Geometry Question. How to do?

Question

In a space, $\Pi$ is a plane and $l$ is a straight line, and they intersect at the point $O$. Suppose $A$, $C$, $E$ are points on $\Pi$ while points $B$, $D$ are points on $l$, such that $AB$ and $CD$ are perpendicular to $l$, while $BC$ and $DE$ are perpendicular to $\Pi$. If $OA=72$, $OE=42$ and $AE=54$, find the length of $OC$.

Answer 1

Since $DE$ and $BC$ are orthogonal to plane $\Pi$, the points $E$ and $C$ are the orthogonal projections of points $D$ and $B$ respectively on the plane $\Pi$. Hence the line $CE$ is the orthogonal projection of $l$ and must pass through the point $O = l \cap \Pi$. Thus, the plane $\Pi^*$ spanned by $l$ and its orthogonal projection $CE$ is orthogonal to $\Pi$.

Let $M$ be the midpoint of segment $AO$. Since $AO = 72$ the two halves $AM = OM = 36$. Construct the sphere $s_M$ centered at point $M$ and of radius $36$. Then points $O$ and $A$ lie on the sphere $s_M$. Furthermore, let $k = s_M \cap \Pi^*$ be the circle formed by the intersection of the sphere $s_M$ with plane $\Pi^*$. Let $H$ be the second intersection point of circle $k$ with the line $CE$, the other one being $O$. Then, by construction $H$ is the orthogonal projection of point $A$ on $CE$, i.e. $\angle \, AHO = 90^{\circ}$ because, say $H \in k \subset s_M$ so $HM = AM = OM = 36$ which means that triangle $AHO$ is right-angled. Since the plane $\Pi$ passes through the center of the sphere $s_M$, its intersection line $OH$, which coincides with $CE$, passes through the center of the circle $k$, so $HO$ is a diameter of $k$.

By assumption, point $B \in l$ is such that $\angle \, ABD = 90^{\circ}$ so the triangle $ABO$ is right angled with $M$ midpoint of the hypotenuse $AO$. Therefore, $BM = AM = OM = 36$, so $B$ lies on the sphere $s_M$ and since it is on the plane $\Pi^*$, it must lie on the circle $k$. Consequently, triangle $HBO$ is right-angled with $\angle \, HBO = 90^{\circ}$. Furthermore, $BC$ is the altitude through $B$ of the triangle $HBO$ and since $D \in l$ and $CD \perp l$ by assumption, triangle $CDO$ is right-angled.

Now the calculations. Let $\angle \, AOE = \beta$. By the cosine formula in triangle $AEO$, $$\cos(\beta) = \frac{AO^2 + EO^2 - AE^2}{2 \cdot AO \cdot EO} = \frac{72^2 + 42^2 - 54^2}{2 \cdot 72 \cdot 42} = \frac{2}{3}$$ Since triangle $AHO$ is right-angled, $$\frac{HO}{AO} = \cos(\beta) = \frac{2}{3}$$ so $$HO = \frac{2}{3} \, AO = \frac{2}{3} \, 72 = 48$$ Therefore, $HE = HO - EO = 48 - 42 = 6$. Denote by $x = CE$. Then $HC = HE - CE = 6 - x$ while $OC = CE + EO = x + 42$. Now, in the right-angled triangles $HBO$ and $CDO$ the following identities hold: $$BC^2 = HC \cdot OC = (6-x)(x+42) \,\,\, \text{ and } \,\,\, DE^2 = CE \cdot OE = x \cdot 42 = 42 x$$ By the similarity of triangles $DEO$ and $BCO$, $$\frac{DE}{BC} = \frac{OE}{OC}$$ which when squared on both sides becomes $$\frac{DE^2}{BC^2} = \frac{OE^2}{OC^2}$$ $$\frac{42x}{(6-x)(x+42)} = \frac{42^2}{(42+x)^2}$$ Cancel out the common terms $$\frac{x}{6-x} = \frac{42}{42+x}$$ When you solve it, you get $$x = 12 \sqrt{14} - 42$$ which means that $$OC = OE + EC = 42 + 12 \sqrt{14} - 42 = 12 \sqrt{14}$$