Very difficult for most students
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- Опубликовано: 24 окт 2024
- A square with side length 1 has a circle in passing through one corner that is tangent to the opposite two sides. What is the area contained by the square and circle not including the overlapping area between the square and circle?
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i had a completely different approach to start. if you move the blue parts that are outside of the square inside, you end up with a square in a square. the math ends up being about the same from here, since my inner square is the same size as 2 purple triangles
I also complete this problem in this way
it's only a slight variation, let's be real.
Nice solution.
aaah very nice observation
That illustrates well why there's no "pi" in the answer.
Every time I see a 3, 1, and 4 together, my mind goes: **Pi**
Your subconscious is programmed
Due to repeated contemplation of it that
"3.14...= π"
Ans is 0.π
@@Priyanshu_Shukla we can write it as π/10
@@tanmaywatnani1874 Yeah, Ofcourse !!
@@Priyanshu_Shukla k
I went along the same lines, but failed to note that 2 semicircle and the circle cancelling out early on..... Otherwise I used the same technique - Presh you've taught me well over the years! When I saw terms cancelling out at the end I was wondering whether I could have done it more efficiently! I was stoked to nail an exact answer in any case. Your solution was more elegant I must say!
Comment made 6 days ago... I'm still trying to figure out th@ conundrum ¿🤔😉😕😏
☆The Video only just released 7 minutes ago eh lol 😆
@@steveclark2205 He's a patreon, so he get estou access to the videos. It takes a little while for Presh to make a video public.
@Nilo Neto lol appreciate that 😉🙄🤔🤣🆒️
6 days ago?
To cancel out the circle, you have to assume the semicircle passes through the circle's exact center point; an assumption you have no way of coming too unless you eyeball it or already have the trigonometric proof of that.
i used to watch the problems on this channel and shudder two years ago when i first started watching, but now i feel way more confident and can even get the answers to many of them! thank you for making such wonderful videos!
So we only need to practice?
I moved the outer blue shapes inside the circle, this made a smaller black square within the bigger square. Let the black square have length 'x', then the area of the blue shapes is 1-x^2. I found the radius in terms of x then used an equation I dervied before that relates the length of a chord based on a distance along the diameter from the edge of the circle to that chord. I substituted in my radius and solved a quadtratic equation giving x=2(sqrt(2)-1), substituted x back into 1-x^2 and got the answer. First time I ever got one of these problems correct. 😁😍
Moving the outer blue parts like that is a clever way of making the blue area as a whole easier to understand! You managed to skip having to realize and prove the part about semicircle.
You can simplify the setup a lot by rotating the semicircle 180 degrees to show a square with a side length of one with a square cut out of it of a side length of... well that's the part you had to calculate, whice is more difficult. Doesn't help with the calculation, but does make the setup simpler.
It’s a great technique with area sums to move things around like a puzzle!
If you rotate the circle 180 degrees about its center, the two shaded segments will nest into the shaded portion of the square. You are then left with a shaded area that is equal to the area of the large square minus the area of the newly formed small square: (1 m)² - ((2 - √2 m)(√2))² = 1 m² - (12 - 8√2 m²) = 8√2 - 11 m²
That is the way I found
I figured the big and small squares, I just couldn't come up with the side of the smaller one
How are u sure it will nest? (It does but) It is not given that the arc are same in length yet. We need a step proving it first.
@@Errors404 Yes, because if I draw vertical and horizontal secant lines that pass through the circle and square's points of intersection, the shape I form is a square inscribed within a circle. The four segments of any circle that are bound by chords of an inscribed square are congruent because the arcs are bound by congruent central angles. I still have to find the side length of the new square by calculating the radius of the circle using 45-45-90 triangles, but the calculation at the end is easier. I don't have to subtract the overlapping space twice; I just need to find the area of the smaller square and subtract it from 1 m².
@Errors404 it should because the chord that points of interaction make is the diameter (the triangle it makes is right triangle)
Expand the big square to the left and down. We will have two squares, one circumscribed by the circle and the other inscribed in the circle.
We mark with x half of the side of the inscribed square. We mark with y the small piece that makes the difference between the small and the large square.
We will have:
x+y=R
x=R*√2/2
This results in y=R*(2-√2)/2
But 2x+y is the side of the big square, which is 1. So 2R√2/2 + R(2-√2)/2 = 1
It results that R= 2-√2.
