Very Nice Geometry Problem | Circle and semicircle inside a quarter circle | 2 Methods

MY OTHER CHANNELS
••••••••••••••••••••••••••••••••
Calculus Booster : www.youtube.com/@CalculusBoos...
Math Hunter : www.youtube.com/@mathshunter/...
--------------------------------------------------------------------------------
Join the channel to become a member
/ @mathbooster

[@piyathilakamuthunayaka7291]

Nice problem. It is a good mind exercise .Thanks your channel.

[@Irtsak]

Construct PQ and let N the point of contact of the small circle and the small semicircle. Construct OP , ON.
Triangle OPQ is obviously isosceles cause ON is height and median at the same time.
So ON is angle bisector => *∠QON= ∠NOP* (1)
ON,OM are tangent sections => *∠NOP= ∠POM* (2)
(1),(2)=> ∠QON = ∠NOP = ∠POM = 30° (because their sum is ∠MOQ=90°)
In orthogonal triangle ONQ : ∠QON=30° => OQ=2QN = 2•2=> *OQ=4*
Radius of the quadrant is R=4+2=6 => *R=6*
So you can estimate the shaded area , as Math Booster did.

[@prime423]

QN is the radius. No algebra needed!!

[@quigonkenny]

But we don't know that at the time. You can't assume that just because PN appears to line up with the circumference of the semicircle that QN is the radius. The figure may not be drawn to scale.

[@shivachaturvedhi8840]

full circle enter pases through co-tangent, so O to full circlecneter = r-2
also, vertical axis , r-2=2+2cos(t) + 2sin(90-t)=2+4cost
in RIght triangle (O to center of full circle ) , sies are
r-2=2+4cost, 4sint, 2
so, divide all by 2 and use pythogorous
(1+2cost)^2-(2sint)^2=1^2
=> 1+4(cos^2t-sin^2t)+2*2cost=1
=> 4cos2t=-4cost
=> t=60
=> r-2=2+4cost=2+2=4
=> r=6
=> blue area = full area = 1.5 area of circle = pi/4 * 36 - 1.5 * pi*4 = 9pi-6pi=3pi

[@santiagoarosam430]

PO y SO son ejes de simetría y los ángulos MOP, POS y SOQ son iguales y trisecan el cuadrante. Los ángulos en O son 30+30+30=90°. Si PM=SQ=2 entonces PO=QO=2*2=4. Radio del cuadrante =PO+2=4+2=6. Área rosa =π[(6^2/4)-(2^2*3/2)]=3π.
Gracias y saludos.

[@michaeldoerr5810]

Are there any other methods that do not make use of symmetry? I thought that the first method was much simplifed.

[@AbdulJaleel-rd5ul]

I have a doubt. As per my understanding no need of this calculation.if N is end of diameter of semicircle,we can easily get diameter of semicircle plus radius of full circle (4+2=6) is 6.but how can we assume vertical from P is end of diameter of semicircle?

[@AbdulJaleel-rd5ul]

Expert people please explain.

[@quigonkenny]

If you're talking about during the first method, there's no particular significance to N being one end point of the semicircle diameter. It just lines up that way because R = 3r (where r is the radius of the semicircle and circle). The significance of N is that it is the point on OT where PN is perpendicular to OT, so that ON = PM = 2. And since QT = 2 and R is at that point unknown, it lets us define QN as R-4.
Once we determine that R = 6, we can see that QN = 2 as well, which means that N is indeed in the position of the end of the semicircle diameter, even if that's not why we labeled N or why we drew PN. Like I said, it just happens to line up that way.

[@MarieAnne.]

@@AbdulJaleel-rd5ul If you check video at 3:35, you'll find that he has equation:
ON + QN + QT = R
Then he substitutes known values:
2 + QN + 2 = R
Now if he'd assumed that N was at the end of diameter, then he would have just plugged that into the equation above to get R = 6 straight away. But at this time we cannot make such an assumption, and neither did the solution shown in video. So there is no problem.

[@manjunathaba9747]

Nice

[@ulrichgraf2094]

Method 2: the reason why the common tangents intersect at point O is missing. Symmetry is not sufficant to conclude this.

[@FreeFire-hj6sm]

9.42

[@johnbrennan3372]

So since R= 6, this implies that the point N is an end point on the diameter of the half circle

[@MarieAnne.]

True, but until we find that R = 6 we cannot make that assumption.

[@danilopapa3853]

Very simple problem, with non need for elaborate graphs, complicated calculations, rigonometric formulae and :obnoxious stuff
First of all, we extend the figure by repeating four times the pattern shown at the beginning of the video.
So we get an outer circle with six small circles inscribed in it,
The six circles are mutually tangent (and also tangent with the larger circle).
The centers of the internal little circles form a regular hexagon (by the way, the same is true for the tangency points with the outer circle).
So, the angle between the centers of two adjacent small circles is 60°.
Now, join the center of the outer circle with the center of the little circle we can see on the right in the figure at the beginning of the video.
We obtain a 90°-60°-30° triangle , whose minor side (the vertical one) has a length of 2.
Such an object is half of an equilateral triangle, whose vertical half sides is 2.
So, the sides of this equilateral triangle must be 4 in length.
One of the sides is the hypotenuse of the above mentioned 90°-60°-30° triangle
We deduce also the radius of the major circle, wich is 4+2=6.
The continuation is very easy:
- the area of the outer circle is Pi6*6 = 36Pi
- A quarter of them has area = 9Pi
- the area of a single little circle is Pi2*2= 4Pi
- the area of a circle and half in the beginning of the video is 4Pi+2Pi=6Pi
So, the shaded area is 9Pi-6Pi=3Pi

[@quigonkenny]

Label the following points:
Center of the quarter circle: O
Center of the semicircle: P
Center of the circle: S
Point of tangency between circle and baseline: T
Point of tangency between circle and quarter circle circumference: V
Point of tangency between circle and semicircle: L
Vertices of the quarter circle: A (top), B (right)
Let the radius of the quarter circle be R. Draw PS. It will be length 2(2) = 4, or the sum of the radii of P and S. As points of tangency between circles are colinear with their centers, PS passes through L. Drop a perpendicular from S to OA at M. OM = 2, as ST = 2 and SM is parallel to OT. OP = R-2 as PA = 2 and OA is a radius of the quarter circle. Thus PM = R-4.
Triangle ∆SMP:
PM² + SM² = PS²
(R-4)² + SM² = 4²
R² - 8R + 16 + SM² = 16
SM² = 8R - R²
Draw OV. As O and S are colinear with their point of tangency, OV passes through S. Thus OS = R-2. As ST and OM are parallel and both perpendicular to SM and OT, SM = OT.
Triangle ∆OTS:
ST² + OT² = OS²
2² + (8R-R²) = (R-2)²
4 + 8R - R² = R² - 4R + 4
2R² - 12R = 0
R² - 6R = 0
R(R-6) = 0
R = 0 ❌ | R = 6
The Shaded Area is equal to the area of quarter circle O minus the areas of semicircle P and circle S.
Shaded area:
A = πR²/4 - πr²/2 - πr²
A = π(6²)/4 - π2²/2 - π2²
A = 9π - 2π - 4π = 3π