Distance from one vertex of a rectangular prism to the plane determined by three other vertices, without using vectors or calculus

Question

The objective is to find the shortest distance from the point $H$ to the plane $BDE$. The prism $ABCD.EFGH$ has $AB=AD=5\sqrt{2}$ and $AE=12$. I think that these numbers are badly selected by the author.

The following shows my steps to solution but I feel it is too verbose, tedious, and time consuming.

It can be solved easily with either vector (dot and cross products) or calculus (minimizing a distance function of two variables), but because this topic is for students who have not learned those subjects, I insist on only using not more than Pythagorean theorem and basic trigonometry. Analytic approach is not allowed!

Finding $DE$, $EP$ and $HP$

\begin{align*} DE^2 &=DH^2+EH^2\\ &=12^2 + (5\sqrt2)^2 \\ &=194\\ DE &=\sqrt{194} \end{align*}

\begin{align*} \frac{1}{HP^2} &=\frac{1}{DH^2}+\frac{1}{EH^2}\\ &=\frac{1}{122^2}+\frac{1}{(5\sqrt2)^2}\\ &=\frac{97}{3600}\\ HP &=\frac{60}{\sqrt{97}} \end{align*}

\begin{align*} DE \times EP &=EH^2\\ EP\sqrt{194} &= (5\sqrt2)^2 \\ EP &=\frac{50}{\sqrt{194}} \end{align*}

Finding $BD$, $\cos E$, $EQ$ and $PQ$

\begin{align*} BD &= \sqrt{CD^2+BC^2}\\ &= \sqrt{(5\sqrt2)^2+(5\sqrt2)^2}\\ &= \sqrt{50+50}\\ &= 10 \end{align*}

\begin{gather*} BE^2+DE^2-2\times BE\times DE \cos E = BD^2 \\ 194 + 194 - 2\times 194 \cos E = 100\\ \cos E = \frac{72}{97} \end{gather*}

\begin{align*} EQ &= EP \sec E\\ &= \frac{50}{\sqrt{194}}\times \frac{97}{72} \\ &= \frac{2425}{36\sqrt{194}} \end{align*}

\begin{align*} PQ &= \sqrt{EQ^2-EP^2}\\ &= \sqrt{\left(\frac{2425}{36\sqrt{194}}\right)^2-\left(\frac{50}{\sqrt{194}}\right)^2}\\ &= \frac{1625}{36\sqrt{194}} \end{align*}

Finding $HQ$

\begin{align*} HQ &= \sqrt{EQ^2+EH^2}\\ &= \sqrt{\left(\frac{2425}{36\sqrt{194}}\right)^2 +\left(5\sqrt{2}\right)^2}\\ &= \frac{5\sqrt{15218}}{72} \end{align*}

Finding the altitude, of $\triangle HPQ$, passing through $P$ and finding $HH'$

As the badly chosen numbers make the calculation a bit complicated. The process is left behind as your excercise. The altitude is $t = 1500/\sqrt{738073}$.

$HH'$ then can be found by equating the area from two different bases. \begin{align*} PQ \times HH' &= HQ \times t\\ \frac{1625}{36\sqrt{194}} \times HH' &= \frac{5\sqrt{15218}}{72} \times \frac{1500}{\sqrt{738073}}\\ HH' &= \frac{60}{13} \end{align*}

Thus the shortest distance from the point $H$ to the plane $BDE$ is $\tfrac{60}{13}$.

Question

Is there any shorter way to solve it but with neither using vector nor using calculus?

Answer 1

It's worth noting that $H$ and $A$ are equidistant from the plane of $\triangle BDE$. (Consider the symmetry of the points' positions relative to the midpoint of $\overline{DE}$ and then to that plane.) If one can convince oneself of that first, then it might be easier to think-through the problem. But if not, that's okay, too.

