I like them because I can solve them in my head by just considering the extreme case. I.e. assume s=0 for the first problem and assume y=0 for the second problem.
@@crosswingrobots eh, even though you could do that I guess, I'd argue that that'd be less of an authentic answer since it wouldn't actually prove that it would be equal to that.
I like how you can reverse the first problem, and you still have a challenging and interesting problem: “if the area of the rectangle is 25, what is the length of the tangent line?”
For the first problem, you can get the answer in a single step. Consider length as l and breadth as b. Now, if u use secant and tangent theorem, u directly get lxb=(length of tangent)^2=25 Edit: Its called power of point theorem (i forgot the name)
@@MisteribelIF (and this is a major “if”) you are able to assume that the answer is a constant, rather than being an equation in terms of r and/or s. Your solution shows that the area of the rectangle is 25 when s=0, but you have not demonstrated that the area remains 25 when s increases.
@verkuilb correct. I wrote my longer answer in the main comment thread, but I guess it's already buried deeply. It shows the Pythagoras approach on the hypothenuse and that r of the semicircle has no influence, since side 5 remains constant. But, if all you need is an answer, like on a test, I'd forego the proof and just give 25 (btw, the same works with diameter 1, or 2 etc, obviously).
Immediate answer to second problem, before seeing the rest of the video: 25 square centimeters. Because you can make both squares arbitrarily close to the same and the rectangle arbitrarily small without breaking any part of the statement of the problem.
Yeah but that's no proof. Because you assume that the solution has to be a constant. That's a plausible assumption, given that it's a puzzle and those usually have nice answers. But it's an assumption that's not allowed in a proof whatsoever.
@@2dark4noir The questions don't ask for a proof, just the area. Similarly they don't ask for the area in terms of anything else, such as the height of the rectangle. If the exact same problem with the same answer had a named variable I wouldn't have made the assumption the answer is fixed. For that matter on an exam I wouldn't make that assumption either, and would have gone through the full math. But I would first take the limit and note it, to get an idea of what sort of answer I am looking for.
@@corvididaecorax2991 I absolutely agree with the second paragraph. Saying that the question didn't ask for proof is weird to me. That's an implied necessity of mathematical methodology. You never actually answer such a question by just stating the solution and be done with it. If you do, you could've used a algebraic computer system in the first place. Also you can't hope that every problem comes along with named variables if you need some in the solution.
The problems that you solve always inspire me the most. I haven't got such interest in solving the problems before as now but after watching your videos and problems. Each problem that you speak in your videos are unique. Keep your flags higher by posting such videos.
25. Problem one. Since it seems to be reminiscent of an architectural application, I felt compelled to solve this on the drawing board, scaled it up for accuracy, guessed the quarter circle's radius at 3.5, working from there on in centimeters, instead of units, and looked out for a half-circle producing a tangent length of 5. Initially, on the first hastily done primary sketch, I got 24.85cm for the shaded area, further refining the drawing for accuracy, I got 25.55, and eventually 25.05, and after some arithmetic to corroborate, I got 25.025. I settled for 25.
For the rectangle + circular arcs: Taking shorter side as q and radius of semicircle as r: q·(q + 2r) = required area = q² + 2rq As the 5cm line is tangent to the semicircle, it forms a right angle with the radius at point of tangency. thus, (q + r) is the hypotenuse of a right triangle with legs 5 and r. 5² + r² = (q + r)² 5² + r² = q² + 2raq + r² 25 = q² + 2rq = required area. For the squares + rectangles: I think i cannot express this in-line.
what a beautiful result..! It's 5's square, 25. let the length of radius 'a' and 'b', (a is left one's) (a + b)² - b² = 5². using pitagoras. a² + 2ab = 25. and this is also the rectangle's area!
Wow, wow, wow! I was so pleased with myself for solving problem 2 with a convoluted solution involving coming up with the width and height of the rectangle based on angle theta and a cartesian coordinate. I put the origin at the lower left of the rectangle and set the x axis as the horizontal base of the rectangle and theta = the angle between the x axis and the NW side of the big square. Using the formula for a circle and the formula for the tangent to a circle, some trig, and a lot of work I finally got the width to be 5*(cos theta + sin theta) and the height to be 5/(cos theta + sin theta). Then I saw MYD's solution. I should have known there was a more elegant solution using geometry right under my nose!
These are the kinds of problems that (once you learn the answer and the optimum mechanics for deriving said answer) you wind up kicking yourself for, because they kind of trick you into thinking that you have to do more work in solving the given problem than you actually have to
Problema Nº1.- Radio del semicírculo =r. Altura del rectángulo azul =h→ : base del rectángulo =b =h+2r → Área rectángulo azul =b*h =(h+2r)h. Potencia del vértice inferior izquierdo del rectángulo respecto al semicírculo =5²=h(h+2r) =25 =Área rectángulo azul. Problema Nº2.- Respetando las condiciones del trazado original, podemos suponer que el lado del cuadrado grande es igual a la diagonal del cuadrado pequeño; de esa forma aplicando un giro de 45º, el rectángulo resultante también es un cuadrado → Área azul =2*(√25)²/2 =25. Gracias por tan sorprendentes acertijos. Un saludo cordial.
When you are familiar with basic algebra, these problem is just mechanical. It does take some time to do the calculation if you use algebra like this video does. But there is a MUCH simpler way: P1: Use Secant Theorem, takes one second to find the answer. P2: Easily converted to the same shape as P1
Love these math puzzles. They're interesting and I always have something to think about instead of getting bored. For the first problem, I just saw the thumbnail of the video and thought that r = s. After some simple algebra transformation stuff, I calculated the area to be 3 * (5 / sqrt(3))^2, which apparently gives 25. Thanks again for giving our brains something to chew on!
