Can you solve this interesting geometry challenge?
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- Опубліковано 2 лип 2024
- A semicircle is inscribed in a quarter circle, and one of its diameter endpoints is a distance 1 from the center of the quarter circle. What fraction of the quarter circle's area is contained in the semicircle? Special thanks this month to: Kyle, Lee Redden, Mike Robertson, Daniel Lewis. Thanks to all supporters on Patreon! / mindyourdecisions
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I loved that you looked back at the problem and made generalizations in the last minute and a half of the video. So many math teachers, once the answer is gotten, just move on to the next problem without doing this post-solving analysis. That is often when much learning and connections are done. Good job, Presh.
😊
Agreed 👍🏼
Yeah man, he's amazing indeed.
Physics student in chat : Once you asked about the ratio of areas given only a SINGLE lenght, it's immediately obvious the result cannot depend on it (unless the problem is ill-defined and we cannot derive the result), purely from dimensional analysis. If your ratio would depend on this "1" explicitly, there would be no way to ensure the independence of the result on the units of this "1", i.e. lenght have to enter here in ratios as well.
And that is exactly the reason why scaling the problem has no effect whatsoever, it is just like changing units.
I figured the same - assumed you have to be able to shift the position of the semicircle - which instantly gives the solution 1/2 lol
@@S3IIL3CT and that is, in fact, one of the methods of mathematical physics :)
I had a feeling that was going to be the answer, but I had no idea how to prove it. Wow.
9:27 Not only does the semicircle fit perfectly, but the arc rolls perfectly without the tangent point slipping!
Exactly this is why it reminds me a little a planetary gear or a Wankel motor.
Also, points on the smaller circle circumference move in straight lines (all of them, but including the "corner" points as we see). This is a property of the 2:1 ratio and was indeed exploited in the early days of steam engines.
Another interesting point in the final 45 seconds as you move the semicircle within the quarter circle: Line AT will always be parallel to the horizontal side of the quarter circle, and line BT will always be parallel to the vertical side (except, of course, when the semicircle diameter lies along either axis).
Thanks, sir, good work for the brain ..... ,
I solved this using a different approach
Let the bigger quarter circle's centre be at origin and have a radius of R
The points touching the semcircle and the arc's radius are (1,0) and (0,h)
Thus we get the radius to be √(1+h²)/2 and centre at (1/2,h/2) for the semicircle
Now consider a general point (x,y) on the quarter circle,if this touches the semicircle
We can say x²+y²=R²
And (x-1/2)²+(y-h/2)²=(1+h²)/4
Solving this we get x+hy=R²
Which implies that the semicircle touched at (1,h)
This 1+h²=R²
Thus the radius of semicircle is R/2
Thus the fractional area of the shaded region is 1/2
Might seem long but trust me I solved this problem under 5mins with this approach
Love these videos. For this problem, I took an approach that is a bit simpler. Draw a tangent line to the larger quarter circle at the point where the inner half circle is also tangent. Then draw the line from the center of the small circle to the point of tangency and the line from the center of the larger quarter circle to the point of tangency. These lines from the centers both intersect the tangent line at right angles (a well-known geometric theorem), and so the line from the center of the larger quarter circle passes through the center of the inner half circle. You'll then observe that the radius of the half circle is exactly half of the radius of the quarter circle. The length 1 in the drawing is not relevant (as was noted by another commenter).
It reminds me of my undergraduate maths; I watched it without blinking...slap stic if you don't remember the 10 circle theorem...there are two theorems here...but one of them got thrashed by the teacher. I'm still recalling the agony and happiness from 25 years ago.
Several months ago, I met my teacher. I went to touch her feet.
The distance "1" is irrelevant.
What I had in mind is : after noticing that C belongs to the small circle, gently rotate the small circle clockwise around its center until its diameter fits the radius CT of the big circle (T being the tangent point, which doesn’t move !).
Then evidently the red surface is half of the quarter circle’s…
And the information « distance = 1 » has not been used.
Thank you for your interesting videos !
Actually the information about the lower side of the triangle equal to 1 is not necessary. The most important thing is that the line from the centre of the big circle to the point of contact of the small circle and the circumference of the big circle passes through the centre of the small circle. With this we can get
(R-r)sin x = r sin y where R and r are the radii of the big and small circles, and x and y the upper angles of the triangle and the angle between the vertical line and the line from the centre of the big circle to the point of contact of the small and big circles.
