Just out of 7 years of engineering education and watching these videos reminded me of why I chose to follow this path at the first place. Don't forget guys we are not machines enjoy yourselves
Excellent work! I think another way of looking at it is to determine that the half lengths of the square is x/2, and by using the line you drew for the triangle you would quickly find that y actually is 5 - x/2, thus making the equation being: (5 - x/2)squared + (5-x) squared = 5 squared.
I am 30+ years out of algebra, and just stumbled across this in my feed and I still watched the whole thing. That was a terrific explanation. Good teaching and very clear.
x=10 is actually a fairly reasonable result, it just refers to the square which is bounded by the upper arcs of the two circles (not by their lower arcs like the original case we are dealing with) ..
@@hammy2737 I mean with x=2 , the square is in touch with the outer surfaces of the two circles but with x=10 the square is touching the inner surfaces of the circles
For a general solution, using: "r" for Radius; "L" for lenght of the square side; the triangle drawn at 00:26, Use sin^2+cos^2=1. Sin=(r-L/2)/r ; Cos=(r-L)/r. Solving for L, you get a 2nd degree equation in which the solutions are 2r and 2r/5. As you know the side L is smaller than R, the only valid answer is 2r/5. For radius=5, L becomes 2, thus the Area (L²) = 4.
I feel L and r are more representative in this case, but you can use any variable you like. However, if you must always use X when solving implicit problems, I have some bad news for you… 😅
I love the logic in math. When I saw the thumbnail, I couldn’t fathom a way to calculate that. 3 min and 34 seconds later it totally makes sense and it’s not even difficult math. Just logic and reasoning. Right on man!
A simpler solution (what I did) is to draw a perpendicular from the base to the point where the circles meet, so y = 5 - x/2. Then the Pythagoras and quadratic method is just the same but there’s only one variable
Yeah it was in my suggested videos and I just laughed watching him write the steps down Last time I did geometry like this was in 8th grade so it would've taken me a minute to put all those pieces together
@@raidzeromatti knew how to do all the steps too, but you have to be creative and intuitive enough to discover what should be done here. You should of tried the problem on your own first if you're that smart
@@ZiyaB3ast If you know Pythagorean and implicit functions it's just connect the dots Definitely would've taken me a minute to solve but it's not like you're being asked to find the rectangle with the largest area that will fit in that space Geometry doesn't suck until you learn derivatives
that was exciting! I no longer have the brain for hard math stuff like this but watching people solve problems like this is very fun thanks for bringing some joy to my night
Choose a coordinate system such that its origin is at the center of the bottom edge of the red square, y and x axes point towards up and to the right respectively. Since the top right corner (x,y) is on the circle we can write (x-r)²+(y-r)²=r² and if that point (and its mirror image on the other circle) would be moved a bit on the circle, it would distort the red square into a rectangle. The only way it will remain square is when y=2x. Substitute y into the equation of the circle and solve for x: (x-r)²+(2x-r)²=r². x²-2rx+r²+4x²-4rx+r²=r². 5x²-6rx+r²=0. Solutions: x=r and x=r/5. Reject the first and the area is A=2xy=4x²=4r²/25=4
To understand why it's a whole number, note that the triangle you formed first is the 5,4,3 pythagorean triple. We just require that the hypotenuse minus the shorter side is twice the hypotenuse minus the longer side, which of course it is. 5-3 gives the edge of the square.
That was my thought. As soon as he did the 3,4,5 right triangle I was expecting a quick resolution. Then he did polynomial math and I was like wait, why?
@@doormat1 He hadn't established it was a 3,4,5 triangle though. It's not obvious to me why that would be the case, although I'd imagine there is a geometry theorem that could have been called on to show that
Even though I just watched you do it, I still dont know how you did it. I'm slowly starting to understand bits. You're significantly better than any teacher I ever had.
