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    ABC is an isosceles triangle, angle ACB is right.
    M lies inside ABC and is such that AM=\sqrt 3, BM=1, CM=\sqrt 2.
    Find angle AMC.

    Trigonometry failed.
    Please help?
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    Well what trigonometry have you used so far? (You don't need to use anything advanced. I used the sine rule.)

    Also, are you allowed a calculator for this question?

    EDIT:
    Well, you might be able to calculate it without a calculator, using some trig identities I haven't heard of yet (sorry, trig is not my favourite).
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    (Original post by FieldLeftBlank)

    Trigonometry failed.
    Please help?
    What is meant by this?
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    (Original post by dexter -1)
    What is meant by this?
    Presumably that he couldn't arrive at an answer through application of trigonometry.
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    (Original post by DerBoy)
    Well what trigonometry have you used so far? (You don't need to use anything advanced. I used the sine rule.)

    Also, are you allowed a calculator for this question?

    EDIT:
    Well, you might be able to calculate it without a calculator, using some trig identities I haven't heard of yet (sorry, trig is not my favourite).
    Can I get a further hint? I never thought of that. The whole problem is like shouting to use cosines :rolleyes:

    By the way, we're not allowed calculators. For anything.

    PS: As far as I remember it can be solved by some tricky rotations but I didn't take a closer look then.
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    (Original post by FieldLeftBlank)
    Can I get a further hint? I never thought of that. The whole problem is like shouting to use cosines :rolleyes:

    By the way, we're not allowed calculators. For anything.

    PS: As far as I remember it can be solved by some tricky rotations but I didn't take a closer look then.
    Hmmm.... sorry, I have no knowledge of trigonometry beyond (co)sine rule, so I'm unable to say whether you could work this one out the way I did. The answer (I got, at least) is an integer, which is why I think it may be possible to do using identities.

    As for the requested hints:
    As for any geometry question, draw a good big diagram and mark in everything you know, (lengths and angles). Draw a line from C to M.

    Before charging into trig, what can we calculate with what we know? We know AB=[sqrt(3)] +1. Calculate AC=AB using Pythagoras.

    Now, what can we equate using the sine rule?
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    (Original post by DerBoy)
    Before charging into trig, what can we calculate with what we know? We know AB=[sqrt(3)] +1. Calculate AC=AB using Pythagoras.
    Erm... but M is not on AB, it's inside the triangle, but not on AB. And it's given that AC=BC :confused:
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    (Original post by FieldLeftBlank)
    ABC is an isosceles triangle, angle ACB is right.
    M lies inside ABC and is such that AM=\sqrt 3, BM=1, CM=\sqrt 2.
    Find angle AMC.

    Trigonometry failed.
    Please help?
    Do you mean angle ACB is right-angled? Do you have a diagram?

    Is this it ? Corrected
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    (Original post by steve2005)
    Do you mean angle ACB is right-angled? Do you have a diagram?

    Is this it ?
    Yes. And BM=1.
    When it comes to math, my English... :o:
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    Oh... sorry, I thought you meant M was on AB. Never mind.
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    (Original post by FieldLeftBlank)
    Erm... but M is not on AB, it's inside the triangle, but not on AB. And it's given that AC=BC :confused:
    When I said 'Calculate AC=BC' I mean calculate AC, and this value equals BC (or vice versa).
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    (Original post by FieldLeftBlank)
    ABC is an isosceles triangle, angle ACB is right.
    M lies inside ABC and is such that AM=\sqrt 3, BM=1, CM=\sqrt 2.
    Find angle AMC.

    Trigonometry failed.
    Please help?
    where did you get this question from..I can solve it but its quite long..
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    (Original post by rbnphlp)
    where did you get this question from..I can solve it but its quite long..
    It's from a maths competition a couple of months ago.
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    (Original post by FieldLeftBlank)
    ABC is an isosceles triangle, angle ACB is right.
    M lies inside ABC and is such that AM=\sqrt 3, BM=1, CM=\sqrt 2.
    Find angle AMC.

    Trigonometry failed.
    Please help?
    It took me a while to do this:

    Spoiler:
    Show
    Warning!!: as you read through this make sure you draw the sketch with each piece of info ,otherwse you will be lost.

    Construction : draw MD such that MD is perpendicular to AC,ME perpendicular to CB, and MF perpendicular to AB .
    Let AC=x, and AD =y then, DC=x-y
    similarly,CB=x(isosceles), let CE =z, then EP =x-z
    and AB=\sqrt{2}x Pythagoras ,let AF=p, then FB=\sqrt{2}x -p


    Also label the angles of the triangle:
    let angle AMD=a,angle DMC=b ,angle CME=c, angle EMB=d, angle FMB=e, angle FMA=f.

