Please help me identify a problem I was working on.

I usually save the interesting ones in a set for future examination but I didn't this time and I think I'm on the way to solving it.

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TopNewestI'm getting a value of \(r = \displaystyle\sqrt{\dfrac{53 + 6\sqrt{19}}{5}}\),

which equals \(3.9788\) rounded to \(4\) decimal places. Is that what you got? I'll write up my method shortly.

Edit: Label the triangle so that \(OA = 3\) and \(AB = 2\). Also, label the length \(1\) line segment \(AP\) and let \(\angle OAB = \theta\).

Then \(OP = r\) and \(\angle OAP = \frac{3\pi}{2} - \theta\). We can now apply the Cosine Law to triangles \(\Delta OAB\) and \(\Delta OAP\) to obtain the equations

(i) \(r^{2} = 3^{2} + 2^{2} - 2*3*2*\cos(\theta) \Longrightarrow 13 - 12\cos(\theta)\), and

(ii) \(r^{2} = 3^{2} + 1^{2} - 2*3*1*\cos(\frac{3\pi}{2} - \theta) \Longrightarrow 10 + 6\sin(\theta)\).

Equating (i) and (ii) we then have that

\(13 - 12\cos(\theta) = 10 + 6\sin(\theta) \Longrightarrow 1 - 4\cos(\theta) = 2\sin(\theta)\).

Now square both sides and simplify to find that

\(1 - 8\cos(\theta) + 16\cos^{2}(\theta) = 4*(1 - \cos^{2}(\theta)) \Longrightarrow 20\cos^{2}(\theta) - 8\cos(\theta) - 3 = 0\),

which has solutions \(\cos(\theta) = \dfrac{8 \pm \sqrt{64 + 4*3*20}}{2*20} = \dfrac{1}{5} \pm \dfrac{\sqrt{19}}{10}\).

Now from the diagram we are looking for \(\theta \gt \frac{\pi}{2}\), i.e., \(\cos(\theta) \lt 0\), se we take the root

\(\cos(\theta) = \dfrac{1}{5} - \dfrac{\sqrt{19}}{10}\).

Now substitute this value into (i) to find that

\(r^{2} = 13 - 12*(\dfrac{1}{5} - \dfrac{\sqrt{19}}{10}) = \dfrac{53 + 6\sqrt{19}}{5}\),

and so \(r = \displaystyle\sqrt{\dfrac{53 + 6\sqrt{19}}{5}}\).

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Hi, Brian, what's the name of the problem?

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Oh, sorry, you were just wanting a name. I have no idea if it has a specific name, but it seems to be an exercise in the application of the Cosine Law. I can think of one other way of trying to solve it, but I haven't worked out the details of that approach yet.

I'm sorry if I spoiled this question for posting. I just saw it as an interesting problem and immediately had to solve it. You could delete this note and then post the problem after; I would be fine with that. :)

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Uh oh, answer spoiled? D:

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Yeah, I know. I don't think it's a big problem, since this posts has no likes or re-shares. Sorry about this; I had no idea about the question's "history" at the time I posted my solution. :(

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I think I did see this problem somewhere else posted in Brilliant, but I can't think of the name of either the problem or the creator. Brian's answer is correct, though.

Okay, I've found the problem, posted by [Deleted]

[Deleted]

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Well, that made it easy for me to solve [deleted]'s question, then. :) Having found a solution assuming that \(\theta\) was obtuse it is clear that \(\theta\) can be obtuse, but I neglected to prove that \(\theta\) would necessarily have to be obtuse. I see that in response to [deleted]'s solution to his own problem, Ariel Gershon has proved this to be the case.

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lol free rating...

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no free shot to lv5 geometry attempt

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Hi Michael. In light of Math Man's comment below, I'm wondering if it's a good idea having the link to [deleted]'s question posted here. whatever you decide is fine, but I just thought I should give you a heads up. I could delete my answer, but then that would defeat the point of Guiseppi's post.

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Okay, deleted, as I noticed that problem has just now been reshared.

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its title is this: [deleted]

Moderator's note: please do not post spoilersLog in to reply