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# Tough geometry problem

Hello can somebody help me find 'D' in terms of 'r' and 'd' please? I've been trying to get the answer for a while now but i end up with a really really complicated expression which i have a feeling isn't really right.. could use some help. Thanks in advance

Note by Jord W
3 years, 10 months ago

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The radius of the LH circle is $$D+d$$. Consdering the isosceles triangle with angle $$\theta$$ and base $$r$$, we see that $r \; = \; 2(D+d)\sin\tfrac12\theta$ The other two angles in this isosceles triangle are $$90^\circ-\tfrac12\theta$$ and so, looking at the small right-angled triangle on the right, $\sin\tfrac12\theta \; = \; \cos\big(90^\circ-\tfrac12\theta) \; = \; \frac{d}{r}$ Putting these results together we deduce that $r^2 \; = \; 2(D+d)d$ · 3 years, 10 months ago

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found this a different way a short while ago but thanks anyway :) · 3 years, 10 months ago

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You should post your solution! · 3 years, 10 months ago

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ah its here in one of the other comments :) · 3 years, 10 months ago

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Nice and easy to understand. · 3 years, 10 months ago

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Hi, This can be done using trigonometry

$$cos(\theta) = \frac{D}{D + d}$$ --------- (1)

Joining the point of intersection of the two circles at the top to the center of the small circle, a beautiful triangle would be formed with,

$$sin(\theta) = \frac{r}{D+d}$$ --------- (2)

Squaring and adding (1) and (2), we get

$$D^2 + r^2 = (D + d)^2$$ · 3 years, 10 months ago

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in case you're wondering how i got what i think is the right answer: cos theta=D/(D+d) as you found however when you make the other identity, if you use the cosine rule for non right angled triangles, you get that r=(D+d)sqrt(2(1-cos theta)), and so you just put the cos theta=D/(D+d) in from earlier and it rearranges pretty quickly to D=r^2/2d -d :) · 3 years, 10 months ago

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Thanks for pinning out the problem in my solution and thanks for giving the right solution.

I didn't follow you when you used the cosine rule. Please tell me, which triangle did you used to get it and how did you get r=(D+d)sqrt(2(1-cos theta)). · 3 years, 10 months ago

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ah well its the triangle where the three vertices are the centre of the large circle, the centre of the small circle, and one of the intersections of the two circles. if you use this triangle, its got both the large radii subtending the angle theta, and the far side (relative to theta) is the radius of the small circle. Here you can apply the formula c^2=b^2+a^2-2ab cos(C) for which I've found a clearer explanation here: http://drdjw.files.wordpress.com/2012/01/general-triangle.jpg

when you substitute them into that formula for the triangle we're using, you get that r=(D+d)sqrt(2(1-cos theta)), hope that helps :) · 3 years, 10 months ago

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I think theres a problem with your sintheta = r/(D+d) . when you make that triangle, its an isosceles triangle not a right angled one even though it sort of looks like it should be a right angled one. however thanks anyway, your other identity i think has given me the clue and i think i've found the right answer anyway xD · 3 years, 10 months ago

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the radius meets the tangent at the circumference at right angles. · 3 years, 10 months ago

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(r^2)/2 = Dd + d^2 · 3 years, 10 months ago

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let the radius of the bigger circle be R.

let the perpendicular side of the right angle triangle be x. (having angle $$\theta$$)

now, join the center of the smaller circle and the point of intersection between the smaller and bigger circle ( we have two point of intersections, but we will take the upper one), so now we have an another right angle triangle so, applying pythogorus theoram in our new right angle triangle :
$$x^{2}+d^{2}=r^{2}$$

so, $$x^{2} = r^{2}-d^{2}$$___ [1]

now coming back to our bigger right angle triangle

using pythogorus theoram again

we get

$$x^{2}+D^{2}=R^{2}$$

putting value of x^{2} from eq. 1 and value of R (since R = D+d) we have

$$r^{2}-d^{2}+D^{2} = (D+d)^{2}$$

$$r^{2}-d^{2}+D^{2} = D^{2}+d^{2}+2Dd$$

$$r^{2}-2d^{2} = 2Dd$$

$$D= \frac {1}{2} \frac {r^{2}-2d^{2}}{d}$$ ---- answer :) · 3 years, 10 months ago

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Is the line joining the centre of the bigger circle and the point of intersection of the two circles tangent to the second circle? Can it be proved? · 3 years, 10 months ago

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r/tan@ -d · 3 years, 9 months ago

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This is solved easily using pythagoras famous equation. $$c^2=a^2 + b^2$$

the little triangle on the right tells us: $$r^2 = d^2 + h^2$$

the left side triangle shares the same 'h' side, and we can also say that: $$(D+d)^2 = D^2 + h^2$$

Substitute 'h' obtained from the first equation into the second one, and solve fo D in terms of d and r:

$$(D+d)^2 = D^2 + r^2 - d^2$$

$$D^2 + 2dD + d^2 = r^2 - d^2 + D^2$$

$$D = (r^2/2d) - d$$

Not so tough, but very fun. · 3 years, 10 months ago

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https://www.facebook.com/groups/572200892829750/ .... link of this solution check it out final relation is r=2 * (D+d) * Cos (theta/2) · 3 years, 10 months ago

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wait, is D the radius of the larger circle??? · 3 years, 10 months ago

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no, the radius of the larger circle is D+d · 3 years, 10 months ago

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Firstly we are assuming that the line that we can call $$R$$ which is the radius of the larger circle is at a tangent to the perimeter of the smaller circle and a normal to the perimeter of the larger circle (only reason I mention this is because if you zoom on the picture this is perfectly correct) and lets call this point $$P$$. However assuming these, we can draw a triangle from the center of the small circle along the length $$D$$ and then a $$90°$$ angle up to the point $$P$$ which we call length $$x$$, hence the hypotenuse of this triangle is $$R$$. We can then make another triangle which starts from the center of the smaller triangle along the length $$d$$ and then make another $$90°$$ angle up to the point $$P$$, the hypotenuse of this triangle is $$r$$. Further more, we know that $$R=D+d$$, using this information we can model a function of $$D(r,d)$$:

Using Pythagoras's theorem:

$$R^{2}=D^{2}+x^{2}$$

$$r^{2}=d^{2}+x^{2}$$

Hence, when we cancel $$x$$:

$$R^{2}-D^{2}=r^{2}-d^{2}$$ where $$R=D+d$$

$$(D+d)^{2}-D^{2}=r^{2}-d^{2}$$

Which simplifies to, $$r^{2}=2d(D+d)$$

Hence, $$D(r,d)=\frac{r^{2}}{2d}-d$$ · 3 years, 10 months ago

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D = (r - d sin (teta) ) / (sin (teta) ) ???? · 3 years, 10 months ago

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use trigonometry. $$tan\theta=\frac{r}{D}$$

$$\tan\theta\times D=r$$

or,

$$\tan(90-\theta)\times r=D$$ You have more than enough information in this problem. · 3 years, 10 months ago

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Fahad S. I don't think tan(theta)=r/d. join the radius of the smaller circle to the point of intersection of both the circles. you would find that the radius becomes the hypotenuse. hence the opposite side of the triangle containing theta is not equal to the radius. · 3 years, 10 months ago

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sorry, I had made one small mistake. It is $$\frac{r}{D+d}$$ · 3 years, 10 months ago

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Thats in terms of theta as well. my problem is that i can't eliminate theta. btw the 'D' isn't the radius of the circle, but 'D+d' is · 3 years, 10 months ago

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