(R-r)sin (丌/2-x)=rsin (丌/2-y)，which is of course
(R-r)cos x=rcos y
Squaring both equations and summing them you get the ratio R/r=2
I went a different way but I’m not sure if I solved a general case, or just a specific one.
I saw that the horizontal distance was 1, but the vertical distance wasn’t shown. Therefore it doesn’t matter, since the problem has a solution.
I set the vertical distance as 1 as well, so the semicircle was at 45’ angle.
Diameter would then be rt2, radius is (rt2)/2, and would meet tangent point.
Extend this radius through to the centre of big circle, and this length is sqrt(1/2).
Add these 2 to get the large circle radius as 2/(sqrt2)
Square both and *pi
small area = pi/2
Large r^2 = 2pi
Small semicircle = pi/4
Large quarter circle = pi/2
Ratio = 1/2
I got this intuitively without using the theorems you described by imagining the semi-circle rotating clockwise until the diameter was vertical coinciding neatly with the radius of the larger quarter circle. It seemed to me 1 was probably a red herring. Though my method is no proof, it produced the correct answer with a little imagination and a dash of algebra. Crow, crow, crow!
Beautiful problem and explaination!
Another great problem and solution from Presh.
Awesome! I wonder if anyone ever designed a quarter circler room with a movable semicircle platform inside.
That would be so cool!
This is one of those floating parameter problems
So let’s create the simplest scenario keys say point y where the semicircle touche the vehicle radius of the quarter circle is infinitesimally close to the origin of the quarter circle. In this case
1 = diameter of the semicircle = 2r, r=1/2 and semicircle is r/8 * pi
The quarter circle is r=1 so its area is pi/4. The shaded fraction is 1/2
Next, let’s consider y = 1
If y = 1 then the diameter of the semicircle is SQRT(2), this its radius is SQRT(1/2)
Pi* 1/2 / 2 = pi/4
If this is the case the th radius of the quarter circle = SQRT(1/2)+ SQRT(1/2 = SQRT(2)
Pi* SQRT(2)^2 /4 = pi 2/4 = pi/2 thus the fraction of the red circle is pi/4 / pi/2 = 1/2
This does not mean the plot is linear. Let’s Say y = SQRT(3). If that is the case the diameter of the semicircle is 2, this is pi r^2/2 = 1/2 pi. So if the trend were to hold true then the area of the quarter circle would need to be pi, reflecting a circle of area 4pi or radius of 2. So in this 1 would be the radius of the semicircle pointing to 60° the angle at the origin would be 60° forming an isosceles triangle. Thus the two lines would connect at tge origin of the circle 1 + 1 = 2 so this would be true.
Then the final test is what if we set y to infinity, in this case 1 is an infinitesimally small relative to y and the proportions mirror he first case. The answer being 1/2.
The fact that the circumcenter of a right triangle is the midpoint of its hypotenuse can be directly concluded from Thale's theorem because Thale's theorem is an equivalence and not just an implication.
That's my point
That visualization at the end is exactly my solution from just guessing from the thumbnail lol
In response to video title: no
This is linked to a previous problem. A ladder is standing against a wall. The top and bottom touch the wall. The trace of the midpoint of the ladder as it falls away from the wall top down is a quarter circle. If the ladder is pulled out from the bottom, the trace of the midpoint is the same quarter circle.
It was clear to me that the given value of 1 was irrelevant since they were asking for a ratio. Which led me to consider under what circumstances is it even possible for a semicircle's ends to be on the ends of a quarter circle but still have their arcs be tangent at a point. Intuitively, the only way it works is if the diameter of the semicircle is equal to the radius of the quarter circle since you can place it on either edge of the quarter circle and the tangent would be on the corner. From that point it's easy to find the ratio as you demonstrated in the video. The geometric proof was nice though.
I created two circles, small one inside large one, and restricted details till be like the given problem. After it, I deducted R = 2r for all points T and C(Shown in the video).
Because no other dimensions were stated, it was clear from the start that the “1” was irrelevant. Further, the wording of the original problem implicitly indicated that the final answer was constant. Based on this, and knowing the area of the quarter circle is constant, we also know that the area of the semicircle is constant. We can therefore go straight to the case where the semicircle diameter lies along either axis of the quarter circle, and thus conclude the radius of the quarter circle is exactly double that of the semicircle. We can now go straight to calculating the area of each shape, and their ratio of 1/2-without ever having needed to bother with applying any knowledge about tangent circles or circumcenters.
