x}^2 + 2 P_{z}^2 = 1$ but Prof. Niranjan Khanderia showed me a simple way to get there lot quicker! Here is his method:

$P_{x}^2 = (\cos\alpha - \sin \alpha . \cos \beta)^2$

$= \cos^2 \alpha + \sin^2 \alpha(1-\sin^2\beta) - 2\sin \alpha.\cos\alpha.\cos\beta$

$= 1 - P_{z}^2 - 2\sin\alpha.\cos\alpha.\frac{\sin\alpha}{1+\cos\alpha}$

$= 1 - P_{z}^2 - \sin^2\alpha.\frac{2.\cos\alpha}{1+\cos\alpha}$

$= 1 - P_{z}^2 - \sin^2\alpha.\sin^2\beta$

$= 1 - P_{z}^2 - P_{z}^2$

$= 1 - 2.P_{z}^2$

Thus the locus of P in XZ plane is the ellipse $x^2+2z^2 = 1$ revolving which we get the desired surface - the oblate spheroid: $x^2+y^2+2z^2 = 1$ .

Consider it as a sphere unilaterally scaled (compressed) by a factor $\frac{1}{\sqrt{2}}$ thus giving a volume $\frac{1}{\sqrt{2}}\frac{4}{3}\pi.1^3 = 2.96192...$

Surprising how far simple high school math can take us!

This is a cool problem. I think one simpler way of doing this is to exploit symmetry. Let there be the case where A,B,C are vertices of an inscribed equilateral triangle. Applying the given angle rule forms a regular tetrahedron with base length $\sqrt{3}$ and slant length $\sqrt{3/2}$ . We use the Pythagoras theorem to calculate the height, which is $\sqrt{1/2}$ . Now this is the maximum value in the z-axis. By symmetry will lead to the desired equation of the solid ${x}^{2}+{y}^{2}+{\left(\frac{z}{\sqrt{1/2}}\right)}^{2}=1.$

Use the prismoidal formula (common hack for solids like these), we evaluate:

$\frac{2}{6\sqrt{2}}(0+4\pi+0)=2.9619$

Let $AP, BP, CP$ denote the legs of the tetrahedron , $H$ orthocentre of acute $\triangle ABC$ , $O(0,0,0)$ circumcentre of $\triangle ABC$ and $R$ its circumradius, $P'$ projection of apex $P$ onto $\triangle ABC$ .

$AP \perp BP \perp CP \Rightarrow AP \perp BCP \text{ and } BP \perp ACP \Rightarrow$ $AP\perp BC \text{ and } BP \perp AC \Rightarrow ABCP \text{ orthogonal } \Rightarrow P' = H$

$\vec{AP}=\vec{AH} + \vec{HP} \text{ and } \vec{BP}=\vec{BH} + \vec{HP}$

$\vec{AP} \cdot \vec{BP} = 0 \Rightarrow \vec{AH} \cdot \vec{BH} =-|\vec{HP}|^2$

Let $D$ be a foot of altitude from $A$ on $BC$ and $G$ intersection of line of altitude and the circumcircle.
$|\vec{AH} \cdot \vec{BH}| = AH \times HD \Rightarrow$ $HP^2 =AH \times HD$ Using a fact that $HD = DG$ for any triangle and any vertex and by intersecting chords theorem $AH \times HD = (R^2 - OH^2)/2 \Rightarrow HP^2 = (R^2 - OH^2)/2$ $\Rightarrow 2z^2 =1-(x^2 + y^2)$ Using formula for the volume of an elipsoid: $V = \frac{4}{3 \sqrt{2}} \pi R^3 = 2.961921958772$ More info on tetrahedrons and orthocentre

Note:

$HP$ has similar formula as the polar circle

As a "side effect" of this exercise the following formulas can be easily derived:

$AP^2 = AH \times AD$

$DP^2 = HD \times AD = CD \times BD$ where DP is height of $\triangle BCP$

$AB^2 = AH\times AD + BH \times BE$ - another take on Pythagorean. $E$ is a foot of altitude from $B$ on $AC$

$m{_a}^2 = AH \times AD + \frac{a^2}{4}$ where $m{_a}$ is a median from $A$ .

$m{_a}^2 +m{_b}^2 +m{_c}^2 = \frac{3}{2} (AP^2 + BP^2 + CP^2)$

The surface of the locus of $P$ has the equation

${ x }^{ 2 }+{ y }^{ 2 }+2{ z }^{ 2 }=1$

which means it has the volume of $\frac { 1 }{ \sqrt { 2 } }$ of that of an unit sphere, or $2.96192...$ . so that the answer would be $296$ .