2001 AMC 12 problem 22 was nearly the same problem: artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_22 I'm always amazed at the challenging problems assigned to young students! Here are some more videos you might enjoy (list below): A gifted 10 year old student in China solved this in 1 minute: th-cam.com/video/OuJQaxZvlYs/w-d-xo.html Can You Solve Homework For 10 Year Olds In China? th-cam.com/video/a7q3AJXeFuc/w-d-xo.html Viral VERY HARD Problem For 11 Year Olds In China th-cam.com/video/q81fjAYRDo0/w-d-xo.html Adults Stumped By Geometry Problems For 10 Year Olds. Can You Solve Them? th-cam.com/video/M8DrWjEo3IE/w-d-xo.html Can You Solve The Sand Mixing Riddle? (Homework For 10 Year Olds Singapore) th-cam.com/video/viK-Od_7hbQ/w-d-xo.html HARD Math Problem A 13 Year-Old Solved 1 Second! 2017 MathCounts Final Question th-cam.com/video/NzhXipwiz_E/w-d-xo.html Can You Solve This Problem For 12 Year Olds In Singapore? 4 Overlapping Circles Puzzle th-cam.com/video/fpdIdtYg4YE/w-d-xo.html
Will you please solve this question?? A really difficult question. In a triangle ABC, Angle A=84, B=78. Points D & E are taken on the sides AB and CB so that angle ACD=48 and CAE=63. Find angle CDE.
There's an old trick to this kind of problems in case you don't want to do algebra. The fact that the problem doesn't give you any specific lengths or ratios of the rectangle's sides implies that the answer (assumed to exist) must be the same regardless of those conditions. Therefore you can just give the sides whatever lengths and proceed to compute the area without any unknowns. For example you can make it a square with each side's length equal to 1. You can use particular lengths that make the computation way easier. After you've obtained a number which is the area of the triangle under your assumption of the side lengths, and if your rectangle's area is not 10, simply scale the triangle area up by the ratio of your rectangle area to 10. This kind of tricks are common in Chinese middle and high schools that exploit lazy problem making. For example you can solve some very general and difficult problems quickly by making a triangle equilateral or certain angles right angles. The reasoning is that if the ultimate answer is assumed to be true regardless of certain conditions, then the answer obtained by bending those conditions to give you additional information that is not offered by the problem must also be true.
But of course it is not correct to *assume* that the result is independent from the size of the rectangle, just because the question does not mention it. Proving it is independent is part of the problem. However, I think it is obvious why, to a smart 10 years old.
When I try to solve this question , I thought about assume it is a square . More information I got is an angle in the problem , but it is an unusually angle . This will be difficult to find the 3 sides with trigonometry . Then need to solve the area with Heron Formula . This sound very complicated . Use the similar triangle method is way easier.
These are questions in competitive exams, I believe currently in some Asian countries. The questions aren't bad per se, if you are solving purely for the joy of it. But the comments section is fullied with smart alecs from Asia.
I'm 72. I have just solved it after an overnight "cogitate sit" and now about an hour of slowly working out the areas of all the shapes inside the rectangle. I used the same similar triangles but instead of using ratio of their heights ( definitely the more clever way to go ) I used the ratio of their areas. Rather long-winded. I was surprised the area stayed constant no matter the perimeter but assumed that had to be the case given what information we had been supplied to solve the problem. Actually pleased with myself to ot there. Now I can go die in peace.
Hey there, I guess we are similar too just like the triangles cause you did exactly the same way I did, It was toooooo lengthy but it got solved atleast.
I’m a high school math teacher and it took me a half hour to figure it out. 10 year olds that can solve this are exceptionally bright. I used similar triangles like he showed and then used Cavalierie’s principle to find areas of the triangles. The key for me was to subdivide the whole rectangle into smaller rectangles whose areas can be found based on the side length ratios.
Why is this wrong: We start with the area of 10 and then divide by: 3 to trisect the whole rectangle because we are focussing on the middle third of the rectangle 2 because we are only looking at half of that rectangle which is the triangle EFG 4 because the triangle we are looking for is a 4th of the big triangle. So we end up with 10 devided by 24 which is close to this solution but not the same. I somehow feel like the error is in the last step, but why? No matter how the line HI is drawn, it always goes through the center of the triangle and thus should devide the triangle in 1/4 and 3/4 even if it is not parallel to the bottom (or top in this case).