I graduated in electrical engineering Bsc in 2021 and like algebra. But I always felt bad when had to solve at geometry problems. Your videos make me feel the happyness of understading and solving problems again. Thank you
I'm 44 years old, Maths was my subject in school and this came in my feed, I watch the whole thing and was genuinely rivetted. Andy you are a gift my friend
he's explaining what he's doing in the most plain way possible... this is like the most baseline "teaching" ever, there's not much learning of things you can apply to other problems here (i.e strategies for setting up your solution)
At 1:03 , we can define 5 + 5 = 2y + x , which gives us y = 5 - x/2 From here, you can use this result in Pythagoras, (5-x)^2 + (5 - x/2)^2 = 5^2 leading to a straightforward simplification of a quadratic giving the same results as you, 10 and 2. Personally I think this method is easier as you don’t have to deal with y, since you already know what it is, and it is quicker. Although, we all get the same answer in the end.
Could you explain further please I didn't get it like how did you come up with 5+5 = 2y+x like did you took the radius and length of square or like what?
My issue is when he gets to the quadratic equation he says just factor it, it's -10 and -2. That's fine in this simple example but just use the quadratic formula as your proof.
It's really weird feeling that you enjoy looking at math being solved while not the biggest fan of math. Kinda miss when I used to solve those in college
You can also put the origin at the center of the bottom side of the red square and look at the coordinate of top right corner, we have: 1) y=2x from the square 2) (x-5)²+(y-5)²=5² from the right circle equation Putting 1 into 2, we get : (x-5)²+(2x-5)²=25 that give 5x² - 30x + 25 = 0 and then x² - 6x + 5 = 0. 1 is pretty obvious as a root, so it factor in (x-1)(x-5)=0 and since x=5m doesn't fit we have x=1 Right half of red square is 1*2=2m² so answer is 4m² It avoid a bunch of square roots and pythagore
Really enjoyed this video, Andy - awesome! I had to roll up my sleeves, brush off engineering degree from decades ago and give it a shot! I happened to solve it with a much simpler approach. If the x-y axis is drawn right between the circles and at the bottom of the circles. Equation of the circle (x-5)^2 + (y-5)^2 = 25 must satisfy the point (a/2, a) on it, where "a" is the side of the square. Plugging this point into the equation for circle and solving for "a" will lead to same two values Andy reached (10 and 2). Bam!
I did it like this as well, but it became painfully clear how long ago it was i last took a math course. About 20 highschool algebra mistakes later, i got there though ;)
I solved it by considering the horizontal line at the bottom to be the x-axis, and the vertical line between the two circles to be the y-axis, so that half of the red square is in the first quadrant. Then the top right vertex of the square can be the point (x,y). x is half the side length of the square and y is the whole side length, so we have y=2x. Substituting this into the equation of the right circle (x-5)^2+(y-5)^2 = 25 yields a quadratic that can be solved yielding x=1 and x=5, the former corresponding to a square with side length 2 and area 4.
This video made me feel good about myself. It has been more than 15 years since I graduated high school, but basic algebra still sticks with me and this is exactly the solution I would’ve gone with. Thanks for a delightful video.
I'm 𝗰𝗵𝗮𝗹𝗹𝗲𝗻𝗴𝗶𝗻𝗴 you to solve this 𝗺𝗮𝘁𝗵 𝗽𝗿𝗼𝗯𝗹𝗲𝗺 and if you will be able to solve this i'll give you money as a reward. 𝗩𝗶𝗱𝗲𝗼 𝗹𝗶𝗻𝗸-:ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-iFN4DMh8Wto.htmlfeature=shared
I love watching your videos. High school me would’ve devoured these problems but now I have so many things to think about that I can’t even focus and think properly. Yay age.
Try the equation for circle with origin at the lower point of a square that encapsulates the circle. The equation is (x-5)^2+(y-5)^2=25. But from the small square, you also know that at the point of contact of that square with the circle, y=2x. Simultaneous solution is x=1,y=2 or x=5,y=10. If x=1, then y=2, it is the square's side, so area is 4.