From here it's easy.
Finding this channel has been so helpful in understanding maths.
I drew the square and circle in solidworks, made sure it was the proper shape, extruded the proper shaded regions up by 1 meter so its volume = its area. Used evaluate and it says 0.31 m^3/1m = 0.31 m^2. I probably could have more accuracy if I changed a setting.
I reasoned that the middel of the circle would be on coordinate (1-r),(1-r) in respect to the bottom left corner of the square. this means there is now a right triangle in the bottom left of the square with sides 1-r, 1-r and long edge r. From this i used pythagoras to form an equation that solved to r = 2 - sqrt(2)
* middle
I came up with a little bit more complicated solution but it still works.
I drew 2 triangles both inside the square.
The first one goes from the top right corner down towards bottom right where the square intersects with the circle. That small piece i called X.
The other one is the diagonal of the square which sides are then as follows
√2, 1-x, √(1+x²)
Then the areas of those two triangles have to be 1/2. And we solve for x.
Using that we get the r√2 and solve as the video shows.
It unnecessarily complicated because I forgot about the right angle theorem in the circle.
Some have asked why the purple triangle is assumed to be a 45-45-90 triangle. He did not mention why in the video, but we can determine this by looking at the construction of the original figure: Let 2R be equal to the diameter of the circle. The height (sagittae) of the two circular segments in the original figure would then be 2R - 1 (the diameter of the circle minus the side length of the square). Since the two circular segments have an equal radius (they are segments of the same circle) and equal heights (2R - 1), then their arc lengths (2πr/2/2 = 2πr/4), central angles (180/2° = 90°), and chord lengths (we'll call that C) must also be equal. We now know that the purple triangle is a 90° right triangle (given) with two sides equal to C and a hypotenuse equal to 2R (the circle's diameter by rule of the inscribed 90° angle that he mentions). It is therefore a 45-45-90 triangle. When he splits that triangle, we now have two purple triangles, each with dimensions R-R-C (we know they are isosceles). We have already determined one 45° interior angle of each triangle, so they are also congruent 45-45-90 triangles because of properties of isosceles triangles (the angles opposite to the equal sides of an isosceles triangle are also equal) and the Angle Sum Theorem (the sum of interior angles of a triangle is 180°).
thanks
That seems like over complicating it. Both side lengths of the legs of the purple triangle are equal, therefore it must be a 45-45-90 triangle
You don’t have to do all those calculations to find the chord length of the blue areas in the circle, as all that is needed is observable through simple geometry. If the circle meets two sides of the square tangentially, then it must meet the opposite sides at an equal position.
@@SilverKiwi_YT That's pretty much what I said, but with the relevant properties to show why. He never mentions why the legs of the purple triangle are the same length, nor why the two smaller purple triangles are congruent. If we can't state why the chords are equal, then we are essentially making an assumption that they are. And anything derived from that assumption, while correct, has not been proven. Showing that the segments are congruent is the easiest way to show that the chords have equal length, which is what proves that the purple triangle is a right isosceles triangle, and that the two smaller triangles are congruent by means of SSA.
@@cdmcfall but you don’t have to prove that to solve the question. I was under the assumption that you were just trying to help people who don’t understand, and I’m just saying that there is a much simpler way of explaining it.
@MindYourDecisions, let's do the same problem, but instead of a square, use an oblong rectangle with a base of 3 m and a height of 4 m. The circle is still tangent to the top and right sides of the rectangle and intersects the rectangle's lower left corner. The problem is similar, but I think it's slightly more challenging because we eliminate the 45-45-90 triangles. It's pretty neat to see a general case for all rectangles, and how it changes the angles in the purple right triangle you constructed at (1:04). Plotted on a coordinate plane, the circle is tangent to lines x = 3 and y = 4, and passes through the origin. It's also pretty neat to see the limitations on the size of the rectangle, which must have a base:height ratio between 2:1 and 1:2 (the circle is bisected at those two extremes).
Yes my thoughts exactly. Nothing explicitly says the diagonal line is going to the center of the circle
You can also find the radius in this way, r + rx(sqrt2)=sqrt2, because the diagonal inside the square is sqrt2, simplifying gives you r= sqrt2/sqrt2+1, and simplifying further gives you 2 - sqrt2
That's almost π/10. Not quite, but it's only off by 0.14%! I gotta somehow squeeze in 10(8√2 - 11) as an approximation for pi somehow now.
