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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 $\angle BAC$?2012-10-05
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    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?2012-10-05
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    By not much writing i mean not solving equations and writing out numbers above 100.2012-10-05

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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?2012-10-05
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    @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= 4.2012-10-05
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    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 $h\lt k\lt 10$.2012-10-05
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    @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.2012-10-05
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    @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?2012-10-05
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    Think again about $11$.2012-10-05
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    I did already and edited comment :), thanks for helping.2012-10-05
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    @barto I don't think your proof about congruency is conclusive since it assumes the sides are equal rather than in direct proportion. If we are trying to find the case where the two values k and l are not equal ($k < 10 < l$), let m and n represent the missing sides of the respective triangles. The relationships for congruency should be $n = 10\frac{l}{k}$ and $m = 10\frac{k}{l}$. Your cosine equation then looks something more like: $l^2 = 100 + 100\frac{l^2}{k^2} - 10\frac{l}{k}\sqrt{3}$. There are infinite solutions here, but none that matter to your problem.2012-10-05
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    @cheepychappy: that's what i call much writing. Maybe i wasn't clear enough. I suppose that there are $2$ triangles with different value of $|BC|$, that are congruent. That means that $|CA|=|B'C'|$ en $|C'A'|=|BC|$. Thus, all sides are integers. Then i come to a contradiction by applying the cos law on both triangles $\bigtriangleup ABC$ and $\bigtriangleup A'B'C'$.2012-10-06
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    @barto Right. I guess what I was trying to say is that congruency has to preserve the 30-degree angle at A and since the length of AB has to stay at 10, there are no two triangles with different legs BC and AC that are congruent if you compare them with respects to the angles A = A', B = B', C = C'. To compare them with respects to A = B', B = A', C = C' (which is what you have done) means that AB = A'B' which means it requires equality not congruency (and that angle B = A' = 30). The possible relationship for congruency in this problem is A = A', B = C', C = B' which is the case I showed.2012-10-06
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    @barto But there are still no integer solutions so it won't change your answer. I just was responding to your proof. Interesting problem!2012-10-06
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    Ok. I see. The problem itself isn't difficult, but it is hard to find a quick way to solve it. The question is coming from a first round of the mathematics competition of Belgium in 1995. (I'm solving all questions from 1986 till now.) Since there are 30 questions that should be solved in 2 hours, a quick solution is essential.2012-10-06
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    @barto: The contestants would presumably be familiar with the ambiguous case. Since probably it is multiple choice, a few quick sketches would do the job.2012-10-06
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    I consider myself as a contestant too. I was in the finale for the 5th and 6th year of secundary school last year, and this year i'm in 6th year, so probably I get there again. The only problem is, as you can see, i'm not always that regardful so i can make such terrible mistakes.2012-10-06
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    I am on the other side of the fence, involved in making up questions for a couple of contests. Don't like multiple choice, too many times they are setting traps.2012-10-06