Since the 5cm line is a tangent to the external semicircle, draw the radius from the tangent to the centre of the semicircle. let this radius be r. Thee, let the radius of the inscribed quadrant be x. Then the length of the rectangle is (x+r+r) i.e (x+2r). By circle theorem, the angle between the tangent to a circle and the radius is a right angle. Therefore a right-angled triangle is formed between the tangent and the radius f the external semicircle. Using Pythagoras' rule, 5^2 + r^2=(x+r)^2. Simplify this equation. The resulting equation becomes: 25 = x^2 + 2xr 25 = x(x+2r). The right hand side expression is the area of the rectangle. So, area of shaded rectangle is 25cm2
The most elegant solution to problem 1: 1. Show that S is a ratio of R by constructing the radius R and S such that you form 2 similar, right angle triangles with the common 5cm side. 2. Therefore, the solution is true for any value of S:R while S >= 0. 3. The limit of R as S -> 0 is 5 and Base x Height approaches R^2 which is 25cm^2. This can also be shown as the limit of the angle Theta approaches 90 degrees, formed between the left side of the rectangle and the tangent. The vertices of the two triangles form the square! :)
im no good with math, and its my first time seeing these kind of problems. they feel incredibly cursed to me bcz i cant solve for a variable but i can find the answers. i love it. i now see how people can get hooked into it and have fun. thank you for the video, ive been enlightened
For the 2nd problem, my reasoning was that if there is a solution, it cannot depend on the individual dimensions of the square and the rectangle. So, I considered the case where the horizontal distance is a maximum leading to the figure shown at 9:25. It becomes obvious that the area is the same.
for 1st i solved r=radius of semicircle and R=radius of quarter circle area of rectangle = (2r+R)R ------1 draw radius of semicircle to point of contact by Pythagoras theorem r²+5²=(R+r)² =>25=R²+r²+2rR -r² 25=R²+2rR 25=R(R+2r) 25 sq units=area of rectangle ( by eqn 1)
Another way for problem 2 is to assign coordinates (0,0), u = (a,-b), v = (b, a) to the square and solve for c that makes the coordinates of v+cu equal. This gives one piece of info on the rectangle, use the fact u+v is on it to get the 2nd piece.
Problem 2: U can set up the equation system x(x+y)=25 x+y=5√2 -> [x, y ] = [ 5/√2 ≤ x ≤ 5 , 0 ≤ y ≤ 5/√2 ] OR if the area(A) is completly unknown x = √A/√2 ≤ x ≤ √A y = 0 ≤ y ≤ √A/√2
I loved that first problem! It fell out in about 2 minutes, but it was quite a surprise to me! :)) I cheated in the second problem by noting that different sizes of rectangle were possible depending on the angle between the square and rectangle; then assuming there WAS an invariant answer, and then taking the limit case as the rectangle became almost the square, and the area approaches 25. So the area MUST be 25. Of course this approach COULD be used in the first problem too...
They are, basically, the same problem, afterall: the first one involving circle fragments embedded within a rectangle; and, the second one involving a square embedded within a rectangle
What I find very fascinating about problem #1, is the realisation that it is possible, given any random rectangle, to construct (using a straightedge and compass) a line segment, the length of which, when squared, results in the area of the rectangle. I.e. somehow, through such construction, the two independent dimensions of the rectangle (height and width) that define its area, 'collapse' to a single dimension (the length of the line segment) that can be squared to get the area of the rectangle. Makes me wonder - is it possible to do something similar in 3D. I.e. given a cuboid, somehow construct a line, which when cubed, results in the volume of the cuboid? Did anyone see anything like this?
I have such an intense love/hate relationship with problems like these where you can solve it without finding out all the values by cutting corners. They're so cool and at the same time it feels like they're cheating
The first problem was fairly easy. I had a STRONG hunch on the second one about the answer, but couldn’t prove it. I did notice the limit as b->0 though. What a crazy solution!
Problem 1: There is a thing called the "tangent-secant theorem". It states, that if you have a point outside a circle, the tangent to the edge of the circle squared equals the distance to the circle on a secant multiplied by the distance to the far end of the circle on the same secant. In this case, the tangent is 5, the first secant factor is the radius of the quarter circle - and therefore the height of the rectangle, the other factor is the entire length from left to right - and therefore the width of the rectangle. In other words, (5cm)² = 25cm² = the area of the rectangle. Problem 2: The square is 25cm², as such the side is 5cm. All the triangles formed by the square and the rectangles share - by necessity - all their angles, and are therefore what is called "similar". Similar triangles have the same ratios between their sides. Also all of the triangles have a right angle and therefore the Pythagorean theorem can be used on all of them. Those two facts, coupled with fact, that some of the sides of the triangles add up to sides of other triangles, gives a whole lot of equations, that can be combined to solve the problem... I am just to lazy to do it. 🤣
Answer to first problem: Radius of semi circle=s, radius of quarter circle=r Area of rectangle=(2s+r)*r =(2s+r)*r =2sr+r^2 As tangent line is always perpendicular to radius, we can use Pythagorean theorem: 5^2+s^2=(s+r)^2 25=2rs+r^2 Thus, area of rectangle=2rs+r^2 =25 2) For these questions, I first like to answer a specific case that is not technically forbidden by the question. For this one, I choose the case that the width of the rectangle is 0. Therefore, the upper right corner of the 25cm^2 goes through the upper right point of the square. This fulfills all the necessary requirements. Thus, the square and the are we are looking for overlap each other perfectly. Therefore, the area of the square and rectangle together is 25 cm^2. Now, this is not a rigorous proof. However, it is a shortcut to finding the answer we are looking for overall, and useful for Multiple Choice Questions. (You can also do this with the situation where the width of the rectangle is the same as the square, and find the same answer, showing that the answer is probably not variable since it holds for both extremes of the rectangles width.) I can’t work out a more rigorous proof though… I’ll watch the video to find that out I guess.
The first one is not hard at all to be fair, the second one is a little harder, but also doable. At least so I find them. But still, I like the video, you never disappoint. It's always a little bit of mind workout so that it doesn't get rusty ;D
Easier solution for the first problem is to label r+s as x. Height is x - s, length is x + s, and (x+s)(x-s) is x^2 - s^2. Wait, seems familiar. Is there not a right triangle there to find c^2 - a^2 = b^2 ?
