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Question Number 140838 by mr W last updated on 13/May/21

Commented by mr W last updated on 13/May/21

$s o l u t i o n t o Q 140785$
Commented by mr W last updated on 13/May/21

$s a y :$ $O i s o r i g i n a t a n y p o s i t i o n i n p l a n e .$ $\vec{O A} = A$ $\vec{O B} = B$ $\vec{O C} = C$ $\vec{O D} = D$ $\vec{O E} = E (E i s a n y p o i n t i n p l a n e)$ $\Rightarrow \vec{O P} = P = \frac{A + B + C + D}{4}$ $(P i s i n t e r s e c t i o n p o i n t o f b i m e d i a n s,$ $i s a l s o t h e c e n t r o i d o f q u a d r i l a t e r a l)$ $\vec{A B} = B - A$ $\vec{B C} = C - B$ $\vec{C D} = D - C$ $\vec{D A} = A - D$ $\vec{A E} = E - A$ $\vec{P E} = E - P$ $x^{2} =∣ A E ∣^{2} = (E - A) \cdot (E - A) = E^{2} + A^{2} - 2 A \cdot E$ $y^{2} =∣ B E ∣^{2} = (E - B) \cdot (E - B) = E^{2} + B^{2} - 2 B \cdot E$ $z^{2} =∣ C E ∣^{2} = (E - C) \cdot (E - C) = E^{2} + C^{2} - 2 C \cdot E$ $t^{2} =∣ D E ∣^{2} = (E - D) \cdot (E - D) = E^{2} + D^{2} - 2 D \cdot E$ $∣ P E ∣^{2} = (E - P) \cdot (E - P) = E^{2} + P^{2} - 2 P \cdot E$ $P^{2} = \frac{1}{16} (A + B + C + D) \cdot (A + B + C + D)$ $P^{2} = \frac{1}{16} (A^{2} + B^{2} + C^{2} + D^{2}) + \frac{1}{8} (A \cdot B + A \cdot C + A \cdot D + B \cdot C + B \cdot D + C \cdot D)$ $x^{2} + y^{2} + z^{2} + t^{2} = 4 E^{2} + A^{2} + B^{2} + C^{2} + D^{2} - 2 (A \cdot E + B \cdot E + C \cdot E + D \cdot E)$ $x^{2} + y^{2} + z^{2} + t^{2} = 4 E^{2} + A^{2} + B^{2} + C^{2} + D^{2} - 2 (A + B + C + D) \cdot E$ $x^{2} + y^{2} + z^{2} + t^{2} = 4 E^{2} + A^{2} + B^{2} + C^{2} + D^{2} - 8 P \cdot E$ $x^{2} + y^{2} + z^{2} + t^{2} - 4 ∣ P E ∣^{2} = 4 E^{2} + A^{2} + B^{2} + C^{2} + D^{2} - 8 P \cdot E - 4 E^{2} - 4 P^{2} + 8 P \cdot E$ $x^{2} + y^{2} + z^{2} + t^{2} - 4 ∣ P E ∣^{2} = A^{2} + B^{2} + C^{2} + D^{2} - 4 P^{2}$ $x^{2} + y^{2} + z^{2} + t^{2} - 4 ∣ P E ∣^{2} = A^{2} + B^{2} + C^{2} + D^{2} - \frac{1}{4} (A^{2} + B^{2} + C^{2} + D^{2}) - \frac{1}{2} (A \cdot B + A \cdot C + A \cdot D + B \cdot C + B \cdot D + C \cdot D)$ $x^{2} + y^{2} + z^{2} + t^{2} - 4 ∣ P E ∣^{2} = \frac{3}{4} (A^{2} + B^{2} + C^{2} + D^{2}) - \frac{1}{2} (A \cdot B + A \cdot C + A \cdot D + B \cdot C + B \cdot D + C \cdot D)$ $a^{2} =∣ A B ∣^{2} = (B - A) \cdot (B - A) = B^{2} + A^{2} - 2 A \cdot B$ $b^{2} =∣ B C ∣^{2} = (C - B) \cdot (C - B) = C^{2} + B^{2} - 2 B \cdot C$ $c^{2} =∣ C D ∣^{2} = (D - C) \cdot (D - C) = D^{2} + C^{2} - 2 C \cdot D$ $d^{2} =∣ D A ∣^{2} = (A - D) \cdot (A - D) = A^{2} + D^{2} - 2 A \cdot D$ $e^{2} =∣ A C ∣^{2} = (C - A) \cdot (C - A) = C^{2} + A^{2} - 2 A \cdot C$ $f^{2} =∣ B D ∣^{2} = (D - B) \cdot (D - B) = D^{2} + B^{2} - 2 B \cdot D$ $a^{2} + b^{2} + c^{2} + d^{2} + e^{2} + f^{2} = 3 (A^{2} + B^{2} + C^{2} + D^{2}) - 2 (A \cdot B + A \cdot C + A \cdot D + B \cdot C + B \cdot D + C \cdot D)$ $\frac{1}{4} (a^{2} + b^{2} + c^{2} + d^{2} + e^{2} + f^{2}) = \frac{3}{4} (A^{2} + B^{2} + C^{2} + D^{2}) - \frac{1}{2} (A \cdot B + A \cdot C + A \cdot D + B \cdot C + B \cdot D + C \cdot D)$ $\Rightarrow \frac{1}{4} (a^{2} + b^{2} + c^{2} + d^{2} + e^{2} + f^{2}) = x^{2} + y^{2} + z^{2} + t^{2} - 4 ∣ P E ∣^{2}$ $\Rightarrow \frac{1}{4} (a^{2} + b^{2} + c^{2} + d^{2} + e^{2} + f^{2}) + 4 ∣ P E ∣^{2} = x^{2} + y^{2} + z^{2} + t^{2}$
Commented by mathdanisur last updated on 13/May/21

$t h a n k y o u d e a r S i r c o o l$
Commented by BHOOPENDRA last updated on 13/May/21

$N i c e s o l u t i o n s i r$
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