Actually, if you consider the x=10 as a possible solution, you can see that it leads to another square, which has a side along the horizontal line to which both circles are tangent, but now the square is made by the whole distance between the tangent points, and the two diameters. If you imagine the two radii in each circle moving, the two solutions reflect the two possible perfect squares which can be built so that a side is along the tangent (horizontal line): one when the x is 2 and one when the x is 10.
But then y will become 0. But we know that y is not zero. It is already defined so x cannot be 10. That's just a coincidential solution not real solution
@indiankid8601 When they say solution, they are talking about solution to the quadratic equation, not the solution to the entire problem. You can use the exact same quadratic equation to solve two different problems, which is why it gives two different answers. They are only mentioning it because this explains why it results in both x=2 and x=10, not because it is complete nonsense, but because it is the solution to a different problem.
maths is my fave subject back in high school. I really enjoy solving this type of problems. it has been more than a decade since I left school. I missed this :")
You could also think of the circles as a function. Let's say the "ground" line is the x axis and the origin is the first circle's intersection point with the ground line. Cut off the top half of the circles so we have an actual function, defined from -5 to 15 Now we just need to analyze the function a bit, let's call it f, and we need to find where f(x) = 2(5-x) And the are will be f(x)^2
for the longest time i believed i could not do maths at all, but when i found this video i actually felt good knowing that i understood the equation and worked it out. giving me hope fr
If θ is the angle defined by the horizontal radius of the left circle and its common point with the square, then sinθ=(5-x)/5 and cosθ=(5-x/2)/5. Then squaring both sides of the equations and adding them together and because sin^2(θ)+cos^2(θ)=1, solving for x we get x=2.
@@TheBiomedZed the square Is equal on all sides so if we find one side we find them all. anyway the circle is 5 meters in its radius so is you put another line the same length vertical against the square you can pretty much tell that the square is only about 2/5 the height of 5meters meaning that the square is 2 by 2. 2x2=4 so the answer is 4 squared I hope this helped
@@Kingsidtheslothnope it's 5m squared. The square fits between the two circles with perfect symmetry because the circles are the same size right? So the circle touches the square at the 'half way point' between centre and the bottom which means it's half the radius so 2.5m height
I usually dont find math to be actually fun, but this is pretty cool. Plus when i tried to solve it myself, i had the right idea with using triangles (im in a trigonometry class rn, so it was fresh in my mind)
What's neat about the 2 solutions here is the other one also makes sense if the object is to find a square that connects to both circles at it's corners, because the other option puts the corners on the very top of both circles, and is the only other side length capable of maintaining those constraints.
The best problems can have more than one way to solve them. Drop a perpendicular from the midpoint of where the two horizontal radii will meet so that it goes through the center of the square. If you extend a radius to the center of the square you will make an isosceles right triangle with the hypotenuse equal to 5 root 2. The root 2 is half of the square's diagonal so a diagonal is 2 root 2. Since a square is a rhombus it has an area = d1 times d2 all over 2. Using 2 root 2 for each diagonal gives an area of 4.
1:06 From here just trace a vertical line between the 2 circles which also splits the square into two, the left half becoming 1/2x, the distance from the center to that new vertical line is also 5m, to get Y subtract 1/2x. now Using pythagoras you have (5-x)^2+(5-(1/2x))^2 = 5^2 25-5x+1/4x^2+25-10x+x^2 = 25 5/4x^2 -15x +50 =25 ---> 5/4x^2-15x+25 = 0. multiply everything for 4/5 and we get to your equation.
