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Optimizing the Area of a Triangle

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  • #1
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[SOLVED] Optimizing the Area of a Triangle

I've been doing countless other ones, but this one has really stumped me.

"Triangle ABC has AB = AC. It is inscribed in a circle centre O, radius 10 cm. Find the value of the angle BAC that produces a maximum area for the triangle ABC."

I've drawn lots of diagrams to help me get started, but to no avail. I seriously have no clue how to do this one. Could someone please tell me a clue as to what my first step should be? I just need a starting point. Thanks in advance.
 

Answers and Replies

  • #2
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P.S. I'm not looking for an actual answer, I just need a pointer as to how to approach this problem.
 
  • #3
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Here's one approach. You can try to find the area [tex]K[/tex] of the triangle in terms of [tex]r[/tex] (10 cm) and [tex]\theta[/tex], which is half of [tex]\angle BAC[/tex]. Then you can just use some calculus on this. To find this relationship I would let the length base of the triangle ([tex]BC[/tex]) be [tex]b[/tex] and the height [tex]h[/tex]. Now here's a hint: let [tex]M[/tex] be the midpoint of [tex]BC[/tex]. What do you know about [tex]\triangle OMC[/tex]?
 
  • #4
Dick
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As Durt said, use trig to find the area of the triangle in terms of theta. Then use calculus to maximize it wrt to theta. This really isn't that hard a question, start with the first step of using trig. And present some kind of attempt, ok? What's with the 'poll' thing? Did you do that? It's only been a couple of hours since you posted. There should be a third choice. I.e. the poster didn't try hard enough, or show any concrete evidence of having tried at all.
 
Last edited:
  • #5
Defennder
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This is a little tedious. The area of the triangle inscribed in a circle is given by [itex]\frac{1}{2}(AB)^2sin \theta[/itex]. Let x=AB for convenience of notation here. You know how to find the maximum area given the above expression. But x here isn't a constant, so you can't differentiate the function and treat it like one. Instead, if you draw the diagram, you'll find a way to express x in terms of BC. But again BC is also a variable, just like x, since varying [itex]\theta[/tex] will also change BC. But by a certain angular property of circles, you can express BC in terms of r (the radius which is constant) and [itex]\theta[/tex]. Then once you have this expression, substitute back into the expression for the area. Now you can differentiate it properly.
 
  • #6
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Thanks to everyone who helped. I apologize for it seeming like I didn't do any work, but I'm not quite sure how to post pictures of my sketches. I'm just not that good at this whole technical stuff. I'll look into that the next time I have a problem like this. And about the poll...yeah my friend wanted to google something so I gave him the laptop for a sec, and he thought it would be funny to make me look like a jerk on the forum. I apologize for...well, yeah, looking like a jerk. (How do you make polls anyway? And...how do you type those funky math equations out? Those are so much easier to understand than typing it out normally.)


Anyways, sincerest thanks and apologies.
 

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