I came to the same conclusion. the only thing I presumed is that the line dividing the rectangle through AB en CD, cuts of a small triangle on te left and leaves a blanc on the right of the centerline of the small triangle, but those 2 parts are equally big (one in excess fills out the blanc. @MindYourDecisions where did we go wrong?
The last step is obviously wrong. Why do you think area ghi is 1/4th of efg. That doesn't make any sense. And btw, what do you mean by centre of triangle? Never heard of it before. Do you mean centroid ?
@@pankajjain8794 You are right and I can't really say what I was referring to. Since it was wrong I guess there is no right answer to that question ;-)
I represented this on a co-ordinate plane. Recreated this diagram with the following equations: X=5 Y=2 Y=2-2x/5 Y=12x/5-6 Y=6-12x/5 From there I prolly could have solved it using distance formula to get values I needed or something like that but I wanted to have some fun so I used integration by splitting the area in 2 pieces lol. Anyways thought I'd share it its one of the few problems I've been able to solve on this channel 😅
Interesting solution. Another solution is, at 2:19, we can use the area of the equilateral triangle is 1/6 of the area of the rectangle. As we computed the two edges of the target triangle are 3/5, 3/7 of the corresponding equilateral triangle edges, respectively, we get the area of the target triangle is 9/35, which is 3/70 of the rectangle, which is 3/7.
Interesting, the real answer of 3/7 is roughly 0.428. You can get very close to that by realizing that there are exactly 6 of the big triangles in the rectangle, and the area of the blue triangle is almost the same as the area if you rotate that diagonal (cc) on its center to be a horizontal line. Then you get a perfect fourth of that bigger triangle. So I estimated it to be approximately a quarter of a sixth, which is 5/12 which is 0.41666... and I can see visually that real answer is just a hair larger. So if this was a multiple choice answer, I'd be golden. Did anyone else see this relationship?
I think 10-year-old kids are not given the problem cold like we are. They have gone over several problems already with similar principles. They are taught how to solve them and the individual tools show up again and again. So the kid will see this and _know_ to start looking at all the triangles he can find (not just the ones mentioned in the puzzle) and sides that have known lengths. They know to extend lines beyond the drawing because every pair of lines are parallel or intersect _somewhere_ . They know they can add more lines, anywhere they have defined points. They've seen examples before where the individual lengths are not known but a sum or product or difference is known, and can relate that combination to a geometric formula containing the same pattern.
My answer is close: 5/12 [ .41(6) ] compared to 3/7 [ .(42857) ]. This is what I did: 1. Assume S=10=2x5; draw 2 by 5 rectangle on grid. 2. Scale this rectangle by multiplying each side by 6 to draw the lines in new 12 by 30 rectangle with the grid lines inside. 3. Count whole [8] and half squares [>7] inside and divide this number by 36, according to the scale. This method is primitive and more practical, because it requires less calculations. The downside is always precision.
If you draw a horisontal line splitting the rectangel in two narrower rectangles of equal size, then you see a little triangle cut off from the triangle in question. And the little triangle is the exact mirror image of the one missing in order to construct a new triangle of the same area as the original one, but with a horisontal baseline instead of the slanted one in the original triangel. And then it is easy to see that the new triangle has a length equal to 1/6 of the horisontal sides of the rectangle (a) and a height equal to 1/2 of the vertical sides of the rectangle (b). So the triangle has area: A = 1/2 × base × height = 1/2 × (a)1/6 × (b)1/2 = (a × b) × 1/24 = 10/24 = 5/12
Ah, now I see ... The little cut off triangle is NOT a true mirror image of the one missing. It can't be, because everything in the big triangle shrinks the closer you get to the buttom of the rectangle.
Look at the labels along the top and bottom. The intersections at the top are s/3 and 2s/3, while the bottom is s/2. Think of the top (or bottom) edge as a ruler. How long is the base of each triangle?
I'm not 100% sure but i think i got it. So for the first one (to get 2/3x), we draw an imaginary line at the cross intersection. Since we know that the intersection is between the top s/3 and the bottom s/2, we set a value x to one of the height (in this vid x is on s/2). And if you flip the bottom triangle up to line up with the top smaller triangle, we can see/imagine that it is actually the same triangle but different size. That means we can think that the s/2 triangle is a certain amount larger than the s/3 triangle. Once we know this, we can find how to get s/2 to s/3 (so s/3 divided by s/2 is 2/3). So the s/3 triangle is 2/3 the size of the s/2 triangle. Therefore we get 2x/3. Same thing works for the next one.
