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anonymous

  • one year ago

Is the triangle that this question is speaking of unique, or could there be multiple cases? "Let O be the circumcenter of ΔABC. Suppose OA and OB are perpendicular to each other. What is the measure of ∠C?

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  1. anonymous
    • one year ago
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    Solving the problem is easy, but I just am worried that there could be multiple cases, one in which the triangle could be acute and one in which the triangle could be obtuse. I don't think this problem is possible were the triangle to be a right triangle.

  2. phi
    • one year ago
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    |dw:1437855747306:dw|

  3. phi
    • one year ago
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    the angle at C is always 1/2 of the intercepted arc given OA and OB are perpendicular means the central angle is 90 degrees

  4. anonymous
    • one year ago
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    Yep, that would be the case for an acute. I was then worried if it were possible, based on how the problem had to be constructed, that you could have C be inside the angle .

  5. anonymous
    • one year ago
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    |dw:1437855911900:dw|

  6. phi
    • one year ago
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    oh, yes that is possible

  7. ikram002p
    • one year ago
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    -.- i lost my graph with cancel xDDDDDDD

  8. anonymous
    • one year ago
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    Oops, haha. I hate that x_X

  9. anonymous
    • one year ago
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    And yeah, I thought it might be possible but I wasn't sure. If it's not possible then I don't want to claim multiple cases.

  10. ikram002p
    • one year ago
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    |dw:1437856026860:dw|

  11. anonymous
    • one year ago
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    I know the circumcenter is outside the triangle if the triangle is obtuse. But since the circumcenter is the intersection of the perpendicular bisectors, I couldn't conclude whether or not I could still have the perpendicular bisectors intersect while OA still remainded perpendicular to OB.

  12. phi
    • one year ago
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    the other case has <C = 135 deg

  13. ikram002p
    • one year ago
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    the thing is its not unique triangle but a unique angle :D xD

  14. anonymous
    • one year ago
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    Ah. So there only is one case then?

  15. ikram002p
    • one year ago
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    it need not to be a bisector though :O

  16. anonymous
    • one year ago
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    So what are you saying then? xD

  17. phi
    • one year ago
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    your picture with C forming an obtuse angles meets the given conditions. O is the circumcenter OA is 90 deg to OB so it is valid... < C would be 135 in that case

  18. ikram002p
    • one year ago
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    u already set two cases xD

  19. ikram002p
    • one year ago
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    but 90+ obtuse noway

  20. anonymous
    • one year ago
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    No, I agree, phi. But would O still be the intersection of the perpendicular bisectors under the given conditions? |dw:1437856512007:dw| I'm not sure if this picture is possible. I was trying to draw it out on my own and couldnt show that those perpendicular bisectors would still hit the circumcenter O. Sorry if my question isn't clear, lol.

  21. phi
    • one year ago
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    yes, the perpendicular bisectors of the sides will meet at the circumcenter (but outside of the triangle)

  22. anonymous
    • one year ago
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    Okay, so I do need two cases. Yeah, I couldnt prove it to myself that I need the answers of both 45 and 135 degrees, wanted to confirm :)

  23. anonymous
    • one year ago
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    |dw:1437856799729:dw| Two triangles one acute and one obtuse, but both with the ability to fit the given conditions. That's what I was unsure of. But either way, I'll make sure to mention both cases :3

  24. phi
    • one year ago
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    yes

  25. anonymous
    • one year ago
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    Awesome. Thanks

  26. phi
    • one year ago
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    for what it is worth, here is a figure I made using Geogebra (free and useful tool) where the radiuses are perpendicular, and thus the third angle is 135 degrees It shows the perpendicular bisectors of the 3 legs as blue lines that intersect at the circumcenter.

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  27. anonymous
    • one year ago
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    Ah. Cool. I like that :) Yeah, our class has us using Geometer's Sketchpad, but that's not free :( I'll check out Geogebra ^_^

  28. phi
    • one year ago
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    each side of the triangle is a chord of the circumcircle, and the perpendicular bisectors of a chord always goes through the circle's center. in other words, it *has to work*

  29. anonymous
    • one year ago
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    That's interesting. Yeah, when you say that it definitely clears things up. Thanks again :)

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