This is what is known as "Feynman Trick".
Exactly my logic. Then I imagined the line through the centers and it was proven as well by symmetry
0:33 Trick question! The datum provided to us, that a certain line segment has a certain length (= 1), _is _not _needed to solve the problem - it's redundant.
Let O be the center of the (completed) large quartercircle (Q), U the center of the (completed) small semicircle (S), with A & B the endpoints of the diameter of S shown. Let T be the point of tangency of S with Q.
Note that S and Q share the same tangent line L at T, so since radii TO and TU both meet L perpendicularly, TO and TU are collinear. Furthermore, diameter AB of S is the hypotenuse of the right triangle AOB. Therefore UO is a radius of S. So OT is both a radius of Q and also a diameter of S. Thus S has half the radius of Q. Therefore
(Area(S)/2) / (Area(Q)/4) =
((½)²Area(Q)/2) / (Area(Q)/4)
= 1/2. ∎
Do you need to show that the right triangle ACB is used to form the circumcircle (the other half of a smaller circle is indeed passing through the point C)?
Loved it
Quick solution: The y intercept is arbitrary, so why not set it to 0, so the semicircle is now flush with the bottom of the quarter circle. Then the SC a diameter of 1, and the QC has a radius of 1 and (pi*.5^2/2)/(pi*1^2/4) = 50%
Thank you.
Thanks
Nice:)
I have looked at it like a rectangle ABCT inside bigger circle which is always possible to put rectangle in every circle. Then diagonal of this rectangle CT is actually a „R” Ang half of this diagonal is „r”. So R=2r and we can calculate fraction which is 2:1.
Very interesting problem:)
When you move red half-circle in quarter circle it reminds me a little a rotating piston in Wankel motor…
Nice explanation ❤❤❤❤❤
Old subscriber of presh ❤, love preshes knowledge and math
I recall this problem was in my Grade 12 final exam in 1976, still could do it 😊😊
Love your intuitive graphics - easy to follow and comprehend. The only step I was stuck on was when you said (but didn't show) that OA = OC "because the point O is along the perpendicular bisector of AC". I had to go back and figure out that OAC is an equilateral triangle because everything about OAD and OCD are equal ... (geometry isn't intuitive for me yet). I'm still not sure why your statement is true - just one step too technical for me at this point (pardon pun 🙄)
If there is an answer, it can't depend on the (unspecified) height of the vertical segment, so we may as well assume that is 1 also. That means the radius of the larger circle is sqrt(2) and the area of the quarter circle is 𝜋(sqrt(2))²/4 = 𝜋/2. The radius of the smaller circle is sqrt(2)/2, so the area of the semicircle is 𝜋(sqrt(2)/2)²/2 = 𝜋/4. Therefore the fraction is (𝜋/4)/(𝜋/2) = 1/2.
That "1" threw me off.
That BC=1 is redundant information. It is not needed to conclude that the radius of the semi-circle is half that of the quarter circle. BC could be anything, and yet the relationship between the radii will be preserved.
I haven't thought this through, but my intuition has me thinking that the diameter of the semicircle will always be the same the radius of the quarter-circle. This is confirmed at the limits, but can it be proved/demonstrated at all positions?
LETS GOOO I figured it out in my head
I drew the two cases where the semi circle diameter is vertical and at 45°. In both cases diameter equals radius of the quarter circle. So without proving the general case, the answer is 0.5.
Is adjacent really a clear and accurate word for two PERPENSOCULAR RADII?? After all there are an infinite number of adjacent radio you can draw, so would lpl agree adjacent is ambiguiua without the picture showing he means perpendicular radii?
Good to add to coffin problems 😂
As it is obvious that the given "1" is useless, we can rotate the semi-circle toward one side of the quarter circle , and obtain the result 1/2 immediately.
Grt question
@4:27 - 'If two circles are tangent'; should be 'If two circles are tangential'. Otherwise, a great problem to share. Many thanks.
You take the larger circle, and you look at what happens as its radius approaches infinity. If you do so you will start to notice that the inner circle basically has the radius of the larger circle as its diameter as the value of 1 becomes negligible!
That's how I solved it at least!
wow indeed
Given the way that problem is set up (it's underdefined) I am going to assume that the ratio is constant, regardless of the angle the semicircle sits in the quarter.