@@eriklotus132 You know the base lengths of the two triangles. You know they are similar. So, the proportions of the bases is also the proportions of the heights. And they touch in the middle so the sum of the heights is y.
1:25 i would qualify that with a small proof: since the top and bottom lines are parallel, any line which intersects both will meet at the same angle on both. therefore we have that two of the angles, and therefore all three, are the same; this is what it means to be similar.
3/7 HG/HE = 1/2 / 1/3 = 3/2 , so HG/GE = 3/5. IG/IF = 1/2 / 2/3 = 3/4........ so IG/GF = 3/7 Area of EGF = 1/6 of Area of ABCD, so area of GHI = 10 * 1/6 * 3/5 * 3/7 = 10 * 9/35 * 1/6 = 3/70 * 10 = 3/7 This is a straightforward task and general solution is available for any division of AD and BC
Presh, my friend the day I stumbled upon your channel was a golden day. I am in love with your channel. Eventhough I cannot solve these I enjoy math Thank You Bro
I drew an aux line that bisected AB across the length of the rectangle. This created 2 triangles, one containing point H, the other point I. These two triangles would be symmetric. One triangle could be folded into the other to give an isosceles triangle with apex G and its base at the aux line drawn previously. If you imagine the rectangle folded at the aux line you now have a rectangle of area 5. There are now 12 triangles in that half so the area of each one is 5/12. My assumption that the two triangles created by my aux line are identical must not be correct (I did not rigorously prove it). Oh well. Seemed like a neat way to avoid algebra!
Hey I really like your videos and they really help me out but I was thinking if you could make the background black as my phone's screen really makes my eyes strain
A sketch of a co-geom. sol'n: [1] Align the whole figure w/ the Cartesian plane s.t. ~ the corner of the rectangle (called R) where its side marked as bisected meets its marked diagonal is at (0,0); & ~ the 2 sides meeting at the said corner lie on the +ve axes. [2] Let s (resp. 6t) be the length of R's side on +ve x-axis (resp. +ve y-axis), so its endpoints are, ordered anticlockwise, at (0,0), (s,0), (s,6t) & (0,6t). [3] As area of R = 10, 6st = 10 => st = 5/3. [4] Coordinates of the vertex of the blue △ that is on the y-axis is G(0,3t), while those of the "2 trisection points" on the side of R marked as trisected are F(s,2t) & E(s,4t). [5] By slope-intercept form, eqn. of the line containing R's marked diagonal is L₁: y = (6t/s) x, whereas eqns. of the 2 lines, 1 of which containing GE & the other GF, are, respectively, L₂: y = (t/s) x + 3t & L₃: y = -(t/s) x + 3t. [6] L₁ & L₂ intersect at (3s/5,18t/5), while L₁ & L₃ intersect at (3s/7,18t/7). [7] By the "Shoelace Formula", area of the blue △ | 0 3s/7 3s/5 0 | = (1/2) | | | 3t 18t/7 18t/5 3t | = 9st/35 = 9(5/3)/35 = 3/7.
So, in my opinion, there is a much easier way to solve this. (Using the letters added to the diagram for simplicity.) 1. Let x and y be the vertical and horizontal side lengths of the rectangle respectively. Construct the second diagonal of the rectangle from A to C and call the points where it intersects the triangle GFE I' and H'. (where the segment I I' is parallel to AD and H-H') 2. Note: The triangle GIH is equal to the two triangles I I' G and I' I H. Note: both of these triangles have a base I' I. 3. Therefore we can write the formula to find the area as: len(I' I) * height(I I' G)/2 + len(I' I) * height(I' I H)/2 which simplifies to len(I-I')/2 * (height (I' I H) + height ( I I' G)). Where height is measured in the y-direction (vertically). 4. to find (height (I' I H) + height ( I I' G)) which is just the length of the vertical line from AD to H, hereby called h. Look at the similar triangles GDH and BEH we get the following relationship h/(x/2) = (y - h) /(x/3). 4a. this simplifies to yx/2 - hx/2 = hx/3 --> y/2 = h/3 + h/2 --> y = 2h/3 + h --> y = (5/3)h --> h = (3/5)y. 5. to find len(I'I) consider the horizontal distance from AB to I', herby called w, and note that it is the same as the horizontal distance from CD to I. Therefore, the len(I-I') = x - 2w. To find w, we look at the similar triangles GDI and BIF to get the following relationship w/(x/2) = (x - w) / (2x/3). 5a. this simplifies to 2wx/3 = (x^2)/2 - wx/2 --> 2w/3 = x/2 - w/2 --> (2/3 + 1/2)w = x/2 --> (7/3)w = x --> w = (3/7)x 5b. Plugging that into our formula len(I-I') = x - 2w --> len(I' I) = x - 2((3/7))x --> len(I' I) = x/7 6. Finally, plugging that into our equation in (3.) we get (x/7)/2 * (3/5)y --> xy *(1/14 * 3/5) --> (because xy = 10) 10 * 1/14 * 3/5 = 1/7 * 3 = 3/7.
