How To Solve Insanely HARD Viral Math Problem

hey this is pressure Walker and you're watching mind your decisions here's a really fun and challenging geometry problem that's been spreading around the internet start out with this square with a side length of 10 on one side draw a semicircle and on an adjacent side draw another semicircle now between the two corners of the square draw a quarter circle the challenge is to solve for this area shaded in red now just to present it in another way let me label some of the intersection points ABCD is a square with a side of 10 APD and CPD are semi circles and a dqb is a quarter circle the problem is to solve for the shaded area dpq I think everyone who sent me this problem it was a very popular suggestion and in particular I think che danielle from india who was the first person to send me this problem can you figure it out give this problem a try and when you're ready keep watching the video for a solution [Music] so how can we solve this problem I admit it was very challenging I originally solved it by making a coordinate system and using calculus to solve integrals but I suspected this problem could be solved using geometry and trigonometry so I posted it to math Stack Exchange within hours David K Achille Hugh and Syed posted solutions and I thank them for helping me understand how to solve it using elementary methods so in this video I'm going to present three ways of solving the problem I'll first go over the short way using geometry and trigonometry then I'll present a little bit of a longer way and then finally I'll present the calculus solution so let's get started with the short way in order to solve for this area we'll solve it by finding the areas of two shapes that will be easier to solve for so we'll solve for this red area as the area of this blue shape minus the area of this yellow shape so we have this equation the blue and yellow shapes are examples of a shape known as a geometric lens that's the area when two circles intersect so since we need to know the area of a geometric lens let's derive it generally so we'll start out with a semi circle with a radius of a and perpendicular to it we have a semicircle with a radius of B we want to solve for this intersection area now to do that let's connect these two radii to the intersection point and we form a kite now let's draw a line across this kite so we're going to essentially solve for this shape by solving for each of these upper and lower shapes so the area of intersection will be equal to the area of this circular sector minus the area of this triangle and that'll give us the area of this lower half of the lens then we have this circular sector minus this triangle and that'll give us the area of the upper half of the lens now the trick in this problem is that we actually don't need to solve for the areas of the triangles we can actually solve for the total area that they span so to do that consider these two triangles together let's draw the other diagonal of this kite notice it divides the kite into two right triangles now each of these right triangles has legs of a and B so together to solve for the total area of these triangles we need the areas of two triangles who have one leg of a and one leg of B so each triangle has an area of 1/2 a B so together both triangles will be equal to a B so now we need to solve for the areas of these circular sectors and then subtract a B and that'll be the area of this lens so how can we solve for the areas of each of these circular sectors well we need to know the radius of each circular sector that'll be easy it'll be B for one circular sector and a for the other and then we need to know the central angle of each circular sector so let's consider this right triangle let's label this angle as theta the other angle in this triangle will be PI over 2 minus theta if we look at this right triangle which is congruent to the upper right triangle this will also be theta so here we have a central angle of 2 theta corresponding to the radius of a so we can solve for the area of this circular sector it'll be a squared over 2 times 2 theta and the other circular sector we have a central angle that's equal to 2 times the quantity PI over 2 minus theta so it'll be equal to PI minus 2 theta so we'll solve for the area of this circular sector it'll be B squared over 2 times the quantity PI minus 2 theta but what is theta well we'll use trigonometry in the right triangle data will be equal to the inverse tangent or the arctangent of B over a so we can substitute that in for theta and then we can simplify and this will simplify to be the quantity a squared minus B squared times the inverse tangent of B over a plus PI B squared all over 2 and then we need to subtract that a B so this will be the area of a geometric lens so we now need to apply this formula to the two lenses in this blue shape we have a equal to 10 and B is equal to 5 so we substitute those into the formula and then we simplify to