I did it through a bit of approximation. I drew an imaginary rectangle that encompasses both the circles from the outer side. Then I calculated the area of the rectangle which would be 200 meter squared, and the area of circles combined which would be π×(r)^2 = 158 (π value approx 3.14...taken so 157+a small value considered to be 1) , then we get the empty area in the rectangle without the circles which is 200-158= 42. Now if you have imagined the rectangle you already know that there are 6 different empty spaces in the figure and the ones in the corners are half of the ones in the middle. So in total there are 8 spaces as such and the ones we want are the two in the bottom centre, so 8/2 = 4, and 42/4 = 10.5 Then through extreme approximation just by looking at the figure and assuming that the square is surrounded by three triangles and a little excess curved area we can come to the conclusion that the square is about 2 triangles in area + 3 surrounding triangles + excess area, which would be 5 triangles in total and a little excess area(0.25 part ,not area in m) . So since we are concerned about only the square(2 triangles), 2/5.25 × 10.5 = 4 Now I got the answer to be exact 4(which as shown in the video is right), but if the values change in the question still I would have gotten a approx answer, which I could have rounded up to the closest whole number. And if there were options then it becomes even easier. (I was in the train and couldn't get my notebook so had to resort to approximation. Also the method shown in the video is very good, I'll remember that. Also I speculate that since I got the answer exact right maybe the ratio for the area between those two circles might be actually somewhere around 2/5.25 for all circles?, Idk but I'll look into it seems interesting.)
If you're going for an approximate answer I don't see why you'd make it so complicated. You can easily just visualize that the 5 meter line is 2 and a half red squares in length.
@@Prolute that's assuming that the figures are to scale. And if you actually know what you're doing then it hardly takes some minutes to figure out in your head. But I do see that it's a lot of words for explaining such an easy approach.
Another way to solve this problem is to think about the dimensions of the square as two separate functions. The base of the square starts out at 2r and goes to 0 where the two circles touch. This width can be defined as 2r-2rsin(Θ). Similarly, the height of the square would start at 0, and work its way up to r. The height can be defined as r-rcos(Θ). From there this problem can be seen as the unique case where the width and height are the same, and so 2r-2rsin(Θ)=r-rcos(Θ). There's a little bit of trig involved from this point, but r cancels out almost immediately leaving just Θ to solve for, which should end up as 2*invtan(1/2) or roughly 53°. Plugging back into either the width or height equation gives a side length of 2 for the square.
I looked at it graphically and instinctively knew it was 4, because the r=5 setup, now had the r's been wildly different, ya probably would of done it that way.
You can also solve it quite quickly by using the equation of a circle (x-5)^2+(y-5)^2. Here you are cutting the square in half so you know at the point where the square touches the circle y=2x. Plug that in and solve for x, use the smaller value of x, x=1 so the whole side length is 2, 2^2 is 4.
Awesome video, the method was very interesting yet easy to understand! I think you could have divided both sides by (10-x) at 2:21 for further coolness B)
A good rule is "do not divide using the variable", just in case you divide by zero on accident. In this case, since the polynomial does have 10 as solution for x, then by dividing by "10 - x", you'd be dividing by zero
I saw that a single circle defined the length of a side of the square vertically and half the length of a side of the square horizontally, so instead of using an arbitrary y variable, I set it equal to 5 - x (x in this case being half of your x). I then followed the same process you did and got x = 5, 1 (since mine were half what yours were).
Looked at the problem from the RU-vid recommend screen shot and solved in a few seconds in my head. I’m a retired veteran, so, I haven’t done math in since I quit school in grade 10 lol. Neat.
You can make it much easier without calculations: One side of the triangle = 5, one side of the triangle = 5-x, one side of the triangle we can call 5-1/2x (instead of y, because the other half occupies the area next to the other circle). As a conclusion c^2 (=25) = (c-1/2x)^2 + (c-x)^2 And anyone that has calculated the most basic numbers of this formula sees that 5^2= 4^2 + 3^2 . That shows us x= 5-3 = 2 . I m not a math pro but I thought maybe it can make things more visible. Keep going my friend you make a lot of people stimulate their brain. Much love from romania and sorry for my bad English! ❤
genuinely thought rhis was gonna include some sort of formula that i had never heard of in my life, but i was quite surprised when you used simple algebra to solve this
I remember solving such problems easily and relatively quickly during my school days and now after 10+ years it doesn't feel the same. Feeling very slow.
