In tetrahedron ABCO, angle AOB = angle AOC = angle BOC = 90^\circ. A cube is inscribed in the tetrahedron so that one of its vertices is at O, and the opposite vertex lies on face ABC. Let OA = 1, OB = 1, OC = 1. Show that the side length of the cube is 1/3.

Guest Mar 14, 2023

#1**0 **

Hello, I just try see below i hope it's work, My Fortiva

Let the side length of the inscribed cube be $s$. Since $\angle AOB = \angle AOC = \angle BOC = 90^\circ$, the triangle $ABC$ is a right triangle. Without loss of generality, assume $AB > AC$ and $BC > AC$. Let $D$ be the foot of the altitude from $O$ to $\triangle ABC$. Then $AD = BD = CD$ because $O$ is the circumcenter of $\triangle ABC$, and $OD = s\sqrt{3}/2$ because $O$ is the centroid of $\triangle ABD$.

[asy]

unitsize(1.5 cm);

pair A, B, C, D, E, F, O, P, Q, R;

A = (0,1);

B = (1,1);

C = (1,0);

O = (1/2,1/2);

D = (1/2,0);

E = (1/2,1);

F = (0,1/2);

P = (O + F)/2;

Q = (O + D)/2;

R = (O + E)/2;

draw(A--B--C--cycle);

draw(A--O);

draw(B--O);

draw(C--O);

draw(D--O);

draw(O--E,dashed);

draw(O--F,dashed);

draw(O--P,dashed);

draw(O--Q,dashed);

draw(O--R,dashed);

draw(E--F);

draw(P--Q);

draw(P--R);

draw(D--E,dashed);

draw(D--F,dashed);

label("$A$", A, N);

label("$B$", B, N);

label("$C$", C, E);

label("$D$", D, S);

label("$E$", E, N);

label("$F$", F, W);

label("$O$", O, N);

label("$P$", P, W);

label("$Q$", Q, S);

label("$R$", R, NE);

label("$s$", (O + R)/2, NE);

[/asy]

Let $E$ be the point on $\overline{AB}$ such that $BE = s$. Then $DE = BD - BE = AD - AE$, so

\begin{align*}

s^2 &= DE^2 \

&= (AD - AE)^2 \

&= AD^2 + AE^2 - 2 \cdot AD \cdot AE \

&= 3s^2/4 + (AB/2 - s)^2 - 2 \cdot 3s/4 \cdot (AB/2 - s) \

&= s^2/4 - 3sAB/8 + AB^2/4.

\end{align*}Solving for $s$, we find $s = AB/3$. It follows that $s = AC/3$ and $s = BC/3$ as well, so $s$ is the side length of the tetrahedron's other inscribed cube, and the desired side length is $s = AB/3 = AC/3 = BC/3 = \boxed{\frac{1}{3}}$.

rona55 Mar 14, 2023