Very good video, nothing unnecessary and everything necessary explained in simple terms. Love the notation of adding and subtracting areas!
You can do a lot of geometry to simplify the problem, but the hardest part is the algebra at the end to calculate the RADIUS of that darned circle!
0.313708 approx. Thanks Presh for this beauty. Two values only are needed for the solution. 1) the radius r and 2) the sagitta s (height of segment).
1) Construct a diagonal from bottom left hand corner to top right hand apex where the length is r+r*sqrt2 and apply pythagoras yielding (r+r*sqrt2)^2 = 2 and solve for r. r is 0.585786 and the problem is actually almost solved.
2) For completeness "elegantly" slice off both overhanging segments, cut and paste, and slide them into the corresponding circular alcoves top and right hand side, resulting in a smaller square sitting inside a larger square with and L-shaped "frame" to the top and the right, thus making the difference in area/size. With some imagination one deduces that this "frame" resembles the initial "shaded area" of the problem.
3) The width of this "frame" is the 'sagitta s' itself. There is fortunately not much to work out since 's' is simply '1 - c', c being the chord-length [Note: c^2 = r^2 + r^2].
4) The area of the "frame" consequently equals s*1 + s*1 - s^2. ['- s^2' was put in place in order to NOT add the corner of the frame twice]. The initially "shaded area" is therefore approx 0.313708, being perfectly in line with Presh's final formula.
The answer is also approximately: π dm²
I found a better way...if, on top of the original picture, we draw the larger square with the circle exactly inscribed in it (centered in circles center), being r the circles radius, the lenght of the larger square side will be 2r. Then we'll have the original 1m side square overlapping the top right region of the larger 2r side square. In the bottom-left of the picture, there will be an L-shaped region between both squares. If we now "mirror" the shaded areas that are outside the circle, by "reflecting" them in relation to the 2r square diagonal (top-left to bottom-right corners), these areas will exactly fit inside the L-shaped area, filling them entirely, except for two small squares at both extremities. Then our wanted shaded area will be equal to the area of the L-shaped region, excluding both squares, making a shorter "L". The thickness of the L will be (2r-1), while the length of both legs will be 1. Therefore, our wanted area will be 2.1.(2r-1) - (2r-1)^2 = (2r-1)(2-2r+1) = (2r-1)(3-2r). We can evaluate r by using the diagonal "r" of an internal square with side (1-r), therefore r=(1-r)sqrt(2) => r=2-sqrt(2). By replacing it in (2r-1)(3-2r) we finally get the area = 8.sqrt(2) - 11 = approx. 0,3137 m²
The area of the purple triangle can also directly seen to be r² if you rearrange the two smaller triangles so that they form a square with side length r.
For the diagonal of the 1x1m square you could also juste use the pythagor theorem…
My brain filled in the pieces outside the square into the inside, then died
I moved the 2 round portions inside to create a square inside a square. Then i used angle bisector theorem to find radius and it was easy from there.
If you overlay an octagon over the top of the circle, you know the height from the top to the solid horizontal line below = 1. Compute the area by the ratio of height of each dotted line.
__ __
/........
/..........
|...........
|...........
\ ________
\ __ __
Put the arc areas inside the circle and find the area by: 0.171x1+0.171x0.828 = 0.3137. WAY easier for my brain!
I got the answer by finding the equation of the circle using the points of tangency. From those points and the fact that the lines connecting the center of the circle to the tangent points on the top and right of the circle have directions aligned with my axes, I was able to obtain a condition on the x-coordinate of the circle's center. We denote the center as being at (h,h) due to the planar symmetry of the problem extending through the line x=y. The distance from (0,0) to (h,h) is sqrt(2)*h. The distance from (1,h) to (h,h) is 1-h. Those distances must be equal, so h must be 1/(1+sqrt(2)). Then the radius is simply 1-h. From this, I got my equation for the circle, then plugged in x=0 to find the location of the intersection point of the circle and the left wall of the square. Neglecting the y=0 solution since that one is already obvious, I could then easily see the distance between the two intersections on the left of the square is 2/(1+sqrt(2)). From this, I compute the area of the triangle formed between the left and bottom intersection points. I also compute the area of the semicircle defined by those two points in the upper right of the figure. Since I know the radius of the full circle, the area of the square, and the area of the entire white part now, I can simply add the circle and square areas, then subtract two times the area of the interior white part I found. My exact answer is 1-4/(3+2*sqrt(2)) which is around 0.3137.
radius r can be calculated differently.
