2

Proof that$$\dfrac{\|\vec{CA}\|}{\|\vec{AX}\|} = \dfrac{\|\vec{CB}\|}{\|\vec{BX}\|}$$

having $$\dfrac{\vec{CA}}{\|\vec{CA}\|} + \dfrac{\vec{CB}}{\|\vec{CB}\|} = \alpha\vec{CX}$$
I have tried to inject the second formula in the first, but I wasn't able to figure out the necessary algebra, if that is the correct way.

I hope this triangle helps too!

Solution:

Using Shifrin's notation (see best answer below), we try to rearrange the sides to make easier to see the coefficients of $\vec{a}$ and $\vec{b}$, which must be equal on both sides of equation: $$ \dfrac{\vec{a}}{\|\vec{a}\|} + \dfrac{\vec{b}}{\|\vec{b}\|} = \alpha\vec{a} + \alpha t (\vec{b} - \vec{a}) $$ $$ \dfrac{\vec{a}}{\|\vec{a}\|} + \dfrac{\vec{b}}{\|\vec{b}\|} = (\alpha - \alpha t)\vec{a} + \alpha t \vec{b} $$ Hence, $\dfrac{1}{\|\vec{a}\|} = (\alpha - \alpha t) \iff \|\vec{a}\| = \dfrac{1}{(\alpha - \alpha t)}$ ,where $ \|\vec{a}\| \neq 0 $
$\dfrac{1}{\|\vec{b}\|} = \alpha t \iff \|\vec{b}\| = \dfrac{1}{\alpha t}$ ,where $ \|\vec{b}\| \neq 0 $
From these, $\cfrac{\|\vec{CA}\|}{\|\vec{AX}\|} = \cfrac{\cfrac{1}{(\alpha - \alpha t)}}{\|t(\vec{b} - \vec{a})\|} = \cfrac{1}{\|\alpha t(1 - t)(\vec{b} - \vec{a})\|}$

and $\cfrac{\|\vec{CB}\|}{\|\vec{BX}\|} = \cfrac{\cfrac{1}{\alpha t}}{\|(t - 1)(\vec{b} - \vec{a})\|} = \cfrac{1}{\|\alpha t(1 - t)(\vec{b} - \vec{a})\|}$

Carlos Oliveira
  • 301
  • You should definitely include the comment I made about setting coefficients equal. (In fancy linear algebra language, this follows from linear independence.) One serious error is that you cannot divide by a vector — of course, you meant to have $|\vec b-\vec a|$ in your denominators. You can make this a bit more concise by getting $$\frac{|\vec a|}{|\vec b|} = \frac{\alpha t}{\alpha(1-t)} = \frac{t}{1-t}$$ and then recognizing the ratio of the lengths. – Ted Shifrin Sep 27 '17 at 20:17
  • I hope I have got things right. I am sorry if I am not completely understanding. I think linear independence is the next chapter of my analytic geometry course. – Carlos Oliveira Sep 27 '17 at 22:11
  • No need to apologize. This is great stuff. – Ted Shifrin Sep 27 '17 at 22:44

2 Answers2

1

Of course you left out the crucial hypothesis that $\overrightarrow{CX}$ bisects $\angle ACB$. (But that's what your displayed equation is equivalent to.)

HINT: You wrote down one condition. There's also another: $X$ lies on $\overline{AB}$. Let $\vec a = \overrightarrow{CA}$ and $\vec b = \overrightarrow{CB}$ and write both equations in terms of $\vec a$ and $\vec b$.

Because $\vec a$ and $\vec b$ are not parallel, if you get $s\vec a + t\vec b = u\vec a + v \vec b$, what can you deduce about the scalars $s,t,u,v$?

Ted Shifrin
  • 115,160
0

Hint:$$\dfrac{\vec{CA}}{\|\vec{CA}\|} + \dfrac{\vec{CB}}{\|\vec{CB}\|} = \alpha\vec{CX}\\\vec{e_{CA}}+\vec{e_{CB}}=\alpha\vec{CX}$$ note that $$|\vec{e_{CA}}|=1\\|\vec{e_{CB}}|=1\\\vec{e_{CA}}+\vec{e_{CB}}=\sqrt{1^2+1^2+2\cos({\widehat{\vec{CA},\vec{CB}}})}=\\2\cos(\frac{\widehat{\vec{CA},\vec{CB}} }{2})$$

Khosrotash
  • 24,922