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March 23rd, 2019, 02:12 PM   #11
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Originally Posted by skipjack View Post
If about 5.6 were the smallest value, the smallest integer value would be 6.
True...but wouldn't "fit in" the triangle;
anyhoo, I completely misread the damn problem!
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March 23rd, 2019, 02:24 PM   #12
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Quote:
Originally Posted by Denis View Post
True...but wouldn't "fit in" the triangle;
anyhoo, I completely misread the damn problem!
Do you kiss your Mother with that mouth?

Or, ummmm... hold hands with her with those fingers that you type with... or something like that.

-Dan
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March 23rd, 2019, 02:54 PM   #13
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Or, ummmm... hold hands with her with those fingers that you type with...
I only use 1 finger to type: left index...
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March 23rd, 2019, 03:34 PM   #14
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Originally Posted by skeeter View Post
$\dfrac{4}{\cos(2\omega)} = |BD| \implies \dfrac{4}{|BD|} = \cos(2\omega)$


$0 < \cos(2\omega) < 1$

substitute $\dfrac{4}{|BD|}$ for $\cos(2\omega)$ in the above inequality

$0 < \dfrac{4}{|BD|} < 1$

multiply every term in the inequality by the positive value $|BD|$ ...

$0 < 4 < |BD|$
Geez! How come I could have overlooked this? Sorry skeeter. Now it makes sense. I was a bit cautious on how to understand the inequality between $0<\cos 2 \omega<1$ as the angle is double. But then I remembered that the constant before the angle only changes its amplitude, not the boundaries set by the trigonometric function, so it will always be between $0$ and $1$. The rest was just simple; the smallest whole number is $5$. Thanks!

Last edited by skipjack; March 23rd, 2019 at 09:48 PM.
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March 23rd, 2019, 03:55 PM   #15
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Originally Posted by skeeter View Post
Working it out, $|BD| =5 \implies \omega \approx 18.4^\circ$

therefore, $\angle{A} \approx 36.9^\circ \text{ and }\angle{BDA} = 4\omega \approx 73.7^\circ \implies \angle{ABD} \approx 69.4^\circ$

Further, $\angle{ABC} = 124.7^\circ \text{ and } \angle{C} = \omega \approx 18.4^\circ$
I didn't followed that problem into that far. But by checking on Wolfram Alpha it seems that is all okay!. There I got $18.43494...$. The rest just checks accordingly. $\angle{ABC}=180-(3*18.434)= 124.6951...$.
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March 23rd, 2019, 04:09 PM   #16
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Originally Posted by Denis View Post
As I told you, I ASSUMED that the diagram can be rearranged,
with BD perpendicular to AC.
This results in w = 22.5 degrees.
Triangle ABD becomes a right triangle with hypotenuse AB = 8,
and AD = BD, thus isosceles; AD = BD = ~5.6568

Looks like that was NOT the intent of your problem.

WHAT really does "smallest whole number that BD can attain" mean?
Is that a translation from another language?
I could only see that $AB \cong BE$ but couldn't get $AD=BD$. How exactly you do establish this?. From this the rest is simple. But you already noticed that it wasn't the intend of the problem. Interesting observation though. Regarding to your curiosity it was translated and wrote from the source. What I believe it meant was to find the least integer value that $BD$ can have given the conditions stated. I hope this makes it more clear.
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March 23rd, 2019, 09:56 PM   #17
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I could only see that $AB \cong BE$ . . .
There is no point $E$ in the diagram. A triangle with angles of 90 degrees and 45 degrees must have 45 degrees as its third angle, so it's an isosceles triangle.
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