Suppose the center of the circle is at (a,a) and the circle passes through the origin and is tangent to the lines x=1 and y=1
So:
(x-a)^2 +(y-a)^2=2*a^2; r^2=2a^2; a>0
after simplification we have
x^2+y^2-2a(x+y)=0
for x=1, y=a (for y=1 x=a) we have
1+a^2-2a(1+a)=0
1+a^2-1a^2-2a=0
a^2 +2a-1=0
we are interested in the positive solution, so
a = sqrt(2) - 1
r=a*sqrt(2)=2-sqrt(2)
Naming A to the upper left blue area, B to the upper right blue area and C to the lower and left blue areas, the total area will be 2A + B + 2C
1- one can calculate the radius using simmetry and one can see that y=x is the simmetry axis, hence (knowing that the side of the square is 1), r=0.5858. Centre of the circle is (0,0)
2- "A" can be calculated as the area of a rectangle of (1-0.5858) x 0.5858 = 0.4142 x 0.5858 = 0.2426, minus the area of the circle from x=-0.4142 to X=0. Writing the circle as y = sqr(r^2-x^2) and integrating, S=0.2205. The difference is 0.2426 - 0.2205 = 0.0221, hence 2A=0.0442
3- B is the area of a rectange r^2 minus a quarter of circunference, hence B=0.3432-0.2695=0.0737
4- knowing that a 90º secant generates a triangle that its base includes the centre of the circle, 2C is half circunference minus a triangle 2r as base and r as high, hence 2C=0.5389-0.3432=0.1957
5- Area = 2A+B+2C=0.314 m2
The common area is easy to calculate with the radius, so the area itself is easy to calcukate with the radius. To find the radius, let x be the distance between the top left corner and the upper tangent point and you'll find that the diameter is 2 sqrt(2) x but also 1 + x²
say that the line inside the circle that is intercepted by itself and the circumference is a. That means you can construct a square such that one side is a and the side perpendicular to it is the same (by symmetry of right angled quadrilaterals inscribed in circles). Thus, we can flip the opposite sides of the circle that are leaving the square into these slots, which have width 1-a. By further symmetry, the bottom area of the circle that leaves the square will also have a height 1-a. That means the diameter is 1-a + a + 1-a, or the height of the square plus 1-a, giving 2-a. Since the diameter is also a times the square root of two (by 45,45,90 triangle theorem), a times square root two equals 2 minus a. Moving a to the left, we get that a√2 + a = 2, factoring out a we get a(√2 + 1)=2. Isolating a, we get that 2/(√2+1)=a, and since a^2 is the total unshaded part, the shaded area will be 1-a^2, or 1- (4/(3+2√2) by simplification, no rationalization yet), which gives 0.31370849899. I love your videos and this one took me particularly long, in contrast to the ones that I can't do at all. But I'm still 14 and will learn a lot along my way. Thanks for being the universal extra teacher of students!
When the circle cancels, the lightbulbs in my head started shining. Thank you, very good explanation like always.
All I did was realize that all the shaded areas can be broken up and rearranged into 3 rectangles, except the square in the top right which is easy to find the area of. Once I did that I had to find the radius, 2-sqrt2, then the overlap length of the sides of the circle and subtract those from 1 to get the 2 sides of the rectangles I found. That works out to 2(2r+2)(2r-1). I then added the square of the short sides of the two rectangles, 2r-1, to get 2(2r+2)(2r-1)+(2r-1)^2 which once you plug in the radius r and simplify you get 8sqrt(2)-11. Took me a while to figure it out.
My solution: Draw a diagonal through the point of intersection of the square and circle to the opposite corner of the square (call this AB). Then drop a perpendicular from the corner of the square (C) to AB (call this D). Drop a perpendicular from D to line CB. EB is then 0.5m and therefore DB is root 2 = 0.7071m. Therefore the circle radius is 0.7071m, giving an area of 1.5708sqm.