I solved the second one completely differently. Let us suppose that the left side of the 5x5 square forms an angle θ with the base of the shaded rectangle. Let "s" be sin(θ) and "c" be cos(θ). Assuming that the lower-left corner of the shaded rectangle has coordinates (0, 0), then the coordinates of the top-left corner of the 5x5 square are (5c, 5s), and the points along the top of the 5x5 square have coordinates (5c + ns, 5s - nc), where "n" is the distance from the upper-left corner. We can use the coordinates (5c + ns, 5s - nc) to find the width and height of the shaded rectangle. The width will be the x-coordinate where n=5 (the upper-right corner of the 5x5 square), which is 5(s+c). The height will be the y-coordinate where the top of the 5x5 square intersects the shaded rectangle, which happens where it intersects the line x=y. Setting 5c+ns = 5s-nc and solving for n, we find that this intersection happens where n = 5(s-c)/(s+c). Substituting this value of "n" into 5s-nc, we get 5(s^2 + c^2)/(s + c), which further reduces to 5/(s+c). Multiplying the width 5(s+c) times the height 5/(s+c), we're left with 25.
The first question was easy-peasy. The second question is a LOT harder. I'm still working on it. You can tell the 2nd question is a LOT harder just by how Presh presents it. The first question, he just says who submitted it, but the second one he gives not only the submitter BUT also a solution by someone else. Presh wouldn't do that if he figured it out on hs own. Human nature. So, if you get the second problem, apparently you might have gotten a problem that Presh didn't and there's always some kind of satisfaction in that.
Problem 1: Draw a radius from the center of the semicircle to the point of tangency of the 5cm line segment. Let r be the semicircle radius and R be the quarter circle radius. Note that the height of the rectangle is R and the width is R+2r r² + 5² = (R+r)² r² + 25 = R² + 2Rr + r² R² + 2Rr = 25 R(R+2r) = 25 cm² Problem 2: The side length of the large rotated square is √25 = 5. Let the side length of the smaller square be s and the width of the rectangle be w. Note that the area of the square and rectangle will be s(s+w). Draw two diagonals from the point of rotation, one to the opposite vertex of the large square (where it meets the right side of the rectangle), and one to the opposite vertex of the smaller square (where it meets the intersection point between the square and rectangle. This will create two right triangles, with the diagonals as the hypotenuses. For the larger right triangle, the long leg is s+w and the hypotenuse is 5√2. For the smaller triangle, the long leg is 5 and the hypotenuse is √2s. As both squares have a right angle at the rotation point, the angle between the left sides of the two squares and the angle between the bottom sides of the two squares are the same. As the angle btween the left side of the smaller square and its diagonal and the angle between the bottom side of the larger square and its diagonal are both 45°, the two triangles share the same smaller angle, and so the triangles are similar. (s+w)/5√2 = 5/√2s √2s(s+w) = 25√2 s(s+w) = 25 cm²
Hey, Presh! I have a good challenge! Can you determine how the banker from 'Deal or No Deal' makes the offers? Their offers are never the average, and I want to see how your train of thought would be if you were the banker in the game show.
Not seen the vid yet, but problem #2: if we make the rectangle approach zero width, the tilted square and the base square overlap exactly. It follows that the rectangle + base square always equal the tilted square: 25cm². Same can be done by making the rectangle the max size: a square, where the hypothenuse of its diagonal has length 5 (tilted square is now tilted 45°). Edit: I see now that the vid took the long route around, which is also interesting to see.
@davidwillis7991 true. But the question implies an answer, not a range. If on a test, I'd go for the fast deduction, unless there's ample time. But I agree it's more fun to find out why it is the way it is.
So you just have two different ways of measuring the same thing between the two problems. The second one I already knew as I had that problem back in geometry. The first one I figured was something similar when I notice there was a tangent relationship with the 5 cm. In the end you have just two methods to solve the same problem because you know the side of the large square and the tangent line are the same.
In this case, for the second problem, i used the principle shown at the very end of the video to solve the question at first glance. Since IF there would be any possibility to determine the area of the rectangle from that of the - ARBITRARILY - rotated square, that area MUST be INDEPENDENT from the rotation, thus always equal to what we get with no rotation and thus with no "attached rectangle".
If the answer is independent of a variable (such as the semicircle radius in the first and the rectangle width in the second), setting that variable to zero simplifies the problem.
Interesting: the answer to the second was obvious to me - I could just 'see it'. unfortunately I think 'they will always have the same total area as the square rotates - because they just do...' is not acceptable. Is it odd that I thought of this problem as liquid filling up containers - containers that changed dimensions ( with the liquid volume remaining constant ) couldn't see the answer to the first one in the same way though - even though they are actually the same problem. I guess my brain can spatially fill in a rotating square to elongating rectangle - but cant do it with circles and tangents - as the 'square' is invisible here. I guess what my brain did automatically was solve it for the 0 case and solve it for the 90 degree case and realise the transition was linear? genuinely good puzzle there - especially as they are actually both the same.
From the thumbnail, I (wrongly) assumed the radius of each arc is the same. With that assumption, I calculated the area of the rectangle to be 25. That calculation was trivial (I did it in my head). Intrigued to discover, on watching the start of the video, that the radii were different, I repeated the calculation for hypothetical situations where the radius of the large semicircle was a multiple of the radius of the small circle. The area always calculates to 25 (which chimes with the second part of the video). Interesting problem. thanks.
I see both of these as minimal-information problems, for which the paucity of information is our friend. Spoiler alert. Both problems have the same answer, which is 25 cm². For the first, the semicircle can be imagined, without loss of generality, as having a radius of zero. For the second, the partially rotated square can be imagined as either corresponding to the smaller square, or rotated to 45 degrees, so that its uppermost vertex corresponds to the top-right vertex of the smaller square. I'm not saying that these are the best ways of approaching the problems, but they are certainly the fastest.
I wonder if it would be "proper" in such questions, knowing all varients yield the same solution, to just choose the simplest? eg in the 2nd question choose angle=0°.
Only if you are able to assume the answer is constant-which is hinted at in the problem statement, but not overtly stated. Without that, to have a proper solution, you have to either demonstrate that the answer does not depend on the variables, or if it does, you have to define that dependency.
