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In a triangle $\bigtriangleup ABC$ is $\widehat A=30^{\circ}$, $|AB|=10$ and $|BC|\in\{3,5,7,9,11\}$.

How many non-congruent trangles $\bigtriangleup ABC$ exist?

The possible answers are $3,4,5,6$ and $7$.

Is there a quick solution that doesn't require much writing?

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$\widehat{A}$ is angle $\angle BAC$? –  EuYu Oct 5 '12 at 18:38
    
Can you further explain what you mean by not much writing? Is it okay if it's extremely complicated, difficult, and hard to understand, as long as there isn't much writing? –  Graphth Oct 5 '12 at 18:40
    
By not much writing i mean not solving equations and writing out numbers above 100. –  barto Oct 5 '12 at 19:04
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1 Answer

up vote 1 down vote accepted

Hint: Use the Sine Law: $$\frac{\sin C}{10}=\frac{\sin A}{k},$$ where $k$ is one of our numbers $3$, $5$, $7$, $9$, $11$. Since $\sin A=1/2$, one of our $k$ is problematical. Another yields a triangle we all know and love. For the others, we are dealing with possibly the "ambiguous" case. A couple of sketches will give the answer, or knowledge about when we really are in the ambiguous case.

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@barto Is this classified as not much writing? –  Graphth Oct 5 '12 at 18:43
    
@Graphth: you could say so, yes. Nicolas: $k=3$ gives $\sin C>1$, which is nonsens. To have $2$ congruent triangles, the two other sides must be two different values of $7,9,11$. ($5$ gives a $90^{\circ}$ and $5\sqrt{3}$ as other side.) The two values (say $k,l$) have to satisfy $k^2=100+l^2-10l\sqrt{3}$ and $l^2=100+k^2-10k\sqrt{3}$ by the cos law. But then $10-l\sqrt{3}=k\sqrt{3}-10$, which is impossible since $k$ and $l$ are integers. That means no congruent triangles. Answer= 4. –  barto Oct 5 '12 at 19:00
    
The problem does not say that all sides are integers. You dealt fine with $3$. For $5$ there is a single triangle. For the others, drop a perpendicular from $B$ to side $AC$. Call the length of this $h$. A sketch should take care of the rest. Or use the fact that we are in the ambiguous case if $h\lt k\lt 10$. –  André Nicolas Oct 5 '12 at 19:08
    
@barto you can also just always solve for angle $B$ for each triangle by $180 - A - C$. If any of your triangles have the same triplet of angles, they are congruent. Just use the law of sines to solve for $C$ first. –  cheepychappy Oct 5 '12 at 19:09
    
@AndréNicolas: yes, my mistake. Of course there are 2 options for $7,9$. That makes $6$ in total. (Since according to my previous comment no two triangles with different $k$ can be congruent.) Right? –  barto Oct 5 '12 at 19:15
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