multiple choice question

[logistic_guy]

here is the question

Look at the diagram and find \(\displaystyle RS\).

(a) \(\displaystyle -2\sqrt{15}\)
(b) \(\displaystyle 4\)
(c) \(\displaystyle 2\sqrt{15}\)
(d) \(\displaystyle 8\)




my attemb
i can draw a line from the center \(\displaystyle Q\) to \(\displaystyle S\) or from the center \(\displaystyle P\) to \(\displaystyle R\) to get a right triangle
but this still won't help me as i'll have two sides unknown
i don't see other idea

 

[jonah2.0]

Beer drenched reaction follows.

here is the question

Look at the diagram and find \(\displaystyle RS\).

(a) \(\displaystyle -2\sqrt{15}\)
(b) \(\displaystyle 4\)
(c) \(\displaystyle 2\sqrt{15}\)
(d) \(\displaystyle 8\)

View attachment 38933


my attemb
i can draw a line from the center \(\displaystyle Q\) to \(\displaystyle S\) or from the center \(\displaystyle P\) to \(\displaystyle R\) to get a right triangle
but this still won't help me as i'll have two sides unknown
i don't see other idea

Here is the hint.
Well, the hint is somewhere in that video.

 

[Aion]

here is the question

Look at the diagram and find \(\displaystyle RS\).

(a) \(\displaystyle -2\sqrt{15}\)
(b) \(\displaystyle 4\)
(c) \(\displaystyle 2\sqrt{15}\)
(d) \(\displaystyle 8\)

View attachment 38933


my attemb
i can draw a line from the center \(\displaystyle Q\) to \(\displaystyle S\) or from the center \(\displaystyle P\) to \(\displaystyle R\) to get a right triangle
but this still won't help me as i'll have two sides unknown
i don't see other idea

My hint is to extend the tangent line as shown above. Since the radius is perpendicular to the tangent at the point of contact, the angles [imath]\angle QRZ[/imath] and [imath]\angle PSZ[/imath] are right angles. Hence, the triangle [imath]\Delta QRZ \sim \Delta PSZ[/imath], because they both share [imath]\angle Z[/imath] and a right angle. Using similarity, you can calculate [imath]y[/imath]. Then, using the Pythagorean theorem, you can calculate [imath]t[/imath] in terms of [imath]y[/imath]. Finally, applying the Pythagorean theorem again on the larger triangle allows you to solve for [imath]x[/imath].

 

[logistic_guy]

Beer drenched reaction follows.


Here is the hint.
Well, the hint is somewhere in that video.

thank
i watch the whole video
it don't change the circles. they're not tangent in the whole exampel

View attachment 38935
My hint is to extend the tangent line as shown above. Since the radius is perpendicular to the tangent at the point of contact, the angles [imath]\angle QRZ[/imath] and [imath]\angle PSZ[/imath] are right angles. Hence, the triangle [imath]\Delta QRZ \sim \Delta PSZ[/imath], because they both share [imath]\angle Z[/imath] and a right angle. Using similarity, you can calculate [imath]y[/imath]. Then, using the Pythagorean theorem, you can calculate [imath]t[/imath] in terms of [imath]y[/imath]. Finally, applying the Pythagorean theorem again on the larger triangle allows you to solve for [imath]x[/imath].

thank
i think i see your idea
\(\displaystyle \frac{r_1}{r_1 + r_2 + y} = \frac{r_2}{y}\)
\(\displaystyle r_1y = r_1r_2 + r_2^2 + yr_2\)
\(\displaystyle r_1y - yr_2= r_1r_2 + r_2^2\)
\(\displaystyle y(r_1 - r_2) = r_1r_2 + r_2^2\)
\(\displaystyle y = \frac{r_1r_2 + r_2^2}{r_1 + r_2} = \frac{5(3) + 3^2}{5 + 3} = \frac{15 + 9}{8} = \frac{24}{8} = 3\)

\(\displaystyle y^2 = t^2 + r_2^2\)
\(\displaystyle t^2 = y^2 - r_2^2 = 3^2 - 3^2 = 0\)

 

[Aion]

thank
i watch the whole video
it don't change the circles. they're not tangent in the whole exampel


thank
i think i see your idea
\(\displaystyle \frac{r_1}{r_1 + r_2 + y} = \frac{r_2}{y}\)
\(\displaystyle r_1y = r_1r_2 + r_2^2 + yr_2\)
\(\displaystyle r_1y - yr_2= r_1r_2 + r_2^2\)
\(\displaystyle y(r_1 - r_2) = r_1r_2 + r_2^2\)
\(\displaystyle y = \frac{r_1r_2 + r_2^2}{r_1 + r_2} = \frac{5(3) + 3^2}{5 + 3} = \frac{15 + 9}{8} = \frac{24}{8} = 3\)

\(\displaystyle y^2 = t^2 + r_2^2\)
\(\displaystyle t^2 = y^2 - r_2^2 = 3^2 - 3^2 = 0\)

[math]\frac{y+r_2}{y+2r_2+r_1}=\frac{r_2}{r_1}[/math]
Substitute [imath]r_1=5[/imath] and [imath]r_2=3[/imath]