What are the programs you use in creating such videos ? Do you use powerpoint then you present this on a video ? I think powerpoint is not enough, specially when that moment of limiting S to zero came. What is the software you use ? Thanks. ❤
I'm not bad at math but I'm also not good. On the second problem I tried to find the missing angles. So because I knew the square angles that meant the first triangle angle was "a" and the other was "90-a' after going around all angles are proven to be "a" or "90-a" . My problem arrived when I was met with the formula 90-90-a=90. I always end with the angle equally zero. Does this some how prove the rectangle area equals the square?!?
dear & respected sir, I still have objections about some problems. So in the problem with squares. I tried to build it and despite many attempts I failed. I would be grateful if you would be so kind as to suggest me a construction solution for the shapes. Thank you very much!
Math contest trick brain took over for problem 1: clearly the radii of the quarter and semicircle aren't fixed so the answer has to be the same for any pair of radii, just assume the semicircle has radius 0 then it's just a square with side length 5
I dont know how, for the first question I did I found out length of the quarter circle in terms of the tangent and then using quadratic formula twice along the way, I came at the correct answer
The second problem seems to have been represented. The tilted square seems to have to be larger than the square connected to the rectangle. The problem implies that they are the same size. Am I the only person who seems to have a difficulty?
So you're telling me I did like 15 lines of working and already had the answer on line 6 😅, lol me going to lengths to find out r and s, and only ending with equations where they cancel out completely and im left with only numbers and no letters lol, that 25 was in most lines tho
In the first problem, I joint the point at which tangency meets and the centre of the semicircle, next i used the Pythagorean theorem and Sine Rule in that triangle and got the radius of both the circles and henceforth got the answer, But yes, it seemed impossible at the first glance 😅
@@verkuilb In the 2nd question it was, but in the first question it is fixed, because in you increase or decrease size of the semicircle, the length of tangent will increase and decrease respectively; but in the question it is given 5cm fixed
@@add1ct683 Incorrect. In the video, rewatch the animation at the end of the solution of the first problem. It demonstrates that if you keep the tangent line fixed at 5 but rotate its angle, the radii of BOTH arcs change, the length and width of the rectangle co, but the area of the rectangle remains constant at 25.
Why can't You just start with simplifing the shapes (semicircle s->0 so r->5, rectangle y->0), and just skip all the calculations? Isn't it valid approach?
Only if you are able to assume that the answers are constant. The problem statements sort of hint at that, but don’t really overtly state that, so the semantics can be argued both ways. Your proposals do indeed determine that the area in the first problem is 25 when the semicircle radius is zero, and that the area in the second problem is zero when the angle of rotation is zero-but it doesn’t prove that those values are constant, independent of the radii in the first problem or the angle in the second.
For the second problem, I figured out the solution had to be 25, without doing any of the algebra. I just visualized rotating the given area=25 square, and the effect that this would have on the size and shape of the smaller square and rectangle-and realized that as you did that, the smaller square would increase in size until it limited out at 25 sq cm, and the rectangle would diminish in size until it simultaneously limited out at zero. (This is demonstrated in the final seconds of the video-but it’s not presented as a means to solve the puzzle. It should be.) Note that this method does not necessarily prove that the total combined area of the small square and rectangle is constant-but that is stated by the phrasing of the problem statement, so there is no need to prove this. Actually, the same method can be used to solve the first one as well-although I’ll admit I didn’t realize that until after the fact.
![Lp. Сердце Вселенной #60 РОЖДЕНИЕ ЛОЛОЛОШКИ [Финал] • Майнкрафт](http://i.ytimg.com/vi/YoR0pAV9FVQ/mqdefault.jpg)
I always love these problems where the answer is static, even though many of the seemingly critical variables remain variable.
Me too
I like them because I can solve them in my head by just considering the extreme case. I.e. assume s=0 for the first problem and assume y=0 for the second problem.
@@crosswingrobots eh, even though you could do that I guess, I'd argue that that'd be less of an authentic answer since it wouldn't actually prove that it would be equal to that.
@@lox7182 I agree that it feels like a cheap trick. But it would save a bunch of time and get you full marks on a multiple choice test.
Me to 0:43
I like how you can reverse the first problem, and you still have a challenging and interesting problem: “if the area of the rectangle is 25, what is the length of the tangent line?”
I like how his voice gets higher overtime when we're so close into getting the answer, iconic.
AND THAT'S THE ANSWER, PRESH!!😊
I had to LOL at that comment. Classic Nerd effect.
4:37 here I have used the tangent secant relation to find the area
For the first problem, you can get the answer in a single step. Consider length as l and breadth as b. Now, if u use secant and tangent theorem, u directly get lxb=(length of tangent)^2=25
Edit: Its called power of point theorem (i forgot the name)
Cool. Or just use Diameter 0 for the semicircle, which doesn't require any trigonometry knowledge, and gives 5x5 instantly.
@@MisteribelIF (and this is a major “if”) you are able to assume that the answer is a constant, rather than being an equation in terms of r and/or s. Your solution shows that the area of the rectangle is 25 when s=0, but you have not demonstrated that the area remains 25 when s increases.
@verkuilb correct. I wrote my longer answer in the main comment thread, but I guess it's already buried deeply. It shows the Pythagoras approach on the hypothenuse and that r of the semicircle has no influence, since side 5 remains constant.
But, if all you need is an answer, like on a test, I'd forego the proof and just give 25 (btw, the same works with diameter 1, or 2 etc, obviously).
best solution of the problem
Actually, you could say, the solution to this problem explains why the power of point theorem works in this type of tangent case in the first place
Immediate answer to second problem, before seeing the rest of the video: 25 square centimeters. Because you can make both squares arbitrarily close to the same and the rectangle arbitrarily small without breaking any part of the statement of the problem.
You can do the same method on the first problem. Looking for limits and extremes is always a good way to start a problem.
@@clayton97330
Yep. I just didn't spot that one before watching the rest of the video. For some reason it was less obvious to me.
Yeah but that's no proof. Because you assume that the solution has to be a constant. That's a plausible assumption, given that it's a puzzle and those usually have nice answers. But it's an assumption that's not allowed in a proof whatsoever.