I used the tangent-secant formula (1-r)^2 = (1-r^2)(1) to get r = 2-sqrt(2), then proceeded with a similar way to get the shaded areas.
As others have done, I solved it via the square within a square, by solving for a tangent and the external segments of two secants. Though really just needed one of the external segments, all three distances are interrelated.
For findind radius I took the point which intersect with circle and corner of the square as (0,0) and wrote equation of circle as (x-a)^2+(y-b)^2=r^2 then put the (0,0) in the circle to give a^2+b^2=r^2 also applied distance formula of line and point where line is the tangent x=1 and y=1 and point being the centre of circle (a,b) which gives two equations more that is a=1-r and b=1-r and we also have one previuos equation a^2+b^2=r^2 solve it and get the radius
I didn't know how to solve for r, but I did know that the shaded region is 1 - 2r^2 because if you draw a diagonal line from the center of the circle to the top right corner of the square, label the point where intersect the side of the circle x. Then draw a point from x down to where the circle intersect the bottom side of the square [label that point a], and another point from x to the point where the circle intersect the left side of the square [label that point b], you've got a smaller square inscribed in the circle with a side length of r(root2). Then, you can rotate the area of the circle that's outside the big square around point a and point b into the square so that the shaded area is exactly the area of the big square minus the area of the smaller square.
Sorry if this explanation is too long or confusing
To cancel out the circle, you have to assume the semicircle passes through the circle's exact center point; an assumption you have no way of coming too unless you eyeball it or already have the trigonometric proof of that.
2√2 -2
If we move two blue pieces FROM outside the circle INTO the circle, the blue area would be the difference between two squares, bigger(original) square with side 1 and area 1 and smaller, inner square with side equal of R√2, and area R²/2 where R is the radius of the circle.
R can be found from R + R/√2 = 1 so R = 2 - √2
Hense blue area is 1 - (6 - 4√2)/2 = 2√2 - 2
I've used the inner square (within the circle), and since it's a square within the circle, the inner square side would be h = 2 * (r * sqrt(2)/2) = r (sqrt(2)). From there it's easy to figure out the radius using the bigger square, but I've completely forgot about the conjugates and the srqt(2) popped up everywhere and I calculated things wrong. I am "most students" :P
My problem was that I didn't understand that the center of the circle is on the diagonal of the square. The problem does not say that the circle cuts off equal parts of the square. And with your eyes you can't really tell if it's true or not. I think the whole difficulty lies in the understatement of the problem.
this is easily visible at the fist glance, as you can see if cut off both sections of the circle that are out of the square, and use them to fill the gaps in the circle, you get the right triangle square left.
Let a be the length that the circle extends beyond the square, a = 2r - 1. Then the answer is 2a - a^2. Because you can take the left and bottom segments, inflect it, and put it to the right and top to make a ┐ shape with sidelength a and 1.
Radio de la circunferencia =R =√2 -R√2 → R=2-√2 → El cuadrado inscrito en el círculo tiene una diagonal d=2R → Área cuadrado inscrito =d²/2 =2R² → Diferencia de áreas entre cuadrado de 1x1 y cuadrado inscrito =(1x1)-(2R²) =8√2-11 = 0.3137 = Área Azul
Si llamamos S a la superficie de cada uno de los cuatro segmentos circulares exteriores al cuadrado inscrito; llamamos F a la superficie diferencia entre el cuadrado inscrito y el cuadrado de 1x1; llamamos E a la superficie suma de las esquinas azules superior izquierda y derecha e inferior derecha → Área azul = 2S+E =2S+(F-2S) =2S+F-2S=F → Resulta que la superficie azul buscada es la diferencia de áreas entre el cuadrado propuesto de dimensiones 1x1 y el cuadrado inscrito en el círculo.
Interesante problema. Gracias y saludos cordiales.
Idk about in your country, but in Brazil we dont use calculators during classes or during tests, so we leave the answer in the rational form (such as 8 sqrd(2) - 11 in the video), but when I moved to Portugal, they do allow calculators and all the answers must be given in decimal form. Its so weird to me.
The way you see that circle and square answers the question
Just cut the outer arcs and paste inside the square. You are left with an unshaded square with side length r√2, so the total shaded area 1-2r².
