What is the area? A nice problem from Singapore

Can you solve this challenging problem from the Mathematical Olympiad of Singapore? Thanks to Truong from Vietnam for the suggestion!

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Press

[thejoeyjason]

I love when he says "and THAT's the answer". I love how he says THAT with conviction and energy. I look forward to it.

[PuzzleAdda]

Riddle : You go at red, but stop at green. What am I?

[rishianand]

Finally solved the first problem of this channel correctly..

[vinisherdaotaku]

Once you get to this point 1:59, you can also just multiply this shape by 2, and then subtract the area of a full square (two of the triangles). I think this is the same exact thing as the video, but is more nicer on the numbers.

[sgngozlerimialyorbebegim]

Although he explains the questions in English, I understand because the language of mathematics is the same everywhere. That's why math is universal

[RezaHadi_A]

2:49 " And that's the answer" i like how he say it

[LostWonderer]

Thanks for sometimes posting some more approachable puzzles as this one. It encourages me to try the harder ones.

[turbflat]

I'm from Singapore, and I did it in a fairly similar way. The way I concluded that triangle was right was by looking at the two congruent triangles that make it up - which also turn out to be right isosceles triangles themselves based on the symmetry of the inner circle. Then I just used the formula for segment area (with radian measure) and doubled it.

[kanizfatima]

This same question was asked in NATA on 10th April 2021 in India
Thank you sir

[lolxdmekaisemaanlu]

2:45 my teacher will dedect 1 marks for not writing "sq. unit"

[arpansen]

Its beautiful to hear your voice after a long time presh

[-igor-]

To make the answer correct, one should not forget to add the units (units of length squared).

[billy.]

Many students consider plain old geometry is the most difficult math topic in secondary school. Geometry proofs are tough nuts to crack.

[WoodyC-fvhz]

Hi Presh,
14.27 . Due to the symmetrie appointed by the inscribed circle, it's obvious that the "fooball" fits or lines up exactly 4 times inside the circumference of the larger circle. This results in an angle of 90° with respect to the the centre of the large circle and both "triple intersections".
Applying 2*r*sin(theta/2), I get 7.071 for the secant, which also denotes the longer symmetrical (here horizontal) axis of the "football". It's height is given by Pythagoras at 3.5355. Next I compute my Quater-circle-sector minus "triangle within" to get the segment resembling "half the football" and finally double the value ... like so: (19.6349 - 7.0711 * 3.5355 /2) * 2 = 14.2699.

[nesjohguei]

Excellent teacher grettings from Venezuela

[MottBotMinecraft]

I took this to another level... I found equations for the circle and the two semi circles by centering the top semi circle and the full circle around an unknown variable k and solving for it to find the centers of both. I then calculated the intersection points of the two semi circles and used these as limits of integration for the bottom semicircle minus the top semi circle. This was tedious but I guess it worked!

[quigonkenny]

Inscribe a diamond (a square rotated 45°) inside the central circle, such that the top and bottom vertices are at the centers of the two semicircles and the left and right vertices are at the points of intersection btween the central circle and the two semicircles (the ends of the blue area). We can see that the sides of this diamond are radii of the two semicircles, and thus length 5, and that despite not being given a height for the surrounding rectangle, it must be of a set height, as if it were any shorter or taller, the vertices of the blue area would not be on the circumference of the central circle, but rather outside or inside of it.

Draw a line between the two points of the blue area. This will be a diagonal of the diamond, and create two isosceles right triangles between the centers of the semicircles. This diagonal also creates two circular segments, one with each semicircle, of arc length 90°, being formed from corners of a diamond (square).

The area we seek is the area of these two circular segments, each of which is equal to the area of an enclosing sector minus the area of one of the isosceles right triangles:

A = 2(90/360)π(5²) - 2(5²/2)
A = 25π/2 - 25 = 25(π/2 - 1)
A ≈ 14.270

[henry]

Great puzzle thanks. Please include geometry problems from time to time.

[zackwhite]

I managed to solve this problem in a different way, consider the circles graphs and the unknown radius of the circle r to obtain: y = sqrt(25-x^2), y = -sqrt(25-x^2)+2r, y = +and-sqrt(r^2-x^2)+r. Solving these equations you obtain x = (+and-sqrt(5))/2, y = r = 5/2sqrt(2). Then you can do a definite integral between (+and-sqrt(5))/2 of 2sqrt(25-x^2)-5sqrt(2), And finally, provided your calculator is in radians, you reach the correct answer.

[pkmkb]

This or a similar question was in my 10th coaching material in india