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 March 21st, 2019, 05:44 PM #1 Member     Joined: Jun 2017 From: Lima, Peru Posts: 97 Thanks: 1 Math Focus: Calculus How to find the smallest length in a triangle when the angles aren't known? I am confused on how to find the answer for this problem. So far what I believe would apply is the triangle inequality but I'm not sure on how to use it. The figure $ABC$ is a triangle so as $BDC$. It is known that the length of $AB = 8$ inches and $\angle BAD = 2\,\angle BCD$ and $\angle\,DBC = 3 \angle BCD$. Find the smallest whole number that $BD$ can attain. The alternatives given are: $\begin{array}{ll} 1.&7\,\textrm{inches}\\ 2.&4\,\textrm{inches}\\ 3.&8\,\textrm{inches}\\ 4.&5\,\textrm{inches}\\ 5.&6\,\textrm{inches}\\ \end{array}$ So far the only thing I could come up with was to "guess" it might be $4$ since, $\angle BDA = 4\omega$ could it be that since $BD$ is opposing $2\omega$ hence that side might be $4$ inches. But that's how far I went. Needless to say that I'm unsure if that's a right way to go.
 March 21st, 2019, 07:11 PM #2 Math Team     Joined: Jul 2011 From: Texas Posts: 2,941 Thanks: 1545 Using the Law of sines ... $\dfrac{8}{\sin(4\omega)} = \dfrac{|BD|}{\sin(2\omega)}$ $\dfrac{8}{2\sin(2\omega)\cos(2\omega)} = \dfrac{|BD|}{\sin(2\omega)}$ $\dfrac{4}{\cos(2\omega)} = |BD|$ $0 < \cos(2\omega) < 1 \implies |BD| > 4$ I'd say minimum of 5"
 March 22nd, 2019, 09:43 AM #3 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,580 Thanks: 1038 BD is shortest when perpendicular to AC. So triangle ABD is right/isosceles: do the work and you'll end up with ~5.656.... thus 5 as per Skeeter.
March 22nd, 2019, 06:17 PM   #4
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Quote:
 Originally Posted by skeeter Using the Law of sines ... $\dfrac{8}{\sin(4\omega)} = \dfrac{|BD|}{\sin(2\omega)}$ $\dfrac{8}{2\sin(2\omega)\cos(2\omega)} = \dfrac{|BD|}{\sin(2\omega)}$ $\dfrac{4}{\cos(2\omega)} = |BD|$ $0 < \cos(2\omega) < 1 \implies |BD| > 4$ I'd say minimum of 5"
I've also tried to follow this method but the part where I got stuck was at:

$\frac{4}{\cos 2\omega}= BD$

How does this equality becomes into an inequality?

The range you given as:

$0 < \cos 2\omega < 1$

How does it imply this?

$BD > 4$

I don't know how to translate the words "minimum value" from those equations:

So far I think that if they ask the smallest value for $BD$ then does it mean $\cos 2\omega$ should be close to $1$ so as when it is inserted in the equation it will not increase the factor of $4$ in the numerator. But from there I'm still lost. What am I missing?. Can you explain me more, please?

March 22nd, 2019, 06:23 PM   #5
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Quote:
 Originally Posted by Denis BD is shortest when perpendicular to AC. So triangle ABD is right/isosceles: do the work and you'll end up with ~5.656.... thus 5 as per Skeeter.
I wonder how did you obtained 5.656? But yes it makes sense that it will be the smallest when perpendicular. It took me a while to find out where the isosceles triangle you were referring to is, but I found it. But from there I'm still stuck. Maybe can you offer an additional hint?

Last edited by skipjack; March 23rd, 2019 at 07:46 AM.

March 23rd, 2019, 05:23 AM   #6
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Quote:
 Originally Posted by Chemist116 I've also tried to follow this method but the part where I got stuck was at: $\frac{4}{\cos 2\omega}= BD$ How does this equality becomes into an inequality? The range you given as: $0 < \cos 2\omega < 1$ How does it imply this? $BD > 4$ I don't know how to translate the words "minimum value" from those equations: So far I think that if they ask the smallest value for $BD$ then does it mean $\cos 2\omega$ should be close to $1$ so as when it is inserted in the equation it will not increase the factor of $4$ in the numerator. But from there I'm still lost. What am I missing?. Can you explain me more, please?
$\dfrac{4}{\cos(2\omega)} = |BD| \implies \dfrac{4}{|BD|} = \cos(2\omega)$

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

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 $|BD|$ ...

$0 < 4 < |BD|$

March 23rd, 2019, 06:30 AM   #7
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Quote:
 Originally Posted by Chemist116 Find the smallest whole number that BD can attain.
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,

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?

 March 23rd, 2019, 07:53 AM #8 Math Team     Joined: Jul 2011 From: Texas Posts: 2,941 Thanks: 1545 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, $\angle{ABC} = 124.7^\circ \text{ and } \angle{C} = \omega \approx 18.4^\circ$
March 23rd, 2019, 08:45 AM   #9
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Quote:
 Originally Posted by Denis ~5.656.... thus 5 as per Skeeter.
If about 5.6 were the smallest value, the smallest integer value would be 6.

The smallest integer value for $|BD|$ is 5, as demonstrated in the diagram below.
Figure1.PNG

 March 23rd, 2019, 09:37 AM #10 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,580 Thanks: 1038 Ahhh....gotcha! And angle ADB = 4w, of course...

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