I did:
AB is the diameter of the semicircle, therefore AB/2 is its radius.
Given the "tangent collinear" rule, AB is also the radius of the quarter circle.
So the area of the quarter circle is 1/4 * pi * AB^2.
Area of semicircle is 1/2 * pi * (AB/2)^2, which simplifies into 1/8 * pi * AB^2.
So the ratio is 1/2.
That number 1 is pointless in the problem because there are no other numbers to put it to scale, so it's the same with the diameter of the half circle of it were tilded anywhere against the side of the quarter circle. So tilt it all the way up or down so that it's diameter is that of the quarter circles radius. So ( pi*r^2 / 2) / (pi*(2r)^2 / 4 ) = 1/2.
This is the first problem I solved on my own
You don't have to use all that geometry to get the answer.
The radius of the quarter circle is arbitrary so the distance of 1 could be any proportion of the bottom side. If you imagine the bottom side is like a million or whatever, the semicircle would basically just fit against the left side of the shape. At that point you know the radius is half
Technically 1/2 so long as the radius of the large circle is greater than or equal to 1.
Let P be the point where the half circle touches the bigger quarter circle, O the center of the big circle, and Q, R the two intersections of the half circle with the axes. QP and PR are orthogonal because of the property of the circle. Hence OPQR is a rectangle and QR equals the radius of the big circle. But QR is also the diameter of the small circle. Hence the radius of the small circle is half the bigger radius, hence a quarter of the big circle is twice a half of the small circle.
Why you took the ratio semi over quarter. Why not quarter by semi ?
✨Magic!✨
Why did you go through all that trouble at the beginning when we already know by definition of the circle that O is the same distance away from A and B?
(When your excitement doesn't match up with the teacher's)
the percentage does not depend from l2, see line 20:
10 print "mind your decisions-can you solve this interesting geometry challenge"
20 l1=1:sw=l1/10:l=sw:xm=l1/2:l2=2.2:ym=l2/2:goto 70
30 ri=sqr(l1^2/4+l2^2/4):xs=l/ri*l1/2:ys=l/ri*l2/2
40 ra=sqr(xs^2+ys^2):
50 dgu1=xm^2/ra^2:dgu2=ym^2/ra^2:dgu3=(ri-ra)^2/ra^2
60 dg=dgu1+dgu2-dgu3:return
70 l=sw:gosub 30
80 dg1=dg:lu1=l:l=l+sw:if l>100*l1 then 120
90 lu2=l:gosub 30:if dg1*dg>0 then 80
100 l=(lu1+lu2)/2:gosub 30:if dg1*dg>0 then lu1=l else lu2=l
110 if abs(dg)>1E-10 then 100
120 print l;"%";ri;"%";:ra=sqr(xs^2+ys^2):print ra
130 mass=850/ra:move 0,0:move ra*mass,0:plot 165, 0,l2*mass
140 gcol 9:move l1/2*mass,l2/2*mass:move l1*mass,0:plot 181,0,l2*mass
150 aa=pi*ra^2/4:ai=pi*ri^2/2:perca=ai/aa*100:print "die rote flaeche=";perca;"% des viertelkreises"
160 print "run in bbc basic sdl and hit ctrl tab to copy from the results window"
mind your decisions-can you solve this interesting geometry challenge
2.41660919%1.2083046%2.41660919
die rote flaeche=50% des viertelkreises
run in bbc basic sdl and hit ctrl tab to copy from the results window
>
Not even gonna bother pausing the video. I know my limitations.
Brillante
Cuemath❤❤
Regarding not using the "1" dimension: it should be obvious that the "1" is not relevant because there is no other dimension given, & because problem is requesting a ratio which is dimension independent. So either there is not enough information to solve problem because another dimension is missing (not true in this case), or that the answer is "independent" of the dimension (this case).
You have presented other problems where the initial figure is what I call "dimensionless". The most notable of these pertain to cones (the cone glass "how far is it filled" problem (ua-cam.com/video/g88VAJccotI/v-deo.html) & the filling the cone problem (ua-cam.com/video/0vAeXhtSR6U/v-deo.html). For the "how far is it filled" problem, ratios are sought so no dimension is needed; volume is proportional to linear dimension cubed. (Like this problem, you could confuse things by adding arbitrary dimension, such as diameter of the top or depth.) For the 5cm/10cm/15cm problem, until the cone is "cut" by the waterline, it has no dimension. Changing the dimension is equivalent to scaling the whole figure up or down, so volume scales by the cube of the linear dimension.