I'm glad I tried it myself first!
I know that the distance from the middle of the circle and the upper right corner of the square has to have a fixed ratio to r. Let's call the distance from the edge of the circle to the corner of the square =: x.
With a square with a side length of 1 and a circle inside of it with r = 0.5, I know that r+x = 1/2*sqrt(2). Thus x = 0,4142*r
After simplifying, the area is equal to A = 1 - (1/2)*(2*r)^2 || Here I had the same approach with 2 semi-circles and 2 triangles.
I know that 2*r + x = sqrt(2), as it's the diagonal of the square.
Thus r = sqrt(2)/2.4142 = 0.586 || After inputting 0,4142*r for x.
Now just input r in the initial formula, and:
=> A = 1 - 0.686 = 0.314
Weeeee! Took a little while before I saw the ratios but it was a good fun looking for the solution. Thanks!
As someone that hasn't done complicated math in a decade and was just "okay" at it...I don't remember 'how' to do the things I was thinking of, but I had my concept for how to get it done. My idea was that there has to be a "formula" for making a square "outside" a circle, and then another "formula" for making a square "inside" the circle. Just fill it in the radius with X or, more simply, a number. Use that to figure out the differences between the inside and outside square's lengths and cut that in half. Congratulations, you've just created a pathway to reaching "this" square's length. Backtrack along that path by filling in the 'correct' length of this image's square to find the radius of the circle. Tada?
I made it into a square taken away from a square then computed it like this, 1-((1/((2+(1/sqrt(0.5)-1))/2))^2) it is worth noting however that this is not simplified. I also may not have written everything down right but checking it against the end value of the video shows that it works.
Base of the purple triangle = r + r = 2r
Height of the purple triangle is r
Area of the purple triangle = 1/2(2r)(r) = r^2
Is this the correct way of thinking?
My approach was entirely different. I drew a square surrounding the whole circle. Also drew a small square inside the circle. I added so many more things. It was soo complicated but I could solve it eventually
Please bring more of those problems, i felt it was out of my scope at the beginning and barely tried, great video though
In fact, it is necessary to find the difference between a large square and an inscribed in a circle
You could do this with integration starting from the square left and tracing until the square right, then subtract the bottom right shaded area as another integral. This is how I thought to do it. Your solution is maybe easier though.
A simpler solution is to realize the 2 blue parts in the circle can be rotated 180 degrees and fit nicely with the blue parts in the square to transform the green part into a small inner square.
The answer is simply the area of the big square: 1^2, minus the inner square area: 2*r^2 = 8*sqrt(2) - 11.
99.8565% of pi/10
If the answer was pi/10 I would have absolutely lost it
Solution:
diagonal of the square = √(1²+1²) = √2 = √(r²+r²)+r = r*√2+r = r*(√2+1)
r = √2/(√2+1)
Black area
= black semicircle with the radius r +
black isosceles right triangle with the hypotenuse 2r and the leg length a
= π*r²/2+a²/2
with a²+a² = (2r)² ⟹
2a² = 4r² ⟹ a² = 2r² ⟹
Black area = π*r²/2+r² = r²*(π/2+1) = r²*(π+2)/2
Blue area = square - black area + circle - black area
= square + circle - 2*black area
= 1²+π*r²-2*r²*(π+2)/2 = 1²+π*r²-r²*(π+2) = 1-2r²
= 1-2*[√2/(√2+1)]² = 1-2*2/(2+2*√2+1) = 1-4/(3+2*√2)
≈ 0.3137[m²]
Wow!!!! Very good explanation!
Could you please talk more about why certain tricks can be applied and what they are?
That. Is. Glorious.
Should there be parentheses around 8*sqrt(2)-11? As it stands now, the square meters is only reflective of the 11…
Rather than this intuitive method, I solved a = square - circle + 2 * arc
Arc = circle slice with a triangle cut off, so circle / 4 - triangle
Triangle = r^2 / 2 (after simplifying unit circle shenanigans
I ended up with the same answer via an entirely different method, and that's one of the reasons I love math
I first brute forced with cartesian coordinates. Painful.