[math]\frac{y+3}{y+11}=\frac{3}{5}[/math]
Solve for [imath]y[/imath]

[math]y=9[/math]

 

[logistic_guy]

[math]\frac{y+r_2}{y+2r_2+r_1}=\frac{r_2}{r_1}[/math]
Substitute [imath]r_1=5[/imath] and [imath]r_2=3[/imath]

[math]\frac{y+3}{y+11}=\frac{3}{5}[/math]
Solve for [imath]y[/imath]

[math]y=9[/math]

if \(\displaystyle y = 9\)
then \(\displaystyle y^2 = t^2 + r_2^2\)
\(\displaystyle t^2 = y^2 - r_2^2 = 9^2 - 3^2 = 81 - 9 = 72\)
\(\displaystyle t = \sqrt{72}\)
\(\displaystyle \frac{r_1}{r_2} = \frac{x + t}{t}\)
\(\displaystyle \frac{r_1 t}{r_2} = x + t\)
\(\displaystyle x = \frac{r_1 t}{r_2} - t = \frac{5(\sqrt{72})}{3} - \sqrt{72} = 10\sqrt{2} - 6\sqrt{2} = 4\sqrt{2}\)
not one of the choices

 

[Aion]

if \(\displaystyle y = 9\)
then \(\displaystyle y^2 = t^2 + r_2^2\)
\(\displaystyle t^2 = y^2 - r_2^2 = 9^2 - 3^2 = 81 - 9 = 72\)
\(\displaystyle t = \sqrt{72}\)
\(\displaystyle \frac{r_1}{r_2} = \frac{x + t}{t}\)
\(\displaystyle \frac{r_1 t}{r_2} = x + t\)
\(\displaystyle x = \frac{r_1 t}{r_2} - t = \frac{5(\sqrt{72})}{3} - \sqrt{72} = 10\sqrt{2} - 6\sqrt{2} = 4\sqrt{2}\)
not one of the choices

Sorry, I think you've misinterpreted what [imath]y[/imath] is in this context. Apologies maybe I should've labelled more points to be more precise. You can't apply the Pythagoras theorem this way. The right deduction is

[math](y+r_2)^2=r_2^2+t^2[/math]Hence
[math]t=\sqrt{(y+r_2)^2-r_2^2}[/math]

 

[Aion]

Sorry, I think you've misinterpreted what [imath]y[/imath] is in this context. Apologies maybe I should've labelled more points to be more precise. You can't apply the Pythagoras theorem this way. The right deduction is

[math](y+r_2)^2=r_2^2+t^2[/math]Hence
[math]t=\sqrt{(y+r_2)^2-r_2^2}[/math]

After you've calculated [imath]t[/imath] you don't necessarily need to use the Pythagoras theorem again, you could also use the Transversal Theorem, then we have

[math]\frac{x}{t}=\frac{r_1+r_2}{r_2+y}[/math]

 

[Aion]

After you've calculated [imath]t[/imath] you don't necessarily need to use the Pythagoras theorem again, you could also use the Transversal Theorem, then we have

[math]\frac{x}{t}=\frac{r_1+r_2}{r_2+y}[/math]

If you consider the general case then one can show that

[math]t=\frac{2r_2\sqrt{r_1r_2}}{r_1-r_2}[/math] and
[math]y=\frac{2r_2^2}{r_1-r_2}[/math]
Hence [math]x=\frac{t(r_1+r_2)}{r_2+y}=\left(\frac{2r_2\sqrt{r_1r_2}}{r_1-r_2}\right) \cdot \left(\frac{r_1+r_2}{r_2+ \frac{2r_2^2}{r_1-r_2}} \right)=2\sqrt{r_1r_2}[/math]

 

[logistic_guy]

Sorry, I think you've misinterpreted what [imath]y[/imath] is in this context. Apologies maybe I should've labelled more points to be more precise. You can't apply the Pythagoras theorem this way. The right deduction is

[math](y+r_2)^2=r_2^2+t^2[/math]Hence
[math]t=\sqrt{(y+r_2)^2-r_2^2}[/math]

let me try it one more time

\(\displaystyle (y + r_2)^2 = r_2^2 + t^2\)
\(\displaystyle t^2 = (y + r_2)^2 - r_2^2 \)
if \(\displaystyle y = 9\) then
\(\displaystyle t^2 = (9 + 3)^2 - 3^3 = (12)^2 - 9 = 144 - 9 = 135\)
\(\displaystyle t = \sqrt{135}\)

After you've calculated [imath]t[/imath] you don't necessarily need to use the Pythagoras theorem again, you could also use the Transversal Theorem, then we have

[math]\frac{x}{t}=\frac{r_1+r_2}{r_2+y}[/math]

i'll use the transversal theorem
\(\displaystyle \frac{x}{t}=\frac{r_1+r_2}{r_2+y}\)
\(\displaystyle x=\frac{(r_1+r_2)t}{r_2+y} = \frac{(5 + 3)\sqrt{135}}{3+ 9} = \frac{8\sqrt{135}}{12} = \frac{2\sqrt{135}}{3} = \frac{6\sqrt{15}}{3} = 2\sqrt{15}\)

i get choice (c)
is my answer correct?