@@2dark4noir
The questions don't ask for a proof, just the area. Similarly they don't ask for the area in terms of anything else, such as the height of the rectangle. If the exact same problem with the same answer had a named variable I wouldn't have made the assumption the answer is fixed.
For that matter on an exam I wouldn't make that assumption either, and would have gone through the full math. But I would first take the limit and note it, to get an idea of what sort of answer I am looking for.
@@corvididaecorax2991 I absolutely agree with the second paragraph.
Saying that the question didn't ask for proof is weird to me. That's an implied necessity of mathematical methodology. You never actually answer such a question by just stating the solution and be done with it. If you do, you could've used a algebraic computer system in the first place. Also you can't hope that every problem comes along with named variables if you need some in the solution.
The problems that you solve always inspire me the most. I haven't got such interest in solving the problems before as now but after watching your videos and problems. Each problem that you speak in your videos are unique. Keep your flags higher by posting such videos.
25. Problem one. Since it seems to be reminiscent of an architectural application, I felt compelled to solve this on the drawing board, scaled it up for accuracy, guessed the quarter circle's radius at 3.5, working from there on in centimeters, instead of units, and looked out for a half-circle producing a tangent length of 5.
Initially, on the first hastily done primary sketch, I got 24.85cm for the shaded area, further refining the drawing for accuracy, I got 25.55, and eventually 25.05, and after some arithmetic to corroborate, I got 25.025. I settled for 25.
I like how he gets really excited when he arrives at the answer
Well, _I_ got really excited when _I_ found the answer!!!! ;)
I would be to :)
For the rectangle + circular arcs:
Taking shorter side as q and radius of semicircle as r:
q·(q + 2r) = required area = q² + 2rq
As the 5cm line is tangent to the semicircle, it forms a right angle with the radius at point of tangency.
thus, (q + r) is the hypotenuse of a right triangle with legs 5 and r.
5² + r² = (q + r)²
5² + r² = q² + 2raq + r²
25 = q² + 2rq = required area.
For the squares + rectangles:
I think i cannot express this in-line.
Ummm…yes…but why are you simply repeating the same solution he presented in the video?
@@verkuilb oh right I should have clarified, I typed this out _before_ watching the video
what a beautiful result..! It's 5's square, 25.
let the length of radius 'a' and 'b', (a is left one's)
(a + b)² - b² = 5². using pitagoras.
a² + 2ab = 25. and this is also the rectangle's area!
Pythagoras. But indeed, exactly that 😊.
Wow, wow, wow! I was so pleased with myself for solving problem 2 with a convoluted solution involving coming up with the width and height of the rectangle based on angle theta and a cartesian coordinate. I put the origin at the lower left of the rectangle and set the x axis as the horizontal base of the rectangle and theta = the angle between the x axis and the NW side of the big square. Using the formula for a circle and the formula for the tangent to a circle, some trig, and a lot of work I finally got the width to be 5*(cos theta + sin theta) and the height to be 5/(cos theta + sin theta). Then I saw MYD's solution. I should have known there was a more elegant solution using geometry right under my nose!
These are the kinds of problems that (once you learn the answer and the optimum mechanics for deriving said answer) you wind up kicking yourself for, because they kind of trick you into thinking that you have to do more work in solving the given problem than you actually have to
I love how stragely uplifting your videos are!
Problema Nº1.- Radio del semicírculo =r. Altura del rectángulo azul =h→ : base del rectángulo =b =h+2r → Área rectángulo azul =b*h =(h+2r)h.
Potencia del vértice inferior izquierdo del rectángulo respecto al semicírculo =5²=h(h+2r) =25 =Área rectángulo azul.
Problema Nº2.- Respetando las condiciones del trazado original, podemos suponer que el lado del cuadrado grande es igual a la diagonal del cuadrado pequeño; de esa forma aplicando un giro de 45º, el rectángulo resultante también es un cuadrado → Área azul =2*(√25)²/2 =25.
Gracias por tan sorprendentes acertijos. Un saludo cordial.
When you are familiar with basic algebra, these problem is just mechanical. It does take some time to do the calculation if you use algebra like this video does.
But there is a MUCH simpler way:
P1: Use Secant Theorem, takes one second to find the answer.
P2: Easily converted to the same shape as P1
Love these math puzzles. They're interesting and I always have something to think about instead of getting bored.
For the first problem, I just saw the thumbnail of the video and thought that r = s.
After some simple algebra transformation stuff, I calculated the area to be
3 * (5 / sqrt(3))^2, which apparently gives 25.
Thanks again for giving our brains something to chew on!
Since the 5cm line is a tangent to the external semicircle, draw the radius from the tangent to the centre of the semicircle. let this radius be r. Thee, let the radius of the inscribed quadrant be x. Then the length of the rectangle is (x+r+r) i.e (x+2r). By circle theorem, the angle between the tangent to a circle and the radius is a right angle. Therefore a right-angled triangle is formed between the tangent and the radius f the external semicircle. Using Pythagoras' rule, 5^2 + r^2=(x+r)^2. Simplify this equation. The resulting equation becomes: 25 = x^2 + 2xr
25 = x(x+2r). The right hand side expression is the area of the rectangle. So, area of shaded rectangle is 25cm2
Oh man! Absolutely love it when he's close to the answer and gets louder progressively... Check 04:18
The most elegant solution to problem 1:
1. Show that S is a ratio of R by constructing the radius R and S such that you form 2 similar, right angle triangles with the common 5cm side.