Then I saw if you inscribe the circle in a bigger square, and you inscribe a smaller square in the circle, prolongating the lines of this small square, you could move the shaded area at the top to the bottom, completing it into a rectangle. And you could move the shaded area on the left to the right, completing it into a rectangle of the same size. That leaves a small shaded square at the top right. It's not hard to calculate the lengths involved and you get:
Area = 4*(sqrt(2)-1)^3+(sqrt(2)-1)^4 = 8*sqrt(2) - 11
A full proof requires to calculate the lengths and explain the symmetries so you can move the shapes around.
Presh's proof's more elegant.
Haven't watched the video yet, but you should be able to flip over the semicircle pieces and put them in the square, so the total shape is the 1m square with a smaller square inside of it, where you only use the area of the large square minus the smaller one. So the formula for the area is (1 - ((2-d)^2)) sq m, where d is the diameter of the circle.
i don't know how to calculate d, but it must be less than the square root of 2 given that the line from the bottom left corner of the square to the top right one has a length of the square root of 2 and the circle is obviously smaller than that, and it obviously has to be more than 1. It probably has something to do with pi though, given what I know about circles.
I took this approach. Radius of circle is r. Side of smaller square is 2r/sqrt2, call that x. The width of the semicircle is (2r-x)/2. Then just solve for r, starting with the side of the larger square (1), subtracting the width of the semicircle to get x, then solving for r.
It's so simple. There are tow squares .You just need find out the small one's size(r^2*2).
How do you prove that those two little triangles are congruent?
I'm 11th grade student and in a normal level math book, i saw that question and yes we supposed to solve all of em like that.
My brain won't be like this forever......
I also got with the other version which I think is the slower way.
I start with the inner square with X as length, then I get X√2 as the 2r. So we can see the r is X/√2. Then the r√2 will be X so the √2 = ( X + X/√2 )
Then I got X = ( 4 / 3 + 2√2 ) and then come to answer with: 1 - X^2 which is 1 - ( 4 / 3 + 2√2 )^2
I got the right answer but I solved for the radius of the circle in a slight different way.
I also didn't solve the area in blue as clever as you did it , but I did notice the way I did it, the 'pi' expressiond cancelled out.
i've cut your purple triangle in half once more and then calculated the height of this triangle because r+h=1 and so i got to r*sin(60)+r=1 which also get you r=2-r(2). And you know that hole thing with circles canceling each other out i didnt really needed that cause i could just calculate the area of the purple triangle with the height and then i could calculate the area of the circle and so on...
The answer works out at 0.313708m^
I did a slightly different method, started off the same, lets call the radius r, but i split the bottom triangle in 2, using pythagoras I was able to find the distance from the centre to the edge of the square, this being sqrt0.5r.
Therefore we have 1=r+sqrt0.5r or rearrange to find the value of the radius as 1/(1+sqrt0.5).
Long story short i ended up with the area being equal to 1 - 2/(1.5+2sqrt0.5)
yea did it just like you know it felt a little bit intuitive by starting at the bottom and working to the top up, i dont really like how he did it because if you start on the top on like him it can always lead you anywhere, this is definitly the safer option.
The hard part about this problem is figuring out the radius of the circle
The area = [ ( half circle - triangle ) + [ square - (half circle + triangle) ] = 1 -2*triangle = 1 - 2*(2*Rsq/2) = 1 -2*Rsq
R = 2 - 1.41 = 0.5857. So Area = 0.3137
Shape algebra is neat. I didn't think of that.
I don't know why I should consider that the diagonal inserted at 1:03 must pass through the center of the circle. I do not know this property and it was not mentioned.
I'm not great at explaining it, but because the problem involves a square and not a rectangle, the two lines that form the 90° angle of the purple triangle can only lead to this property, as they meet at the edge of the circle and are equal in length. They form two sides of a square whose corners would all touch the edge of the circle, and whose center would be the same point as the center of the circle. Thus its diagonals would also pass through this center point.
3:37 As a physics student, that last equation is giving me a real bad headache
Thank you Presh. Your channel is awesome
Just solved a same type of problem to the students. Thanks for extreme technical knowhow
Thank you for your step by steps!
I just had to check real quick that 8root2 -11 doesn’t equal pi/10
Just to be sure
lol I gave up kind of early watching the thumbnail and then was really mad when I saw the construction. Pretty clever. Cool video.