 

[Aion]

let me try it one more time

\(\displaystyle (y + r_2)^2 = r_2^2 + t^2\)
\(\displaystyle t^2 = (y + r_2)^2 - r_2^2 \)
if \(\displaystyle y = 9\) then
\(\displaystyle t^2 = (9 + 3)^2 - 3^3 = (12)^2 - 9 = 144 - 9 = 135\)
\(\displaystyle t = \sqrt{135}\)


i'll use the transversal theorem
\(\displaystyle \frac{x}{t}=\frac{r_1+r_2}{r_2+y}\)
\(\displaystyle x=\frac{(r_1+r_2)t}{r_2+y} = \frac{(5 + 3)\sqrt{135}}{3+ 9} = \frac{8\sqrt{135}}{12} = \frac{2\sqrt{135}}{3} = \frac{6\sqrt{15}}{3} = 2\sqrt{15}\)

i get choice (c)
is my answer correct?

Yes, that's correct! I apologize if I shared too much information upfront. I should have given you the chance to solve it before posting the general solution.

 

[logistic_guy]

thank Aion very much

you explained it nicely

Yes, that's correct! I apologize if I shared too much information upfront.

you don't have to

i appreciate your effort very much

i've question about your diagram. how do you come with idea to draw a point outside the two circles?

 

[Aion]

thank Aion very much

you explained it nicely


you don't have to

i appreciate your effort very much

i've question about your diagram. how do you come with idea to draw a point outside the two circles?

To be honest, it wasn’t my original idea. I came across a similar problem in a book a couple of years ago. Once you’ve seen the method, it becomes easy to apply the concept to other similar problems.

 

[logistic_guy]

To be honest, it wasn’t my original idea. I came across a similar problem in a book a couple of years ago. Once you’ve seen the method, it becomes easy to apply the concept to other similar problems.

still it's very beautiful idea. i'll write it down in my notebook to remember next time

thank very much

 

[jonah2.0]

Beer drenched reaction follows.

thank
i watch the whole video
it don't change the circles. they're not tangent in the whole exampel

You say you watched the whole video. Unfortunately, you failed to appreciate the hint (or idea) mentioned at 1:36 of that video. You could have arrived at the same answer with less work if you only paid attention to that particular hint. Consider the following diagram that says pretty much the same deal as in 1:36 of that video.


So, in a similar fashion (including Aion's vision) we have

 

[Aion]

Beer drenched reaction follows.

You say you watched the whole video. Unfortunately, you failed to appreciate the hint (or idea) mentioned at 1:36 of that video. You could have arrived at the same answer with less work if you only paid attention to that particular hint. Consider the following diagram that says pretty much the same deal as in 1:36 of that video.
View attachment 38942

So, in a similar fashion (including Aion's vision) we have

View attachment 38941

Well, that is a more elegant solution! If [imath]d[/imath] is the distance between centers [imath]Q[/imath] and [imath]P[/imath] then [imath]x^2+(r_1-r_2)^2=d^2[/imath]. In this case [imath]d=r_1+r_2[/imath], therefore

[math]x=\sqrt{(r_2+r_2)^2-(r_1-r_2)^2}=2\sqrt{r_1r_2}[/math]

 

[jonah2.0]

Beer induced reaction follows.

[math]x=\sqrt{(r_2+r_2)^2-(r_1-r_2)^2}=2\sqrt{r_1r_2}[/math]

Minor oversight.
That should have been
[math]x=\sqrt{(r_1+r_2)^2-(r_1-r_2)^2}=2\sqrt{r_1r_2}[/math]

 

[logistic_guy]

what an idea

i can't see this in a million years

Beer drenched reaction follows.

You say you watched the whole video. Unfortunately, you failed to appreciate the hint (or idea) mentioned at 1:36 of that video. You could have arrived at the same answer with less work if you only paid attention to that particular hint. Consider the following diagram that says pretty much the same deal as in 1:36 of that video.

i don't see this idea because i was focus for the vp to show two tangent circles

\(\displaystyle (r_1 + r_2)^2 = x^2 + (r_1 - r_2)^2\)
\(\displaystyle (5 + 3)^2 = x^2 + (5 - 3)^2\)
\(\displaystyle 8^2 = x^2 + 2^2\)
\(\displaystyle x^2 = 64 - 4 = 60\)
\(\displaystyle x = \sqrt{60} = 2\sqrt{15}\)

thank jonah very much
also
thank Aion very much

the combination of your ideas written in my notebook. your mastery is stolen

 

[Agent Smith]

[imath]RS^2 + 2^2 = 8^2[/imath]

 

[logistic_guy]

thank

[imath]RS^2 + 2^2 = 8^2[/imath]

\(\displaystyle RS^2 + 2^2 = 8^2\)
\(\displaystyle RS^2 = 64 - 4 = 60\)
\(\displaystyle RS = \sqrt{60} = 2\sqrt{15}\)

i think it's same my calculation, i only use \(\displaystyle x\)