Isoperimetric points and Tetrahedrons 2.

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In A B C , A C = 13 , A B = 14 \triangle{ABC}, \overline{AC} = 13, \overline{AB} = 14 and B C = 15 \overline{BC} = 15 and point P P is the isoperimetric point .

Let P Q PQ . be the height of the tetrahedron above.

If the volume V V of the tetrahedron is V = 224 V = 224 , find the measure of the angle(in degrees) between the two triangular faces B Q A BQA and B Q C BQC . to six decimal places.


The answer is 66.900021.

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1 solution

Rocco Dalto
Nov 5, 2018

Refer to isoperimetric point .

The perimeter P = 13 + b + c = 14 + b + d = 15 + c + d b = d + 2 c = d + 1 P^{*} = 13 + b + c = 14 + b + d = 15 + c + d \implies b = d + 2 \implies c = d + 1

Using Heron's formula A A P C = 42 ( d + 8 ) ( d 5 ) , A A P B = 48 ( d + 8 ) ( d 6 ) , A B P C = 56 ( d + 8 ) ( d 7 ) A_{APC} = \sqrt{42(d + 8)(d - 5)}, A_{APB} = \sqrt{48(d + 8)(d - 6)}, A_{BPC} = \sqrt{56(d + 8)(d - 7)}

and

A A B C = 84 84 = d + 8 ( 42 ( d 5 ) + 48 ( d 6 ) + 56 ( d 7 ) ) A_{ABC} = 84 \implies 84 = \sqrt{d + 8}(\sqrt{42(d - 5)} + \sqrt{48(d - 6)} +\sqrt{56(d - 7)})

Solving for d d we obtain: d = 80 11 A P = 102 11 , C P = 91 11 d = \dfrac{80}{11} \implies \overline{AP} = \dfrac{102}{11}, \overline{CP} = \dfrac{91}{11} and B P = 80 11 \overline{BP} = \dfrac{80}{11} .

Let B D \overline{BD} be altitude in A B C : \triangle{ABC}:

A A B C = 84 = 13 2 ( B D ) B D = 168 13 A D = 70 13 A_{ABC} = 84 = \dfrac{13}{2}(\overline{BD}) \implies \overline{BD} = \dfrac{168}{13} \implies \overline{AD} = \dfrac{70}{13} .

Let A ( 0 , 0 ) , B ( 70 13 , 168 13 ) , C ( 13 , 0 ) A(0,0), \:\ B(\dfrac{70}{13},\dfrac{168}{13}),\:\ C(13,0) and P ( x 0 , y 0 ) P(x_{0},y_{0}) .

Using A P C \triangle{APC} \implies

x 0 2 + y 0 2 = 10404 121 x_{0}^2 + y_{0}^2 = \dfrac{10404}{121}

x 0 2 26 x 0 + 169 = 8281 121 x_{0}^2 - 26x_{0} + 169 = \dfrac{8281}{121}

x 0 = 1026 143 \implies x_{0} = \dfrac{1026}{143} and y 0 = 840 143 y_{0} = \dfrac{840}{143}

Point P ( 1026 143 , 840 143 ) P(\dfrac{1026}{143},\dfrac{840}{143})

The volume of the tetrahedron V = 28 h = 224 h = 8 V = 28h = 224 \implies h = 8 \implies Point Q ( 1026 143 , 840 143 , 8 ) Q(\dfrac{1026}{143},\dfrac{840}{143},8) .

B A = 70 13 i 168 13 j + 0 k \overline{BA} = -\dfrac{70}{13}\vec{i} -\dfrac{168}{13}\vec{j} + 0\vec{k}

B C = 99 13 i 168 13 j + 0 k \overline{BC} = \dfrac{99}{13}\vec{i} -\dfrac{168}{13}\vec{j} + 0\vec{k}

B Q = 256 143 i 1008 143 j + 8 k \overline{BQ} = \dfrac{256}{143}\vec{i} -\dfrac{1008}{143}\vec{j} + 8\vec{k}

u = B Q X B A = 1344 13 i 560 13 j + 8736 143 k \vec{u} = \overline{BQ} \:\ X \:\ \overline{BA} = \dfrac{1344}{13}\vec{i} -\dfrac{560}{13}\vec{j} + -\dfrac{8736}{143}\vec{k}

v = B Q X B C = 1344 13 i + 792 13 j + 4368 143 k \vec{v} = \overline{BQ} \:\ X \:\ \overline{BC} = \dfrac{1344}{13}\vec{i} + \dfrac{792}{13}\vec{j} + -\dfrac{4368}{143}\vec{k} u v = 749952 121 \vec{u} \cdot \vec{v} = \dfrac{749952}{121} and u = 112 157 11 , v = 24 3221 11 |\vec{u}| = \dfrac{112\sqrt{157}}{11}, \:\ |\vec{v}| = \dfrac{24\sqrt{3221}}{11} .

Let λ \lambda be the measure of the angle between the two triangular faces B Q A BQA and B Q C BQC

cos ( λ ) = 279 505697 λ 66.900021 \implies \cos(\lambda) = \dfrac{279}{\sqrt{505697}} \implies \lambda \approx \boxed{66.900021} .

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