Explain how two congruent triangles formed having each hypotenuse r√2
I wish he had proven this in the video. If we call the diameter of the circle 2R, then the sagitta (height) of each of the shaded segments is 2R-1. Since those shaded segments belong to the same circle, they also have the same radius, the same arc angle, and (most importantly) the same chord length. Those two chords with equal length are the legs of the purple triangle, so the purple triangle is both right (given)and isosceles (shown here), thus 45-45-90. When he draws the median of the purple triangle, we know that the legs of the two smaller triangles each have a length equal to R. We also know that each of these triangles has one 45 degree interior angle that it shares with the larger triangle. So by SSA, these triangles are also congruent.
Not sure it this is the best method, but here's how I solved it. Edit: Ok yeah it seems there would have been much easier way to prove that BC cuts the circle in half. Also, I didn't notice that r could be simplified further so my answer is not in the simplest possible form.
Draw a line from the center of the circle (O) to the point where the circle touches the corner of the square, call this point A. This line OA is the raduis of the circle, call it r. Now, draw lines from O to the points where the circle intersects the square (let's call them B and C), these also have the length r. So now we have two identical isoceles triangles. Because the triangles are identical, the angles BAO and OAC are the same, and their sum is a right tringle, therefore both are 45°. Because the triangles are isoceles, the angles OBA and ACO are also 45°, and therefore the angles AOB and COB are both 90°. This means that the line BC goes through O and therefore cuts the circle perfectly in half.
Therefore, we can calculate that the black area that is included in both the square and the circle is a combination of half the circle and a right trianle with a base of 2r and height of r. The blue area in turn is the combined area of the square and the circle minus two times the black area:
A = 1^2 + pi*r^2 - 2(2r*r/2 + pi*r^2/2) = 1 + pi*r^2 - 2r^2 - pi*r^2 = 1 - 2r^2
Now, we have to find r. Let's call one of the points at which the circle is tangent to the square D. Because the circle is tangent to the square at D, DO is parallel to two of the sides of the square and has the length r. In turn, we know that OA is at 45° angle to the side of the square and also has the length r. If we choose a point E so that it's at the halfway point between A and B, it will create a right isoceles triangle AEO where OA is the hypotenuse. Also, OE will be a continuation of DO, and DE has the length 1. From pythagorean theorem we find that OE = OA/sqrt(2) = r/sqrt(2). So:
DE = DO + OE
1 = r + r/sqrt(2)
r(1+1/sqrt(2)) = 1
r = 1/(1+1/sqrt(2)) = sqrt(2)/(sqrt(2)+1)
Now, we can calculate A:
A = 1 - 2r^2 = 1 - 2*(sqrt(2)/(sqrt(2)+1))^2 = 1 - 2(2/(2+2sqrt(2)+1) = 1 - 4/(3+2sqrt(2))
Which is approximately 0,314.
This feels like the kind of exercise an ancient Greek would consider impossible.
good problem i solved it pretty much the same way, i loved this type of geometry problem. subtraction of areas is nice
Well.. since there is a semi circle inside the square, we can draw another triangle(similar to the first) inside that semi circle . Then we see that the shaded areas can also be found within the square itself by subtracting the smaller square from the larger square ...But it does not make a huge difference. Just saying ....
The circle cancellation is really bright
Excellent problem. I didn't manage to solve this one.
Which application do you use to make graphs and elaboration.
If you could plz share
Because I used to see your all geometry solution in very effective manner and I am also eager to start up my channel.
What kind of students is this for?!? I didn’t even know where to start with this ;-;
Thanks bro, sometimes I forget little/more advanced knowledge to solve the puzzle. I hope I'll be better with the help of your content :))
Good job! That was a simple and beautiful solution, very easy to understand.
The method is very difficult to come to at the first place anyway. I did not get that myself (I've got a working approach though before watching the solution).
When the two lines are drawn from center of the circle to the tangents, why is it assumed they create a square?
1:00
Why is it two times the overlapping area, and not just the overlapping area?
You don't to put square meters in. You can simple assume the square has a side length of 1 unit length.
You can imagine the same circle with its center in the center of the square... the area is still the same. Then we can imagine a circle thats touching inner walls and a circle touching its corners. When we average these 2 we will get the r of our circle. (0.5+(square root of 2):2) :2) this is the r of our circle.
Amazing how the result is so similar to pi!