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

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[piyathilakamuthunayaka]

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

[AbdulJaleel-rdul]

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?

[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.

[santiagoarosam]

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.

[michaeldoerr]

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

[shivachaturvedhi]

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

[ulrichgraf]

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

[johnbrennan]

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

[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

[danilopapa]

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

[RondoCarletti]

Minute 3:30: You can not know that ON+QN+QT = R. Why? You have not prove it in this moment. That's a mistake.