First look was oh it is fixed and independent from 1, oh half.
There's a slightly easier way to prove that C is on the reconstructed smaller circle by using the central angle theorem.
I found your proof that the two circumcentre triangles were similar was weak.
Hey mind your decisions, i have a question. Please solve it. This is my question:-
a^(a+1)=(a+1)^a solve for a
Haselbauer-Dickheiser test, question 14 baby
I solved it by eyeballing it and it looked like 50% ? So that’s what I went with . Took me only about 10 seconds .
How can you prove, that the other half goes through C?
Bout half 😊
Is there any solution with use of that given one?
It's an irrelevant piece of information. The given measurement of 1 unit, could've been 10 units, and the answer would be exactly the same.
Since it's the only measurement given, it tells you nothing about the relative detail of the shape. You need at least two measurements for the measurements to govern anything other than the scaling of a shape.
@@carultch so it is not. Thank you.
LOL, at first I thought the thumbnail contained not enough information, but after thinking about it and solving it, it turns out that the image in the thumbnail contained _redundant_ information! (the bottom side of the right triangle doesn't need to be 1.)
Solution: 50%
Consider the tangential intersection point of the semi-circle arc of the red half disk, and the quarter-circle arc of the quarter disk. The perpendicular line through that point passes through the center of the semi-circle arc of the red half disk, but also through the center of the quarter-circle arc of the quarter disk. Since the center of the semi-circle arc of the red half disk is also the midpoint of the hypotenuse of the right triangle, it's easily seen that the radius r of the semi-circle arc of the red half disk is half the radius R of the quarter-circle arc of the quarter disk. So the area of the red half disk is (1/2)πr² , and the area of the quarter disk is (1/4)πR² , and with R = 2r, their ratio becomes
[(1/2)πr²]/[(1/4)πR²] = [(1/2)πr²]/[(1/4)π(2r)²] = [(1/2)πr²]/[πr²] = 1/2 = 50%
Is it manim?
Couldn’t you just more generally prove for the first part that since AB would have to be the diameter of the circle, thus the midpoint of the line would have to be the midpoint of the circles diameter and thus the circumcenter?
I don’t mean this to nitpick I’m genuinely curious
Solved!! Before Solution. hurray
My answer that I got to was 19.625 over 4.90625 equaling 4. About to see if it’s right
I was wrong
As a thought experiment this is fun, but what is a practical use for being able to solve this problem?
It's an allied property that, in the case of a circle rolling inside a circle of twice the diameter, ALL points on the circumference of the smaller circle move in straight lines. We can see this with the two "corner" points moving on the quadrant radii. This was used as a practical mechanism in the early days of steam engines, for converting linear to circular motion.
What relevance does the measurement of 1 have in the statement of the question when the question itself asks for the fraction - whatever the measurement it will always cancel out.
No relevance. Since no other dimension is given, it does nothing to establish the proportions of the figure, so he could've given any dimension there, or no dimension at all, and the solution would still be the same.
@carultch and he does explain that, I wrote that before watching, just saw thumbnail - it JUMPED out at me, wasn't thinking it was an intended red herring 😉.
@@fifiwoof1969 It's a bit of a red herring. If it were my choice, I'd assign the outer circle's radius as 1 by default, rather than an intermediate measurement like this.
It's a common strategy to assign unit length where convenient, since it simplifies the number of variables you have to handle.
@@carultch why bother when it's LITERALLY a red herring (unless to act as a red herring, great way to test students on an exam paper - separates the Bs from the A+s) - can also leave it out and have no measurements at all. Potential for extra credit by saying measurement is red herring in answer on exam.
@@fifiwoof1969 It's not like you can go wrong by using it. A true red herring would be a piece of information that leads you to the complete wrong answer.
If the big circle is 1/4 of a circle and the smaller circle is 1/2, couldn't we just easily have said that the ratio is 2:4 = 1:2 = 1/2? As the proof said the answer is true for all semicircles in a quadricircle.
Maths is very beautiful.
1/2
eu amo invariantes
or, just:
- rotate the ABC triangle 180' around D
- observe that ATB is a right-angle (flipped C) and that this is true for _any_ distance BC (between 0 and r).