get 75 the inverse tangent of 0.5 plus 12.5 PI minus 50 in the yellow shape we have a and B equal to 5 so we substitute those in and then we simplify we now need to subtract the second equation from the first when we do that we can cancel out these terms of 12.5 pi and we get the answer of 75 the inverse tangent of 0.5 minus 25 and that's approximately nine point seven seven so that's one way to get the answer but let's suppose you didn't realize you need to calculate the area of a geometric lens how would you solve the problem well let's go over the longer geometric way to solve it it's essentially going to be the same method but it'll be useful to go over the steps to reinforce what we just went over in the short method so let's get started so let's solve for the area of this yellow shape to do that let's get rid of some irrelevant shapes we'll draw a line down the middle dividing it into two equal circular segments so we just need to solve for the area of one circular segment and then multiplied by two now to do the area of one of these circular segments we'll draw in this isosceles right triangle so this circular segment will be equal to this quarter circle minus the area of this triangle as the radius of this semicircle is five the area this quarter circle is pi times five squared over four and the area of this triangle is five times five over two so the area of this circular segment is equal to 25 pi over four minus 25 over to the area of this yellow shape will equal two times this which will equal 12.5 pi minus 25 and is the same result as we had before in the short method so we now need to solve for the area of this blue shape how can we do that we'll draw a line at the two intersection points and we'll solve for each of these areas this upper area and this lower area and we want the sum of these two areas but we'll solve for them individually so this upper area how do we solve for it well we'll draw in our kite like figure as before and we'll draw in the values of these distances we'll also draw the diagonal this kite so notice this kite now has two right triangles who have legs of five and ten let's label this angle as theta now we want the area of this upper circular segment and it'll be equal to the area of this circular sector minus the area of this triangle we'll put in the dimensions of each of these shapes the area of a circular sector is equal to 1/2 times 10 squared times 2 theta and the area of this triangle will equal 1/2 times 10 squared times the sine of 2 theta now theta in this case is equal to the inverse tangent of 5 over 10 or point 5 we can also solve for the sine of 2 times theta by the double angle identity sine of 2 theta is equal to 2 times the sine of theta times the cosine of theta we can get the sine and the cosine by solving for the hypotenuse and then solving for the sine and cosine of each of these angles and this all simplifies to be 4 over 5 so we substitute these values in to our formula above and we get the area of this blue shape is equal to the following which we can simplify to be 100 times the inverse tangent 0.5 minus 40 now to solve for this lower part of the area we do a very similar calculation we put in our kite and then we need to solve for the area of circular sector - the area of this triangle now the key in this we have the dimension which is a radius of five but we need to know this angle now since one angle is equal to theta in a right triangle the other angle will equal PI over 2 minus theta and this angle will also equal PI over 2 minus theta so the central angle will equal pi minus 2 theta so the area of the circular sector is equal to 5 squared over 2 times the central angle and the area of this triangle is 5 squared over 2 times the sine of this central angle so how can we simplify this we can simplify sine of PI minus 2 theta because sine of PI minus 2 theta is equal to sine of 2 theta and we already figured out sine of 2 theta in the previous calculation is equal to 4/5 so we substitute these values in and then simplify to get that this is equal to 12.5 pi minus 25 the inverse tangent of 0.5 minus 10 so we now can solve for the area this blue shape by substituting in the values we just arrived we then simplify it to be 75 times the inverse tangent of 0.5 minus 50 plus 12.5 pi so now we get the area of this blue shape and we need to substitute that in and the substitute in the area of this yellow shape we then subtract the yellow shape from the blue shape so these 12.5 PI's cancel out and it all simplifies to be 75 times the inverse tangent of 0.5 minus 25 and that's approximately nine point seven seven just as before so you can see if we didn't see the trick about the triangles it's a little bit longer but it can be solved if you work through it now let's suppose that you couldn't figure out the geometric way you may have approached it like I did using calculus so first we'll put in a coordinate system here's the y-axis and here is the x-axis will now label some points