I solved it a bit differently! I saw that the two circles were touching tangentially. If you draw that tangent line, you'll see that since the box is a square, that tangent line must divide the box into two equal slices. I decided to make the width of each half equal to x, so the area of the box was 4x^2. Most of the rest of the setup was the same, except y was set as 5-x. Solving for that was much easier :DD
I saw the figure thought the same thing! But is there a generalized proof or property, or a way of proving this (that the square is divided into equal parts)? I kinda fell short on that.
interesting that you did Pythagoras, I went for trig using the lines provided and a radius going to the centre of one of the circles to the corner of the square. since the square touches the circles and the tangent, the corners touch at the midpoint of the arc. This means that the corner of the square touches at a 45 degree angle and you can use 2 triangles using sin45 and the provided lengths to solve also
thanks for sharing. complicated by adding y for no reason, call square 2x by 2x and your y becomes 5-x, your 5-x becomes 5-2x and then you solve it in one step (by solving triangle of 5-x, 5-2x and 5.
That was a lot of math to help prove x and y, supported by it meaning y=4 so that initial small triangle with a hypotenuse of 5 is a 3-4-5 triangle. When I see a right triangle with hypotenuse of 5, I always check to see if that pythagorean triple counts or helps. Saves a fair deal of algebra, which is great for timed issues and quick math, not so much if you need to show the work like all this.
I got the same when I paused, but just used the quadratic formula to get the roots. I also threw away the mirror image and solved for a rectangle of width x/2 and height x. Keeps this 65 year old brain in shape. 😉
Me ha gustado bastante el problema. Yo lo resolví así: primero sistema de ecuaciones con cos(a)=1-x/10 Cos(90-a)=1-x/5 Elevas al cuadrado ambas ecuaciones y las sumas, y obtienes que 1=(1-x/10)^2 +(1-x/5)^2. Y de aquí llegas a la misma ecuación de segundo grado y listo...
A simpler way is to drop a perpendicular line from the intersection down to the baseline through the square. This line would be the same as the radius so 5m. This point at the baseline would be the centre of the baseline of the square and would form a triangle with the centre of the circle and would be going through the corner of the square. The length of this side can be calculated by the Pythagoras theorem. It would be approx 7m. Now the part of this line within the square ie from the angle of the square to the middle of the baseline side of the square would be 7 minus the radius which would be 2m. A line from the angle of the square to the centre of the side on the baseline would be the same length as the side of the square. Because of you were to consider a triangle within the square with its tip at the middle of the baseline of the square and one side formed by the top line of the square it would be an equilateral triangle since it would always be 60 degrees to reach the midpoint of the side of a square to the angle of the square. Since we know the length of one side of the equilateral triangle is 2m. So we know the side of the triangle that makes the top line of the square is also 2m. So area of 4 sq m I liked the more mathy explanation too.. I guess this was more geometric and intuitive to me. 😊
You cant assume the line intersects at the centre of the square. It works for this scenario but if the example was not to scale then your assumption would lead you to the wrong answer
@@darainsyed7515if the square is between two equal circles and two angles of the square touch the circles then wouldn't a tangent from the intersection always fall at the midpoint of the square?
This was satisfying to think through. I did it slightly differently. I thought of the circles as plotted on the quadratic plane. Think of the right most circle plotted as (x and y being any point of the circle not as in the video) 5 = sqrt((x-5)^2 + (y-5)^2) So one point on this circle must intersect with the top right point of the square. Notice the top right point of the square is such that y = 2x because the square's width is bisected by the y-axis. Substituting in: 5 = sqrt((x-5)^2 + (2x-5)^2) -> 0 = (x-5)(x-1) -> x = 5 and x = 1 or the points (1,2) the point we are interested in and (5,10) the top of the circle So then the area is 4 since y = 2x = 2 is the length of the square.