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Consider tetrahedron $HBDE$. Relative to base $\triangle HDE$, it has height $|AB|$; relative to base $\triangle BDE$, it has height $|HH'|$ (our target length). Expressing volume in two ways, we have

$$\frac13 |AB|\cdot|\triangle HDE| = \frac13|HH'|\cdot|\triangle BDE| \tag{1}$$ Therefore, writing $M$ for the midpoint of $\overline{BD}$, $$|HH'|=\frac{|AB|\cdot|\triangle HDE|}{|\triangle BDE|} = \frac{|AB|\cdot\tfrac12|HD||HE|}{\tfrac12|BD||ME|} = \frac{|AB||AD||AE|}{|BD||ME|} = \frac{|AB|^2|AE|}{|BD||ME|}\tag{2}$$ (Note that you'd get the same formula for $|AA'|$, with $A'$ the projection of $A$ onto the plane, by considering tetrahedron $ABDE$, which has the same volume as $HBDE$.) By Pythagoras, we have $$\begin{align} |BD|^2 &= |AB|^2+|AD|^2 \\[4pt] &= 2|AB|^2 \tag{3}\\[4pt] |ME|^2 &= |BE|^2-\left(\tfrac12|BD|\right)^2 \\ &= |AB|^2+|AE|^2-\tfrac12|AB|^2 \\ &= \tfrac12|AB|^2+|AE|^2 \tag{4} \end{align}$$

Substituting the values $|AB|=|AD|=5\sqrt2$ and $|AE|=12$, we find $$|BD|^2 = 100 \quad\to\quad |BD|=10 \qquad\qquad |ME|^2 = 169\quad\to\quad |ME|=13 \tag{5}$$ (These make me think that the problem's given lengths are not so "badly selected".) Therefore,

$$|HH'| = \frac{600}{130}=\frac{60}{13}\tag{$\star$}$$

which agrees with the value derived by OP. $\square$

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We could have calculated $|\triangle BDE|$ via Heron's formula. Alternatively (but equivalently), we could invoke de Gua's Theorem, the dimensionally-enhanced Pythagorean Theorem for right-corner tetrahedra that deserves to be better-known: $$|\triangle BDE|^2 = |\triangle ABD|^2 + |\triangle BAE|^2 + |\triangle BDA|^2 \tag{6}$$ so that $$\begin{align} |\triangle BDE|^2 &=\left(\tfrac12|AB||AD|\right)+\left(\tfrac12|AB||AE|\right)^2+\left(\tfrac12|AB||AD|\right)^2 \tag{7}\\[4pt] &=\tfrac14\left(|AB|^2|AD|^2+|AB|^2|AE|^2+|AB|^2|AD|^2\right) \tag{8} \\[4pt] &=\tfrac14|AB|^2|AD|^2|AE|^2\left(\frac{1}{|AB|^2}+\frac{1}{|AD|^2}+\frac{1}{|AE|^2}\right) \tag{9} \end{align}$$ and thus, from $(2)$,

$$|HH'| = \frac{1}{\sqrt{\dfrac{1}{|AB|^2}+\dfrac{1}{|AD|^2}+\dfrac{1}{|AE|^2}}} \tag{$\star\star$}$$

which gives the same value of $60/13$.

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An analytical addendum: With $A$ at the origin, and $B$, $D$, $E$ along the $x$-, $y$-, $z$-axes, the plane of $\triangle BDE$ has intercept-intercept-intercept form $$\frac{x}{|AB|}+\frac{y}{|AD|}+\frac{z}{|AE|}=1 \tag{10}$$ Thus, the distance from the plane to $(x,y,z)$ is given by $$\frac{\left|\dfrac{x}{|AB|}+\dfrac{y}{|AD|}+\dfrac{z}{|AE|}-1\right|}{\sqrt{\dfrac{1}{|AB|^2}+\dfrac{1}{|AD|^2}+\dfrac{1}{|AE|^2}}} \tag{11}$$ Substituting $(x,y,z)\to A=(0,0,0)$ or $(x,y,z)\to H=(0,|AD|,|AE|)$ gives $(\star\star)$.