in this diagram now all of these are quarter circles or semi circles so this curve will equal the square root of the quantity 25 minus x squared this upper curve will equal the square root of 100 minus x squared minus 5 and this semicircle will equal 5 minus the square root of the quantity 25 minus the quantity X minus 5 squared so we have three different curves that are bounding this area now we can solve for the final intersection point by equating the two equations this point will be eight comma one so now we can solve for the area of this shape as follows we need the area of the upper curve going between 0 & 8 and then we subtract the area of the lower curve going from 0 to 5 and then we subtract the area of the other lower curve going from 5 to 8 so we want the integrals of each of these as follows so now we need to simplify these integrals well this middle integral will be easy to simplify because it's the area of a quarter circle with the radius of 5 so we don't actually need to do the integral we know that this will be equal to 5 squared times pi over 4 which is 25 PI over 4 so that's one part solved now this next integral will split into two integrals with the integral from 0 to 8 of the square root minus the integral from 0 to 8 of the constant 5 the second integral is easy to evaluate it'll be equal to 40 so now how do we solve for this first integral we use a trigonometric substitution X is equal to 10 sine theta this means DX is equal to 10 cosine of theta d-theta we also need to change the limits of integration from going between 0 to the inverse sine of 0.8 so we substitute that in then we need to simplify that 100 minus 100 sine squared of theta is equal to 100 cosine squared of theta we then simplify this all out now we can solve for this because the integral of cosine squared of theta is very well known so we substitute that in and we evaluate from the limits of integration and it simplifies to 50 times the inverse sine of 0.8 plus 24 and we subtract 40 so this will be 50 times the inverse sine 2.8 minus 16 so that'll be the first integral we now need to do a similar thing for this very last integral we split it up into two different integrals this first integral will be equal to 15 this next integral we can solve first by doing a use substitution let's say U is equal to X minus 5 so we change the limits of integration and we now get the integrand is the square root of 25 minus u squared to solve for this integral we can do a trigonometric substitution that U is equal to 5 sine of theta it'll be a very similar calculation to the first integral that we figured out and I'm just gonna skip some of the steps it all comes down to be negative 6 minus 12.5 the inverse sine of 0.6 so we put this all together and we get that this simplifies to be 9 minus 12.5 times the inverse sine of 0.6 so we now just need to simplify this expression and that's our answer we have 50 times the inverse sine of 0.8 plus 12.5 times the inverse sine 2.6 minus 25 minus 25 PI over 4 but this is quite a bit more complicated than the geometry and trigonometric solution and I wondered how can we reconcile it how can we get it to be using the inverse tangent and just you know two simple terms so let's try and reconcile these two answers well we can do that by drawing this right triangle this 3 4 5 right triangle notice that theta is equal to the inverse sine of 0.6 and alpha is equal to the inverse sine of 0.8 furthermore theta plus alpha is equal to PI over 2 because these two angles sum to be 90 degrees therefore 12.5 times the inverse sine 2.8 plus 12.5 times the inverse sine of 0.6 must be equal to 12.5 times pi over 2 so how can that help us well what we're going to do is this first term where we have 50 times the inverse sine of 0.8 we're going to split it up into thirty seven point five times the inverse sine 0.8 and 12.5 times the inverse sine 2.8 so now we have 12.5 times the inverse sine of point eight plus twelve point five times the inverse sine of point six and this is equal to twenty five PI over four so we substitute that in and it cancels out with this other 25 pi over four so this all simplifies to be thirty seven point five times the inverse sine two point eight minus twenty five and this is approximately equal to nine point seven seven so it's equivalent to our answer it's just two terms but something still bothered me about it because it's the inverse sine and the previous answer was equal to the inverse tangent so how can we reconcile those two components well I looked up this formula if you substitute in X is equal to 0.8 you get that this simplifies to be two times the inverse tangent 0.5 and that gives us our final equivalence that this is equal to 75 times the inverse tangent of 0.5 minus 25 so we verified our solution using calculus did you figure out this problem and which method did you use thanks for watching this video these math videos which can be watched for 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