2. Therefore, the solution is true for any value of S:R while S >= 0.
3. The limit of R as S -> 0 is 5 and Base x Height approaches R^2 which is 25cm^2. This can also be shown as the limit of the angle Theta approaches 90 degrees, formed between the left side of the rectangle and the tangent. The vertices of the two triangles form the square! :)
im no good with math, and its my first time seeing these kind of problems. they feel incredibly cursed to me bcz i cant solve for a variable but i can find the answers. i love it. i now see how people can get hooked into it and have fun. thank you for the video, ive been enlightened
For the 2nd problem, my reasoning was that if there is a solution, it cannot depend on the individual dimensions of the square and the rectangle. So, I considered the case where the horizontal distance is a maximum leading to the figure shown at 9:25. It becomes obvious that the area is the same.
for 1st i solved
r=radius of semicircle and R=radius of quarter circle
area of rectangle = (2r+R)R ------1
draw radius of semicircle to point of contact
by Pythagoras theorem
r²+5²=(R+r)²
=>25=R²+r²+2rR -r²
25=R²+2rR
25=R(R+2r)
25 sq units=area of rectangle ( by eqn 1)
Another way for problem 2 is to assign coordinates (0,0), u = (a,-b), v = (b, a) to the square and solve for c that makes the coordinates of v+cu equal. This gives one piece of info on the rectangle, use the fact u+v is on it to get the 2nd piece.
Algebraically we can get a solution as r = 1 and s =12 which also gives the area of the rectangle as 25
Problem 2:
U can set up the equation system
x(x+y)=25
x+y=5√2
-> [x, y ] = [ 5/√2 ≤ x ≤ 5 , 0 ≤ y ≤ 5/√2 ]
OR
if the area(A) is completly unknown
x = √A/√2 ≤ x ≤ √A
y = 0 ≤ y ≤ √A/√2
I loved that first problem! It fell out in about 2 minutes, but it was quite a surprise to me! :))
I cheated in the second problem by noting that different sizes of rectangle were possible depending on the angle between the square and rectangle; then assuming there WAS an invariant answer, and then taking the limit case as the rectangle became almost the square, and the area approaches 25. So the area MUST be 25. Of course this approach COULD be used in the first problem too...
They are, basically, the same problem, afterall: the first one involving circle fragments embedded within a rectangle; and, the second one involving a square embedded within a rectangle
What I find very fascinating about problem #1, is the realisation that it is possible, given any random rectangle, to construct (using a straightedge and compass) a line segment, the length of which, when squared, results in the area of the rectangle.
I.e. somehow, through such construction, the two independent dimensions of the rectangle (height and width) that define its area, 'collapse' to a single dimension (the length of the line segment) that can be squared to get the area of the rectangle.
Makes me wonder - is it possible to do something similar in 3D. I.e. given a cuboid, somehow construct a line, which when cubed, results in the volume of the cuboid? Did anyone see anything like this?
Really like the animations, Presh!
for the second problem i considered angle a is 30 degrees that doesn't change the area value, then solved it with some algebraic methods.
I have such an intense love/hate relationship with problems like these where you can solve it without finding out all the values by cutting corners. They're so cool and at the same time it feels like they're cheating
The first problem was fairly easy. I had a STRONG hunch on the second one about the answer, but couldn’t prove it. I did notice the limit as b->0 though. What a crazy solution!
changing the shape to simplify is a very neat trick!
Problem 1:
There is a thing called the "tangent-secant theorem".
It states, that if you have a point outside a circle, the tangent to the edge of the circle squared equals the distance to the circle on a secant multiplied by the distance to the far end of the circle on the same secant.
In this case, the tangent is 5, the first secant factor is the radius of the quarter circle - and therefore the height of the rectangle, the other factor is the entire length from left to right - and therefore the width of the rectangle.
In other words, (5cm)² = 25cm² = the area of the rectangle.
Problem 2:
The square is 25cm², as such the side is 5cm.
All the triangles formed by the square and the rectangles share - by necessity - all their angles, and are therefore what is called "similar".
Similar triangles have the same ratios between their sides. Also all of the triangles have a right angle and therefore the Pythagorean theorem can be used on all of them. Those two facts, coupled with fact, that some of the sides of the triangles add up to sides of other triangles, gives a whole lot of equations, that can be combined to solve the problem... I am just to lazy to do it. 🤣
Answer to first problem:
Radius of semi circle=s, radius of quarter circle=r
Area of rectangle=(2s+r)*r
=(2s+r)*r
=2sr+r^2
As tangent line is always perpendicular to radius, we can use Pythagorean theorem:
5^2+s^2=(s+r)^2
25=2rs+r^2
Thus, area of rectangle=2rs+r^2
=25
2)
For these questions, I first like to answer a specific case that is not technically forbidden by the question. For this one, I choose the case that the width of the rectangle is 0.
Therefore, the upper right corner of the 25cm^2 goes through the upper right point of the square. This fulfills all the necessary requirements.
Thus, the square and the are we are looking for overlap each other perfectly.
Therefore, the area of the square and rectangle together is 25 cm^2.
Now, this is not a rigorous proof. However, it is a shortcut to finding the answer we are looking for overall, and useful for Multiple Choice Questions. (You can also do this with the situation where the width of the rectangle is the same as the square, and find the same answer, showing that the answer is probably not variable since it holds for both extremes of the rectangles width.)
I can’t work out a more rigorous proof though… I’ll watch the video to find that out I guess.
For the first problem, Use power of point theorem where 5^2 = r*(r +2s)
Beautiful problems and solutions
The first one is not hard at all to be fair, the second one is a little harder, but also doable. At least so I find them. But still, I like the video, you never disappoint. It's always a little bit of mind workout so that it doesn't get rusty ;D
Easier solution for the first problem is to label r+s as x. Height is x - s, length is x + s, and (x+s)(x-s) is x^2 - s^2. Wait, seems familiar. Is there not a right triangle there to find c^2 - a^2 = b^2 ?
the first one is from the geometry text book. Intersecting secants theorem
I solved the second one completely differently. Let us suppose that the left side of the 5x5 square forms an angle θ with the base of the shaded rectangle. Let "s" be sin(θ) and "c" be cos(θ). Assuming that the lower-left corner of the shaded rectangle has coordinates (0, 0), then the coordinates of the top-left corner of the 5x5 square are (5c, 5s), and the points along the top of the 5x5 square have coordinates (5c + ns, 5s - nc), where "n" is the distance from the upper-left corner.
