Question

Challenge Qu

Answer 1

Wow, this is an amazing challenge.

To start off...


Since it's 9inches apart on the left, I'm gonna say the smaller circle shifted right by 4.5 inches

Answer 2

And since it's supposed to be 4.5 inches, it'd be the same for the top if the smaller circle was at the origin.

But instead, it's 5 inches apart at the top, so it went down by 0.5 inches as it shifted 4.5 inches to the right...

Answer 3

Sorry I can't solve it.

I've made an equation where the little circle radius =bigger circle radius - 4.5

But I can't find a 2nd equation relating the radii to form a system of equation..

Answer 4

cool! the algebraic solution is not exactly inspiring. you're missing the fact that the smaller circle, centred at \((\frac{9}{2},0)\) passes through \((0, R-5)\). work that again and you get an answer.

i will post "the answer" from the book, which is not exactly unhelpful.

it's a Bletchley Park pub which seems to assume that everyone is Alan Turing. Would that be so !!

Answer 5

A solution that is still way more contrived than in the book

If we label the centre of smaller circle as "X", and say R = rad of larger circle, r = rad of smaller, then we can build an isosceles triangle using the chord DE and say that EX = DX !!

\(\implies R - 9 + \dfrac{9}{2} = r \implies r = R - \dfrac{9}{2} \qquad \star\)

Next, if we place a x-y coordinate system at C then we can say of the smaller circle that:

\(\left(x- \dfrac{9}{2}\right)^2 + y^2 = r^2\)

specifically for point E

\(\left(0 - \dfrac{9}{2}\right)^2 + (R- 5)^2 = r^2 \)


Using \(\star\):

\(\implies \left( \dfrac{9}{2} \right)^2 + R^2- 10R + 25 = R^2 - 9R + \left(\dfrac{9}{2}\right)^2 \)

\(\implies R = 25 \) And \(\star \implies r = 20.5\)

Or \(D = 50 \) and \(d = 41\)

This is what the books says:


turning green very slowly :)

Answer 6

A solution that is still way more contrived than in the book

If we label the centre of smaller circle as "X", and say R = rad of larger circle, r = rad of smaller, then we can build an isosceles triangle using the chord DE and say that EX = DX !!

\(\implies R - 9 + \dfrac{9}{2} = r \implies r = R - \dfrac{9}{2} \qquad \star\)

Next, if we place a x-y coordinate system at C then we can say of the smaller circle that:

\(\left(x- \dfrac{9}{2}\right)^2 + y^2 = r^2\)

specifically for point E

\(\left(0 - \dfrac{9}{2}\right)^2 + (R- 5)^2 = r^2 \)


Using \(\star\):

\(\implies \left( \dfrac{9}{2} \right)^2 + R^2- 10R + 25 = R^2 - 9R + \left(\dfrac{9}{2}\right)^2 \)

\(\implies R = 25 \) And \(\star \implies r = 20.5\)

Or \(D = 50 \) and \(d = 41\)

Answer 7

answer in book



terse, pithy,...