It's All Tetrahedrons 2 - Reposted

Geometry Level pending

In the irregular tetrahedron above, find the m B C D m\angle{BCD} that minimizes the triangular face B C D BCD when the volume of the tetrahedron is held constant.

Refer to previous problem


The answer is 45.

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1 solution

Rocco Dalto
Oct 13, 2018

A E C \triangle{AEC} is an isosceles triangle E M \implies EM is the perpendicular bisector of base A C AC M E C \implies \triangle{MEC} is a right triangle and A M = M C AM = MC .

Since A C B \angle{ACB} is a common angle to both right triangle A B C ABC and M E C A B C M E C 2 m m = x + a a x = a MEC \implies \triangle{ABC} \sim \triangle{MEC} \implies \dfrac{2m}{m} = \dfrac{x + a}{a} \implies x = a and the pythagorean theorem in A B C y = 2 m = 3 a \triangle{ABC} \implies y = 2m = \sqrt{3}a .

Let h h be the height of the tetrahedron.

v = 3 a i + a j + 0 k \vec{v} = -\sqrt{3}a\vec{i} + a\vec{j} + 0\vec{k}

u = 3 a i + 0 j + h k \vec{u} = -\sqrt{3}a\vec{i} + 0\vec{j} + h\vec{k}

u X v = a h i 3 a h j 3 a 2 k u X v = a 4 h 2 + 3 a 2 \implies \vec{u} X \vec{v} = -ah\vec{i} - \sqrt{3}ah\vec{j} - \sqrt{3}a^2\vec{k} \implies |\vec{u} X \vec{v}| = a\sqrt{4h^2 + 3a^2} and u = 3 a 2 + h 2 |\vec{u}| = \sqrt{3a^2 + h^2}

d = a 4 h 2 + 3 a 2 3 a 2 + h 2 A B C D = 1 2 a 4 h 2 + 3 a 2 \implies d = \dfrac{a\sqrt{4h^2 + 3a^2}}{\sqrt{3a^2 + h^2}} \implies A_{\triangle{BCD}} = \dfrac{1}{2}a\sqrt{4h^2 + 3a^2}

Let A = 1 2 a 4 h 2 + 3 a 2 A = \dfrac{1}{2}a\sqrt{4h^2 + 3a^2}

The volume V = 1 2 3 a 2 h = k h = 2 3 k a 2 A ( a ) = 48 k 2 + 3 a 6 a V = \dfrac{1}{2\sqrt{3}} a^2 h = k \implies h = \dfrac{2\sqrt{3}k}{a^2} \implies A(a) = \dfrac{\sqrt{48k^2 + 3a^6}}{a} \implies

d A d a = 6 a 6 48 k 2 a 2 48 k 2 + 3 a 6 = 0 a 0 a = 2 k 1 3 h = 3 k 1 3 \dfrac{dA}{da} = \dfrac{6a^6 - 48k^2}{a^2\sqrt{48k^2 + 3a^6}} = 0 \:\ a \neq 0 \implies \boxed{a = \sqrt{2}k^{\frac{1}{3}}} \implies \boxed{h = \sqrt{3}k^{\frac{1}{3}}}

d A d a < 0 \dfrac{dA}{da} < 0 when a < 2 k 1 3 a < \sqrt{2}k^{\frac{1}{3}} and d A d a > 0 \dfrac{dA}{da} > 0 when a > 2 k 1 3 a > \sqrt{2}k^{\frac{1}{3}} \implies minimum at a = 2 k 1 3 a = \sqrt{2}k^{\frac{1}{3}} .

Let m B C D = θ m\angle{BCD} = \theta .

For cos ( θ ) : \cos(\theta):

C D = 3 k 1 3 , B C = 2 2 k 1 3 CD = 3k^{\frac{1}{3}}, \:\ BC = 2\sqrt{2}k^{\frac{1}{3}} and B D = 5 k 1 3 BD = \sqrt{5}k^{\frac{1}{3}}

5 = 17 6 8 cos ( θ ) cos ( θ ) = 1 2 θ = 4 5 \implies 5 = 17 - 6\sqrt{8}\cos(\theta) \implies \cos(\theta) = \dfrac{1}{\sqrt{2}} \implies \theta = \boxed{45^{\circ}} .

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