We can use the coordinates (5c + ns, 5s - nc) to find the width and height of the shaded rectangle. The width will be the x-coordinate where n=5 (the upper-right corner of the 5x5 square), which is 5(s+c). The height will be the y-coordinate where the top of the 5x5 square intersects the shaded rectangle, which happens where it intersects the line x=y. Setting 5c+ns = 5s-nc and solving for n, we find that this intersection happens where n = 5(s-c)/(s+c). Substituting this value of "n" into 5s-nc, we get 5(s^2 + c^2)/(s + c), which further reduces to 5/(s+c).
Multiplying the width 5(s+c) times the height 5/(s+c), we're left with 25.
The first question was easy-peasy. The second question is a LOT harder. I'm still working on it.
You can tell the 2nd question is a LOT harder just by how Presh presents it. The first question, he just says who submitted it, but the second one he gives not only the submitter BUT also a solution by someone else. Presh wouldn't do that if he figured it out on hs own. Human nature. So, if you get the second problem, apparently you might have gotten a problem that Presh didn't and there's always some kind of satisfaction in that.
Problem 1:
Draw a radius from the center of the semicircle to the point of tangency of the 5cm line segment. Let r be the semicircle radius and R be the quarter circle radius. Note that the height of the rectangle is R and the width is R+2r
r² + 5² = (R+r)²
r² + 25 = R² + 2Rr + r²
R² + 2Rr = 25
R(R+2r) = 25 cm²
Problem 2:
The side length of the large rotated square is √25 = 5. Let the side length of the smaller square be s and the width of the rectangle be w. Note that the area of the square and rectangle will be s(s+w).
Draw two diagonals from the point of rotation, one to the opposite vertex of the large square (where it meets the right side of the rectangle), and one to the opposite vertex of the smaller square (where it meets the intersection point between the square and rectangle. This will create two right triangles, with the diagonals as the hypotenuses. For the larger right triangle, the long leg is s+w and the hypotenuse is 5√2. For the smaller triangle, the long leg is 5 and the hypotenuse is √2s.
As both squares have a right angle at the rotation point, the angle between the left sides of the two squares and the angle between the bottom sides of the two squares are the same. As the angle btween the left side of the smaller square and its diagonal and the angle between the bottom side of the larger square and its diagonal are both 45°, the two triangles share the same smaller angle, and so the triangles are similar.
(s+w)/5√2 = 5/√2s
√2s(s+w) = 25√2
s(s+w) = 25 cm²
First question could easily be done by the Tangent Secant theorem
r×l = 5²
Area = 25
The first one is solved easily using secant and tangent theorem. a * b = 5 ^ 2 -> a * b = 25.
Hey, Presh! I have a good challenge! Can you determine how the banker from 'Deal or No Deal' makes the offers?
Their offers are never the average, and I want to see how your train of thought would be if you were the banker in the game show.
Since the tangent to the semi-circle is 5. r( r+ 2s ) = 5^2 =25.
the most fun part is when you do something seemingly unrelated but the answer just suddenly hits you like a truck
Not seen the vid yet, but problem #2: if we make the rectangle approach zero width, the tilted square and the base square overlap exactly. It follows that the rectangle + base square always equal the tilted square: 25cm².
Same can be done by making the rectangle the max size: a square, where the hypothenuse of its diagonal has length 5 (tilted square is now tilted 45°).
Edit: I see now that the vid took the long route around, which is also interesting to see.
All this shows is that is the area is a constant it must be 25. If you haven't proven that the area is a constant then you haven't proven anything.
@davidwillis7991 true. But the question implies an answer, not a range. If on a test, I'd go for the fast deduction, unless there's ample time. But I agree it's more fun to find out why it is the way it is.
So you just have two different ways of measuring the same thing between the two problems. The second one I already knew as I had that problem back in geometry. The first one I figured was something similar when I notice there was a tangent relationship with the 5 cm. In the end you have just two methods to solve the same problem because you know the side of the large square and the tangent line are the same.
In this case, for the second problem, i used the principle shown at the very end of the video to solve the question at first glance. Since IF there would be any possibility to determine the area of the rectangle from that of the - ARBITRARILY - rotated square, that area MUST be INDEPENDENT from the rotation, thus always equal to what we get with no rotation and thus with no "attached rectangle".
If the answer is independent of a variable (such as the semicircle radius in the first and the rectangle width in the second), setting that variable to zero simplifies the problem.
Problem 1 can be easily solved using tangent secant property as tangent^2 = r(r+s)
The res area is 25!. After 2 years of practicing extremely difficult olympiad math. Certain areas are finally becoming easier.😊
Interesting: the answer to the second was obvious to me - I could just 'see it'. unfortunately I think 'they will always have the same total area as the square rotates - because they just do...' is not acceptable.
Is it odd that I thought of this problem as liquid filling up containers - containers that changed dimensions ( with the liquid volume remaining constant )
couldn't see the answer to the first one in the same way though - even though they are actually the same problem.
I guess my brain can spatially fill in a rotating square to elongating rectangle - but cant do it with circles and tangents - as the 'square' is invisible here.
I guess what my brain did automatically was solve it for the 0 case and solve it for the 90 degree case and realise the transition was linear?
genuinely good puzzle there - especially as they are actually both the same.
From the thumbnail, I (wrongly) assumed the radius of each arc is the same. With that assumption, I calculated the area of the rectangle to be 25. That calculation was trivial (I did it in my head). Intrigued to discover, on watching the start of the video, that the radii were different, I repeated the calculation for hypothetical situations where the radius of the large semicircle was a multiple of the radius of the small circle. The area always calculates to 25 (which chimes with the second part of the video). Interesting problem. thanks.
I love ❤ math problems tooooo much and the first one was amazing and its solution was excellently amazing but the last problem was also not bad.
I also preferred the first one, because it looks as if it's going to be very hard, but the answer comes out easily.
I see both of these as minimal-information problems, for which the paucity of information is our friend.
Spoiler alert.
Both problems have the same answer, which is 25 cm².
For the first, the semicircle can be imagined, without loss of generality, as having a radius of zero.
For the second, the partially rotated square can be imagined as either corresponding to the smaller square, or rotated to 45 degrees, so that its uppermost vertex corresponds to the top-right vertex of the smaller square.
