Vectors

by v_Enhance, Dec 21, 2011, 12:48 AM

are useful for length bashing.

Let $ABC$ be a triangle. Let $\vec A$, $\vec B$, $\vec C$ be vectors all with norm $R$. Suppose $\vec D = p \vec B + q \vec C$, where $p+q=1$. Notice that

\[ \begin{align*} DA^2 &= (\vec D- \vec A) \cdot (\vec D - \vec A) \\ &= (p \vec B + q \vec C - \vec A) \cdot (p \vec B + q \vec C - \vec A) \\
&= p^2 R^2 + q^2 R^2 + R^2 + 2pq \vec B \cdot \vec C - 2p \vec B \cdot \vec A - 2q \vec C \cdot \vec A \\
&= (p^2+q^2+1+2pq-2p-2q)R^2 - pqa^2 + pc^2 + qb^2 \\
&= pqa^2 -pc^2 -qb^2
\end{align*} \]

This rearranges to $AD \cdot a \cdot AD + ap \cdot a \cdot aq = c\cdot pa\cdot c + b \cdot qa \cdot b$. If $d=AD$, $m=BD$, $n=CD$, then this becomes \[ man + dad = bmb + cnc. \]

Comment

6 Comments

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aka stewarts.

by dinoboy, Dec 21, 2011, 12:51 AM

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Again.
...

by v_Enhance, Dec 21, 2011, 12:51 AM

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is there a non-ugly proof of stewarts?

by sjaelee, Dec 21, 2011, 1:38 AM

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Haha, you sure like proving Stewart's Theorem xD

by Jinduckey, Dec 21, 2011, 3:37 AM

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I think I've been doing too much geo lately.
Time to switch over to combo.

by v_Enhance, Dec 21, 2011, 4:01 AM

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sjaelee the cleanest proof is probably by law of cosines, stewarts is a somewhat ugly theorem in the first place

by pi37, Dec 25, 2011, 2:59 AM

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