    Drawing this triangle we note the following :
    angle AMD=angle EMB=a (vertically opposite angles)--(1)
    angle DMc=angle FMB=b (vertically pposite angles)--(2)
    angleAMF=angleCME=c (vertically opposite angles)---(3)

    and also let angle MAB=v

    Spoiler:
    Show
    Now consider triangle ADM and traingle MEB

    sin(a)=\frac{y}{\sqrt{3}} also triangle EMB ,angle EMB=a from 1
    therefore sin(a)=\frac{x-z}{1} combining the two equations
    we get (x-z)\sqrt{3}=y---4

    now consider the triangle CMD and FMB
    angle CMD =b from 2
    sin(b)=\frac{x-y}{\sqrt{2}}
    however sinb =\frac{\sqrt{2}x-p}{1} from triangle FMB
    combing the two equations:
    \sqrt{2}(\sqrt{2}x-p)=x-y ---(5)

    And finally In triangle CME and AMF sin(c)=\frac{z}{\sqrt{2}}
    sin(c)=\frac{p}{\sqrt{3}} from triangle AMF

    combinig the two gives =>\frac{p.\sqrt{2}}{\sqrt{3}}=z subbing this into equation (4)

    we get =>y=\sqrt{3}x-p\sqrt{2}
    add this equation with (5)
    we obtain
    x=(\sqrt{3}x-p\sqrt{2})+(2x-\sqrt{2}p)

    rearranging one obtains
    \frac{(1-\sqrt{3})x}{2\sqrt{2}}=p----(A)

    NOw for the final part consider triangle AMB
    using cosine rule also angle MAB=v

    we get 1^2=(\sqrt{3})^2+(\sqrt{2}x)^2-2(\sqrt{6})xcosV
    simplifying we get
    -1=x(x-\sqrt{6}cosv
    However from triangle AMF
    cosv=\frac{p}{\sqrt{3}}
    combing the two eqs one gets
    -1=x^2-\sqrt{2}px
    But we know from (A) that\frac{(1-\sqrt{3})x}{2\sqrt{2}}=p
    we can now find x as x is the only unknown..

    if x is known Angle AMC can be found using the cosine rule on triangle AMC as three sides are known ..
    I wouldnt expect make all of that to make sense ,if you have any doubt just quote me and I shall see if I can resolve it

    im pretty sure there might be a easier way to do this..but which competition is this from(like what level)?
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    (Original post by rbnphlp)
    Spoiler:
    Show
    Drawing this triangle we note the following :
    angle AMD=angle EMB=a (vertically opposite angles)--(1)
    angle DMc=angle FMB=b (vertically pposite angles)--(2)
    angleAMF=angleCME=c (vertically opposite angles)---(3)



    im pretty sure there might be a easier way to do this..but which competition is this from(like what level)?

    But in that way we're basically assuming that M lies on AE, BD and CF, aren't we?

    Well it's a Bulgarian competition and we have lots of them here... it wasn't one of the particularly hard but no one in my region solved this problem. The official solution was quite short, actually, and with no trigonometry. Unfortunately, I don't remember it :p:
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    (Original post by FieldLeftBlank)
    But in that way we're basically assuming that M lies on AE, BD and CF, aren't we?
    yes:o: , dont tell me I didnt try...
    howver I can still see another way to do it , but that is quite long as well so and involves trig,Im not sure
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    (Original post by rbnphlp)
    yes:o: , dont tell me I didnt try...
    howver I can still see another way to do it , but that is quite long as well so and involves trig,Im not sure
    Ah well, thanks for your time anyway
    I'll know on Friday anyway. If I don't get a divine touch or something by then :p:
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    (Original post by FieldLeftBlank)
    Well it's a Bulgarian competition and we have lots of them here... it wasn't one of the particularly hard but no one in my region solved this problem. The official solution was quite short, actually, and with no trigonometry. Unfortunately, I don't remember it :p:
    If you know where it's come from, a search via Google can be fruitful sometimes. I've tracked down the odd question from the Bulgarian Olympiad with answers, in the past.
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    (Original post by FieldLeftBlank)
    Ah well, thanks for your time anyway
    I'll know on Friday anyway. If I don't get a divine touch or something by then :p:
    hey after a while I have thought up of another way..I havent completed the solution but if what I have wrote till now looks right I shall complete it .. feel free to point out the msitakes..

    Spoiler:
    Show
    join M to c ,b and a as questions
    let angle ACM=v,then angle MCB=90-v(since aCB is right angled)
    let Ac =CB=x (isosceles )
    Consider triangle AMC
    using cosine rule
    we get
    (\sqrt{3})^2=(\sqrt{2})^2+x^2-2\sqrt{2}xcosv
    => 1=x^2-2\sqrt{2}xcosv--(1)

    Now consider triangle CMB
    using the cosine
    -1=x^2-2\sqrt{2}xcos(90-v)--(2)
    cos(90-v)=sinv
    dividing (1) by (2)
    \displaystyle\frac{x^2-2\sqrt{2}xcosv}{x^2-2\sqrt{2}xsinv}=-1

    cross multiplying an rearranging
    2x^2-2\sqrt{2}xcosv-2\sqrt{2}xsinv=0
    taking the x's out and diving through by 2
    we get:
    x^2-(\sqrt{2}cosv+\sqrt{2}sinv)x=0
    so the two solutions of the quadratic are x=0 , x=(\sqrt{2}(cosv+sinv)
    x cannot =0, sox=(\sqrt{2}(cosv+sinv) is the only solution..
 
 
 
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