I'm not saying that these are the best ways of approaching the problems, but they are certainly the fastest.
0:43
Hi. Excellent video, as always. How did you create the animations for the video?
let quarter circle radius=h, semi-circle radius=r =>h^2+2hr=area... now since 5^2+r^2=(h+r)^2....=> 25=h^2+2hr=area
Thank you very much for this interesting video. Might I ask what tool you used to create the animation near the end of the video?
Cam somebody tell me why such a simple question is met with silence? Reminds me of the mean-spiritedness of Indian business proprietors.
Never mind. I did it with Geogebra: i.imgur.com/0sZ0REI.jpeg
Never mind. I did it with Geogebra: i . imgur . com / 0sZ0REI . jpeg
The best animation!
I wonder if it would be "proper" in such questions, knowing all varients yield the same solution, to just choose the simplest? eg in the 2nd question choose angle=0°.
Only if you are able to assume the answer is constant-which is hinted at in the problem statement, but not overtly stated. Without that, to have a proper solution, you have to either demonstrate that the answer does not depend on the variables, or if it does, you have to define that dependency.
What are the programs you use in creating such videos ? Do you use powerpoint then you present this on a video ? I think powerpoint is not enough, specially when that moment of limiting S to zero came. What is the software you use ? Thanks. ❤
Animations of limit are very interesting
I solved the first one in a few minutes without pen and paper. Very happy :D
i solved it(prob.1) at my note book within 5 min.
I'm not bad at math but I'm also not good. On the second problem I tried to find the missing angles. So because I knew the square angles that meant the first triangle angle was "a" and the other was "90-a' after going around all angles are proven to be "a" or "90-a" . My problem arrived when I was met with the formula 90-90-a=90. I always end with the angle equally zero. Does this some how prove the rectangle area equals the square?!?
dear & respected sir, I still have objections about some problems. So in the problem with squares. I tried to build it and despite many attempts I failed. I would be grateful if you would be so kind as to suggest me a construction solution for the shapes. Thank you very much!
length = w + 2 b
5^2 = (w + b)^2 - b^2 = 2 w b + w^2 =w (w + 2 b)=AREA OF RECTANGLE
Brilliant. Thanks man.
The first one I got no problem but the second one was kinda hard I didn’t even think to make a the first triangle you made
4:15 oh good lord screw you! lol
Math contest trick brain took over for problem 1: clearly the radii of the quarter and semicircle aren't fixed so the answer has to be the same for any pair of radii, just assume the semicircle has radius 0 then it's just a square with side length 5
If it's not too complicated, When you do animations, I wish you could leave link to just the animation portion.
I dont know how, for the first question I did I found out length of the quarter circle in terms of the tangent and then using quadratic formula twice along the way, I came at the correct answer
How exciting
I just did this based on the opening diagram, and my reaction is: HOLY SH*T!
I solved the first one in 3 minutes and had absolutely no idea how to do the second one lol
No you didn’t
Same lol
The elegance of the solutions here almost made me cry 🥹
Why were you gone this long?
For the first time I solved both in 1min. 😂 I think I am starting to learn something here
Good questions
How exiting!
Shaded area=25 cm^2
The second problem seems to have been represented. The tilted square seems to have to be larger than the square connected to the rectangle. The problem implies that they are the same size. Am I the only person who seems to have a difficulty?
Circular power theorem
The second one I solved in my head.
So you're telling me I did like 15 lines of working and already had the answer on line 6 😅, lol me going to lengths to find out r and s, and only ending with equations where they cancel out completely and im left with only numbers and no letters lol, that 25 was in most lines tho
I just solve it by mentally.....(x+y)^2-y^2=25
x^2+2xy=25
x(x+2y)=25
Breadth*length=25 is the right answer
So basicly if the only known value in a geometry problem is 5, the answer is 25?
In the first problem, I joint the point at which tangency meets and the centre of the semicircle, next i used the Pythagorean theorem and Sine Rule in that triangle and got the radius of both the circles and henceforth got the answer, But yes, it seemed impossible at the first glance 😅
You can’t get the radii of both circles, as they remain variable, even though the solution is constant.
@@verkuilb In the 2nd question it was, but in the first question it is fixed, because in you increase or decrease size of the semicircle, the length of tangent will increase and decrease respectively; but in the question it is given 5cm fixed
@@add1ct683 Incorrect. In the video, rewatch the animation at the end of the solution of the first problem. It demonstrates that if you keep the tangent line fixed at 5 but rotate its angle, the radii of BOTH arcs change, the length and width of the rectangle co, but the area of the rectangle remains constant at 25.
Why can't You just start with simplifing the shapes (semicircle s->0 so r->5, rectangle y->0), and just skip all the calculations? Isn't it valid approach?
Yeah, I guess so.
Only if you are able to assume that the answers are constant. The problem statements sort of hint at that, but don’t really overtly state that, so the semantics can be argued both ways. Your proposals do indeed determine that the area in the first problem is 25 when the semicircle radius is zero, and that the area in the second problem is zero when the angle of rotation is zero-but it doesn’t prove that those values are constant, independent of the radii in the first problem or the angle in the second.
For the second problem, I figured out the solution had to be 25, without doing any of the algebra. I just visualized rotating the given area=25 square, and the effect that this would have on the size and shape of the smaller square and rectangle-and realized that as you did that, the smaller square would increase in size until it limited out at 25 sq cm, and the rectangle would diminish in size until it simultaneously limited out at zero. (This is demonstrated in the final seconds of the video-but it’s not presented as a means to solve the puzzle. It should be.) Note that this method does not necessarily prove that the total combined area of the small square and rectangle is constant-but that is stated by the phrasing of the problem statement, so there is no need to prove this.
Actually, the same method can be used to solve the first one as well-although I’ll admit I didn’t realize that until after the fact.
I got the first one lets goo
I just figured that since the problem me not gonna involve a calculator and that it involves Pythagorean theorem, it’s probably gonna be 25
25cm^2, solved in 30 seconds using limits, easy
0.1 sec, 25
If u realise both the problems are the same ,and the first one can be easily solved by coordinate geometry....🙌
Done in a minute