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# KCC's Quizzes AQQ258 about a partial eclipse Sun-Moon

1. Quote of the week: "People say nothing is impossible; but I do nothing every day!" - Anonymous

2. New quiz: AQQ258 about a partial eclipse Sun-Moon

Some of us (in Dallas) had probably the chance to have observed a total eclipse of the Sun last week.

To celebrate that event in our quizzes, we propose the following related challenge:

The above pictures describe a partial eclipse between the moon (circle B) and the sun (circle R).

At a certain instant, the eclipse is such that points MNO form a quarter area of circle R; meaning arc MN is a quarter circle).

At that moment, point O is the center of R and the segment OM measures 4 cm.

Question:

What is the area of the portion of blue circle B that is situated outside circle R?

Good luck! And try to be among the firsts!

Kuo-Chang

P.S. Don't hesitate to share those weekly quizzes in EZ to colleagues or friends!

• Let P then center of the Moon.
P is the bisector of angle MON, by symmetry, so MOP is 45 degree.
Triangle MOP has two sides equal, to R, radius of the Moon, so angle OMP is also 45 degree.
So, by the triangle OPM, angle OPM is 90 degree.
Diagonal of OMP = 4,  so by Pythagoras, R = square root of 8  ( R2+R2 = 42), or R2 = 8.

The sector of the Moon to the left of OM is A = ¼πR2 – ½R2
(that is the sector radius R of 90 degree  less the area of the right triangle OPM) = 2π - 4

The portion of the Moon hiding the Sun is the sector of radius 4 of 90 degree plus twice the previously computed Area:  As = ¼π42 +2(2π - 4) = 8π-8.
The total area of the Moon is πR2 = 8π, thus we were hunting for an area of 8. (A little bit less than one third of the full Moon).

My answer is 8 square cm.

• Bingo  ! Correct answer with crystal clear development, congratulations! And you are the first!

SInce the circumf. angle is PI/2, the segment connecting M and N is a diameter of B. You can then calculate radius of B by using Pitagora, r= 2*sqrt (2).Let's call C the portion of circle B we're looking for. This is equal to half of the area of B (you can easily calculate that as you know radius r) minus the area A of the circular segment identified by M, N and the diameter connecting these two points. The area A is equal to a quarter of the area of R minus the area of the triangle OMN. With some calculation you get C= 8cm2

• The are of the portion that is outside will be 8

If sun center is o meaning mo is radius that is 4 and on will also become 4, area of triangle will become 1/2*4*4 that is 8 and quarter area of sun is pi*4*4/4 that is 4*pi so the area of arc mn excluding traingle will be 4"pi -8 and area of half circle moon will be 4*pi with radius 2 sqrt(2) and so 4*pi-(4*pi-8)

And hence the arc that is outside with have area of 8

• Without diving into proving all the geometric theorems, using Thale's Theorem, one can determine that line segment MN goes through the center of circle B, so if we can determine the distance of MN, we know the area of circle B.

This is determined to be MN - sqrt(4^2 + 4^2) = sqrt(32) = 5.66

If the diameter of the circle B is sqrt(32), then the area of B is B(area) = pi * (1/2 * sqrt(32))^2 = pi * 1/4 * 32 = 8π

The area of the arc MON = 1/4 * (pi * 4^2) = 4π

All that leaves is the area of the two blue arcs inside circle R (but not inside arc MON).  This can be found by calculating the area of B minus the square created by flipping angle MON onto B (forming a square) and using 1/2 of that.

Area of blue arcs inside R = 1/2 * (B(area) - area of MON square) = 1.2 * (8pi - 4^2) = 4π - 8

Therefore the area of the blue circle outside circle R =

ANSWER = 8π - 4π - (4π - 8)  = 4π - 4π + 8 = 8 cm²

Cheers!

Brian

• The area of the portion of blue circle B that is situated outside circle R is 8 sq cm.

Complete the square inside the circle.  The circle B radius is half the diagonal of the square or sqrt(8).

Each segment of B with cord length of 4cm has an area of (2PI-4) sq cm.

The quarter circle area is 4PI sq cm

The circle B area is 8PI

Area of B outside is equal to 8PI - 4PI - 2*(2PI-4) = 8 sq cm.

• The origin of circle B is half distance of line OM and ON, therefore 2cm from O.  Radius of circle B is a^2+b^2=r^2 = sqrt(8).

quarter of the circle R is 4^2*pi/4=12.566

The areas of arc MO and NO can be found by taking 1/2 area of circle B minus area of triangle MNO.  Therefore = 4.566

The total Intersect area of circle R and B would be  12.566+4.566= 17.133cm^2    <------HERE I REALIZE I CALCULATED THE INTERSECTED AREA

So the Outer Area would be Area circle B minus the Inside Intersect which is 8cm^2

• I think this might be a crude solution, feel free to help me improve my logic:

1. We can form a square on circle B from the fact that OM and ON are same valued sides, then from this subtract it to the area of B, which circumscribes this square. This leaves us with 4 leftover pieces of the circle, of which we need only 2 :

1.a. radius sqaured of B can be acquired from this formed square : 4^2 = 2r^2 -> r^2 = 4^2 / 2

1.b. multiply 1.a. by pi to get area of B : pi*(r^2) = 8pi.

1.c. Subtract 1.b. to 4^2 and divide by 2 to get 2 leftover circle pieces : 9.133.

2.Subtract the area of the square to the quarter Circle made by circle R:

2.a 4^2 - (pi*(4^2)/4) = 3.434

3. Add 2.a and 1.c to get the final result:

A = 3.434 + 9.133

A = 4pi.

• I realized I made a small calculation mistake in 1.c. this should yield 9.133/2 , which means A equals 8.

• Correct Gaetano! Congratulation!