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anonymous

  • one year ago

Let the incircle of ΔABC have radius 2 and let it be tangent to BC at X. Suppose |BX| = 3 and |XC| = 4. What is the exact length of ΔABC? No trig functions should be used.

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  1. anonymous
    • one year ago
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    According to my calculations, the longest side should be 5. As long as ABC is a right triangle.

  2. anonymous
    • one year ago
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    It would be an arbitrary triangle, though, no guarantees whatsoever that it would be a right triangle.

  3. anonymous
    • one year ago
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    Alright. Give me a minute u have this down in my Geometry math journal I will just have to find it.

  4. mathmate
    • one year ago
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    |dw:1437781734957:dw|

  5. anonymous
    • one year ago
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    Right, I agree with the picture. Just wasn't able to come up with something using that. Tried for a while and either nothing worked or something that I thought might work had a bad assumption.

  6. mathmate
    • one year ago
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    The in-circle is tangent to all three sides, so the centre joined to the vertices bisect the vertex angles. Call them \(\alpha, \beta, \gamma\) at vertices A, B and C respectively. Name the centre as O, then \(\angle OBX =tan^{-1}2/3\) = 33.69 \(\angle OCX =tan^{-1}2/4\) =15.86 We can then calculate \(\angle A =2(90-33.69-15.86)\) =80.90 Then you can use the fact that the longest side is opposite the greatest angle of a triangle to find the longest side.

  7. anonymous
    • one year ago
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    Oh, I apologize, I forgot to add in that the answer must be exact.

  8. mathmate
    • one year ago
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    |dw:1437782481651:dw|

  9. mathmate
    • one year ago
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    The answer that you get will be pretty much exact! lol

  10. anonymous
    • one year ago
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    Yeah, but not allowed to do that. Basically we have to avoid trig functions.

  11. anonymous
    • one year ago
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    Completely spaced on that requirement when typing up the question, my bad.

  12. anonymous
    • one year ago
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    I can't find it at the moment. So, since all three sides are all different lengths, we can assume it is a Scalene triangle or a right triangle. The first thing we should do is draw and label a Scalene triangle. (I will post pictures of my work in just a second)

  13. anonymous
    • one year ago
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    @ganeshie8 Any idea?

  14. Loser66
    • one year ago
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    |dw:1437789668137:dw|

  15. Mertsj
    • one year ago
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    |dw:1437791155114:dw|

  16. Loser66
    • one year ago
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    @Mertsj from where you have BX/CX = AB/AC??

  17. Mertsj
    • one year ago
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    Angle bisector theorem

  18. Loser66
    • one year ago
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    It is apply if AOX is a line, that is A, O, X on the same line. This case is not that.

  19. Mertsj
    • one year ago
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    Good point.

  20. Loser66
    • one year ago
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    |dw:1437791652962:dw|

  21. anonymous
    • one year ago
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    I dont suppose there's an easy way to get |AO| then, huh?

  22. ganeshie8
    • one year ago
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    is the perimeter 21 ?

  23. ganeshie8
    • one year ago
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    It seems calculating the area two different ways gives two independent equations which can be solved easily |dw:1437844198135:dw|

  24. ganeshie8
    • one year ago
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    That is first way, we may use \(\href{https:///en.wikipedia.org/wiki/Heron%27s_formula}{\text{Heron's formula}}\) to work the area again : |dw:1437844485456:dw|

  25. anonymous
    • one year ago
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    Isn't 14 + 2d just the perimeter?

  26. ganeshie8
    • one year ago
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    \[14+2d = \sqrt{12d(7+d)} \implies d = 3.5\]

  27. ganeshie8
    • one year ago
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    good observation, it just happens that perimeter = area in our case :)

  28. anonymous
    • one year ago
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    What 2nd calculation for the area gives 14 + 2d?

  29. ganeshie8
    • one year ago
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    just addup the areas of individual triangles keeping in mind that \(2\) is the height in each triangle

  30. ganeshie8
    • one year ago
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    |dw:1437844853560:dw|

  31. anonymous
    • one year ago
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    Ah, gotcha. And then yeah, Heron's formula for the other calculation. Alright, that works perfectly :D

  32. ganeshie8
    • one year ago
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    Ofcourse the price for avoiding trig is that we need to remember Heron's formula

  33. anonymous
    • one year ago
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    Im perfectly fine with Heron's formula and a lot of the other formulas. Been doing too much math to forget it, just geometry is so much of an issue of whether or not you see the trick or where to start, lol. Sometimes the solution is easy, but your running through all the connections and things you know in your head and you just never come up with the one that leads to an easy solution.

  34. anonymous
    • one year ago
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    Thanks once again :)

  35. ganeshie8
    • one year ago
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    Haha ikr! sometimes u just have to wait, think in shower, sleep over it till the right method/idea strikes ;)

  36. anonymous
    • one year ago
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    I don't have that option during exams, though. Bummer! D: Haha.

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