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#1 h Jul 7, 2013, 8:55 AM • 1 Y

Y by Adventure10

The three distinct points $B, C, D$ are collinear with C between B and D. Another point A not on
the line BD is such that $|AB| = |AC| = |CD|.$
Prove that ∠ $BAC = 36$ if and only if $1/|CD|-1/|BD|=1/(|CD| + |BD|)$
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#2 h Jul 7, 2013, 1:11 PM • 1 Y

Y by Adventure10

thryiomt wrote:

The three distinct points $B, C, D$ are collinear with C between B and D. Another point A not on
the line BD is such that $|AB| = |AC| = |CD|.$
Prove that ∠ $BAC = 36$ if and only if $1/|CD|-1/|BD|=1/(|CD| + |BD|)$ .

****Suppose that we have: $\widehat{BAC}=\frac{\pi}{5}$ , and it is well-known that $\cos(\frac{\pi}{5})=\frac{1+\sqrt{5}}{4}$ .
We also denote: $AB=AC=CD=a$ .
By Alkashi formula's, we obtain: $BC^2=AB^2+AC^2-2.AB.AC.\cos(\widehat{BAC})=2a^2(1-\frac{1+\sqrt{5}}{4})$ .
So $BC^2=2a^2(\frac{3-\sqrt{5}}{4})=a^2(\frac{6-2\sqrt{5}}{2^2})=a^2.\frac{(\sqrt{5}-1)^2}{2^2}$ .
Then $BC=a\frac{\sqrt{5}-1}{2}$ .
We conclude that: $BD=BC+CD=a\frac{\sqrt{5}-1}{2}+a=a\frac{\sqrt{5}+1}{2}$ .
We have: $\frac{1}{CD}-\frac{1}{BD}=\frac{1}{a}-\frac{2}{a.(\sqrt{5}+1)}=\frac{1}{a}(1-\frac{\sqrt{5}-1}{2})=\frac{1}{a}(\frac{3-\sqrt{5}}{2})$ .
And: $\frac{1}{CD+BD}=\frac{1}{a(1+\frac{\sqrt{5}+1}{2})}=\frac{1}{a}(\frac{2}{3+\sqrt{5}})=\frac{1}{a}(\frac{3-\sqrt{5}}{2})$ .
We are done!
****Now, if we have: $\frac{1}{CD}-\frac{1}{BD}=\frac{1}{BC+BD}$ .
This condition is equivalent to $BC^2+a.BC-a^2=0$ with the same notation.
We can consider it as an equation in $BC$ , and it's discriminent is $\Delta=a^2+4a^2=5a^2$ .
Then, the equation has two solutions, and we reject the negative one.
So, $BC=\frac{a+a\sqrt{5}}{2}=a.\frac{-1+\sqrt{5}}{2}$ .
We denote: $\widehat{BAC}=\alpha$ , we conclude $\widehat{ABC}=\frac{\pi}{2}-\frac{\alpha}{2}$ because $AB=AC=a$ .
By the sinus law, we get: $\frac{\sin(\alpha)}{BC}=\frac{\sin(\frac{\pi}{2}-\frac{\alpha}{2})}{a}$ so $\frac{2\sin(\frac{\alpha}{2}).\cos{(\frac{\alpha}{2})}}{a.\frac{-1+\sqrt{5}}{2}}=\frac{\cos(\frac{\alpha}{2})}{a}$ .
Then, $\sin({\frac{\alpha}{2}})=\frac{-1+\sqrt{5}}{4}$ .
We have: $\cos{(\frac{\alpha}{2})}=\sqrt{1-(\frac{-1+sqrt{5}}{4})^2}=\frac{1}{4}\sqrt{10+2\sqrt{5}}=\cos(\frac{\pi}{10})$ .
So, $\frac{\alpha}{2}=\frac{\pi}{10}$ , and $\alpha=\frac{\pi}{5}$ .
****References: http://mathworld.wolfram.com/TrigonometryAnglesPi10.html and http://mathworld.wolfram.com/TrigonometryAnglesPi5.html .

This post has been edited 6 times. Last edited by Akhenaten, Jul 7, 2013, 1:56 PM

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#3 h Jul 7, 2013, 1:26 PM • 1 Y

Y by Adventure10

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#4 h Dec 12, 2022, 7:19 PM • 1 Y

Y by Mango247

The three distinct points $B, C, D$ are collinear with $C$ between $B$ and $D$ . Another point $A$ not on the line $BD$ is such that $|AB| = |AC| = |CD|.$ Prove that $\angle BAC = 36^o$ if and only if $\frac{1}{|CD|}-\frac{1}{|BD|}=\frac{1}{|CD| + |BD|}.$

This post has been edited 3 times. Last edited by parmenides51, Dec 21, 2022, 8:04 PM

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