How to find the smallest side of a triangle when the interior angles are unknown?

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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 in my book 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. Can somebody help me with this question?. Please try to include some drawing or sketch of the situation as I'd like to know what can I do to solve this problem relying mostly in geometry?.

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3 Answers

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A possible path without trigonometry.

EDIT Added explicitely the angles in the diagram, as required in comments.

1. Draw line $BE$ so that $\angle EBC = \angle ACB $.
2. $\triangle EBC$ is isoceles (why?); $\triangle BDE$ is isosceles (why?); $\triangle ABE$ is isosceles (why?)

Now use triangular inequality on $\triangle BDE$, to get $2BD > 8$.

EDIT Since $\triangle ABE$ is isosceles, $\angle AEB \cong \angle EAB < 90°$, whereas one of the two between $\angle BDA$ and $\angle BDE$ is greater than or equal to $90°$. Since in any triangle the side opposite to the greater angle is greater, we can state either that $BD < BE$ or that $BD < AB$. So $BD< 8$.

EDIT (Thanks to Lubin for pointing this out in comment)

Once the fact that $4<BD<8$ has been proven, one should also check whether an integer solution between $4$ and $8$ does exist for $BD$. I will try to show, then, that in fact $BD = 5$ is a valid solution. Consider the Figure below.

Start at point $D$ and draw a circle $\gamma_1$ centered in $D$ and with radius $5$; draw then a diameter of $\gamma_1$, and let $E$ be one of its endpoints. Draw a circle $\gamma_2$ centered in $E$ having radius $8$, which intersects $\gamma_1$ in $B$, and the line $DE$ (at the opposite side of $D$) in $C$. Finally draw circle $\gamma_3$ of radius $8$, centered in $B$. This intersects line $DE$ in two points. One is $E$; call the other one $A$. The triangle $\triangle ABC$ satisfies the requirements, with $BD = 5$, as desired.

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Since $\angle ADB=4\omega$, Law of Sines tells you that the length of $BD$ is $\frac{8\sin(2\omega)}{\sin(4\omega)}$. Then, since $\sin(2\alpha)<2\sin\alpha$ for all acute angles $\alpha$, the fraction is always greater than $4$, whereas for $\omega=29^\circ$, $\angle ADB=116^\circ$, and Law of Sines gives the length of $BD$ to be about $7.54$in. This length clearly varies continuously with $\omega$ in the range $0<\omega<30^\circ$, so there are integer values for the length of $5,6,7$in. Your choice then is #4, $5$ inches.

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Drawing Issocoles Triangles was inspired and leads to an easy analysis

$\angle ADB = 4\omega$ as its external to Triangle $BED$

$BD$ at max lies along $AD$ when $\angle ADB =120°$

$BD$ min when $\angle ADB = 90°$

$2\omega = 45° \text{ & } AD = BD\\ 8^2 = 2BD^2\\ \therefore BD = 4 \sqrt 2 ≈ 5.656$

$BD$ again nearly reaches max just before triangle collapses when $AD \implies 16 \text{ and } \angle ADB \implies 0°$

Smallest whole number is $6$

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