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suppose there is given right triangle(AC=6) and circle (radius=5) which goes through A and C and meets AB leg in midpoint.here is picture enter image description here

we are asked to find length of leg.in my point of view ,we can connect center of circle with A and C point,get isosceles triangle we know all sides ,find angle at center,then connect center also to A and D here we know both length(radius)angle will be 180-a(a it is which we got by cosine law) calculate AD by the cosine law and got finally Ab,can you show me shortest way?or am i correct or wrong?please help

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What do we know? We know the radius of the circle and one angle. Do we know any lengths or angles? –  mixedmath Jul 23 '11 at 10:13
    
oops sorry AC=6 i forgot –  dato datuashvili Jul 23 '11 at 10:16
    
in fact it is not homework it is just token from national exam tasks –  dato datuashvili Jul 23 '11 at 10:21
    
@user3196: Sorry for being presumptuous. I think the tag is unnecessary in that case. –  anon Jul 23 '11 at 10:28
    
@anon no dont worry please there is not any problem –  dato datuashvili Jul 23 '11 at 10:31
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2 Answers

up vote 2 down vote accepted

Let $S$ be the center of the circle and $M$ be the midpoint of $AC$.

From the right triangle AMS we can get:

$|MS|=\sqrt{5^2-3^2}=4$.

Now we use the right triangle AMD. (This triangle is right since D is the midpoint of AB - have a look at similar triangles ADD' and ACB, where D' is the point of AC such that DD' is perpendicular to AC. You should see that D'=M.)

This right triangle gives us:

$|AD|=\sqrt{3^2+9^2}=3\sqrt{10}$ and $|AB|=2|AD|=6\sqrt{10}$


What is correct English terminology for this: "D' is the point of AC such that DD' is perpendicular to AC"? If I used word by word translation from my language, it would be "D' is the foot of the perpendicular from the point D to the line AC".


If I try to follow your suggestions and compute the angles then I get:

$\sin\alpha=\frac35$ and $\cos\alpha=\frac45$ ($\alpha$ denotes the angle ASM)

Now I digress a little from your suggestion.

$\frac{|AB|}4=5\cos\frac\alpha2=5\sqrt{\frac{1+\cos\alpha}2}=3\sqrt{\frac52}$

$|AB|=4.3\sqrt{\frac52}=6\sqrt{10}$.

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yes absolutely right thanks very much –  dato datuashvili Jul 23 '11 at 11:08
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I'm no native speaker, but "perpendicular foot" seems to be used in English too. –  t.b. Jul 23 '11 at 11:12
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$$\left\vert AC\right\vert ^{2}+\left\vert BC\right\vert ^{2}=\left\vert AB\right\vert ^{2}=4\left\vert AD\right\vert ^{2}$$

$$6^{2}+\left\vert BC\right\vert ^{2}=4\left\vert AD\right\vert ^{2}$$

$$h^{2}=5^{2}-3^{2}=16,$$

where $h$ is the distance from the center to $AC$. Hence $h=4$ and $$|AD|^2=\left( 5+h\right) ^{2}+3^{2}=9^{2}+3^{2}=90.$$

Thus

$$\left\vert AD\right\vert =3\sqrt{10}.$$

And $|BC|$ is such that

$$6^{2}+\left\vert BC\right\vert ^{2}=4\cdot 90,$$

$$\left\vert BC\right\vert =18.$$

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thank you very much –  dato datuashvili Jul 23 '11 at 11:32
    
@user3196: You are welcome! –  Américo Tavares Jul 23 '11 at 11:39
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