- set BC to 0 (AB is now vertical)
- done
Hi :)
Why didn't you just show that animation in the beginning? It's pretty obvious when the radius of the quarter circle and the diameter of the semicircle are equal that the ratio is 1:2.
I have a question.... In solution to the original problem, How do we know that CDT is a straight line???
It is a diagonal of rectangle ABCT. Imagine you can always do that - put rectangle in every circle. Then you can also slice this rectangle into 2 identical triangles which can help you to see that. This was actually my starting point.
@@michal88gno what I meant is .... How can you prove that this diagonal will surely pass through the point D.
But yeah, upon thinking about your idea- rectangle - this is clear now. Thanks!
why not just make a rectangle where AB is the diagonal?
=> AB = CD and O is the midpoint of both => AO = BO = CO = D0...
ABCD are lying on the circle with Center O...
It always puzzles me (more than the puzzle itself) why do people arbitrarily pick and include irrelevant information, such as CB=1 in this case.
If you know that the bottom length of the triangle is 1, the calculations (really, the logical reasoning) to get the radius of the quarter circle and the radius of the semicircle can be easily expressed using the cosine rule. It's an alternate solution. It's another way of getting the same ratios.
@@benisrood If none of the lengths are given, you can always assume one of them is 1 if needed for your solution. No harm in doing that, especially when we're looking for a ratio, then scale is irrelevant, you can use 1, 10 or 42, doesn't matter.
I'm just imagining. Suppose the given length is -1 instead of the boring 1?
Then we will have a quarter circle inside a semicircle with the common tangent point and the two circle centres being collinear
@@rterentius not really. Technically speaking, there's no such thing as negative length in euclidean geometry.
¼ circle radius = r
½ circle radius = 2r
Ratio: 1/2
This was light
Took me 5 seconds
I feel so SKibidi sigma to finally be so early in ohio
Giving AC=1 is a red herring which has no place in a statement of a problem as it may unfairly send people on the wrong track. Also reading some of the comments here some posters offer solutions of special cases which only prove the result for that special case and is not a solution of the problem as stated. Because it is a solution in that case does not imply it is a solution in the general case . Unfortunately fortunately this is a widespread misconception.
Apologies . Above should read BC=1 ,not AC=I.
No, I can't.
There's a gap in the proof of the first theorem at 2:00 minutes in. Point O is introduced as the centre of the circle and the next 8 minutes are used to show that OA=OB=OC. Well, yes, BUT, the gap in the proof is the assumption that O lies on the line AB. In fact it does do so, but you can't just assume that without proof.
Worse still, all the geometry that follows using similar triangles up to time 10.10 is a complete waste of time, because if O is the centre of the circle and it lies on the line AB (assumed, but not proven), then OA, OB and OC are all radii of the circle and are therefore all equal. You don't need to mess about with similar triangles to show that. But you do need to show that O lies on AB and that wasn't done.
Other than that it's an excellent video.
What is the point of giving that distance = 1 ?
Useless for the problem.
You divided by r, so r can't be zero.
I figured out the answer using cosine theorem. Below, r is the radius of the inner semicircle and R is the radius of the outer quarter circle. Other characters are as in the video.
1. Denote the angle(ABC) as x. From the triangle ABC we get that cosx = 1/2r.
2. Figure out that CD and DT are on the same line, as both DT and CT form the same angle with the tangent line and have a common point. Then, CD=R-r. DT=DB=r.
3. Consider the triangle CDB. From the cosine theorem:
(R-r)² = 1² + r² - 2*r*1*cosx =1 + r² - 2r*1/2r = 1+r² -1 = r²
R² - 2rR + r² = r²
R² = 2rR
r/R = 1/2
The ratio of areas is:
0.5*pi*r² / 0.25 pi*R² = 2 * (r/R)² = 2*1/4 = 1/2
Interestingly... If we change the "1" for any arbitrary, say, "a", then
cosx = a/2r
And then (R-r)² = a² + r² - 2ar*cosx = a² + r² - 2ar * a/2r = a² + r² - a² = r²
And then it's the same ratio if r/R = 1/2 that comes out. So, the length of CB can really be any non-negative number, and it won't change the solution indeed.
Aren’t you supposed to subtract the red area to get the actual shaded area?
No need, the shaded area is the semicircle with radius of r, that's all.
No, the question is what is the ratio. It is always 1/2 regardless of dimensions.