Question If the problem doesn't say anything about the rectangle but the area, can I assume a priori it not depends on the shape and, without loss of generatility, solve for a square?
I used a similar, but more convoluted method: basically, calculate just one height, then the area of triangle BFI, then write a set of equations for these five triangles (the shaded one, and the two pairs of similar ones). At the end, I got the same answer.
Solved it!!! In 15 mins......using similarity and coordinate geometry.......however this was a good sum ....... thanks for uploading such nice probs..... 🙂
I solved this using 3 line equations, and using line-line intersections to solve for the two upper vertices of the shaded triangle. Rect height is h, and width is 6s. Origin is put at bottom tri vertex: L1: y = -hx/s L2: y = hx/s L3: y = hx/6s + h/2 p1 = [ -3s/5 , 3h/5 ] intPt L1&L3 p2 = [ 6s/14 , 6h/14 ] intPt L2&L3 The tri area can be solved using the vector cross product: A_tri = 0.5 * || p1 X p2 || = 0.5* || p1.x*p2.y - p1.y*p2.x|| = 0.5 * || -18hs/70 -18hs/70 || = 18hs/70. As a fraction of the rect area: A_tri_fraction = (18hs/70) / 6hs = 3/70. If the rect has area 10, the the small triangle has area 3/7.
This is essentially an affine geometry problem, as it relies entirely on properties that are invariant under affine transformations: straightness of lines, parallelism, ratios of lengths along parallel lines, and ratios of areas. So ABCD could be an arbitrary parallelogram and it would still work.
Presh said "I was told the question was given to 10-year olds". He did not say that it truly was given to 10-year olds. He merely was told so, whether the claim is true or not is a different matter. Keep up the good work Presh!
Since the solution of the problem does not involve the area until the very end, do I understand right that if we generalize, the shaded triangle will be 3/70 of the rectangle anyways?
@@pratham_4355 I repeated the V to the left and to the right of itself. Then I draw the horizontal median of the rectangle. Then I used this median to mirror the 3 V. Now, we have 24 identical triangles (collate the half triangles found at the left and right edges of the rectangle). Obviously, each one of these triangles has an area of 10/24. The diagonal of the rectangle cuts the two center triangles in the middle of their horizontal common edge. So, the resulting bow tie shape is made of two identical triangles and we can "cut-paste" one side to other. Doing so, the area is still 10/4. But to obtain the blue triangle, we need to add a tiny part from the neighbor triangle. This is how I evaluated that the area of he blue triangle is "a bit more than 10/24". Hopefully, it's much easier to draw than to describe with words...
I think that the area of the green or the first triangle at 2:20 is wrong. The base of the triangle should not be s/2. Please give an explanation to the s/2 expression used for the base.
*I like your solution but I like mine better:* Since we know the shaded region has the same area regardless of the proportion of the side lengths, we can allow the height to be 1 and the base to be 10. Knowing this, we can treat the four corners of ABCD as points A(0,0), B(0,1), C(10,1), D(10,0). Since the bottom side is bisected, the point G is at (5,0). Since the top side is trisected, the points E and F are (10/3,1) and (20/3,1). With these points we can then find equations of GE, GF, and BD to find the coordinates of points H(4,3/5) and I(40/7,3/7). Now we can set up vectors GI= and IH=. Plugging these vectors into a matrix, finding the determinant, and dividing by two gives us 3/7.
Presh Talwalkar. You didn't understand him well. He told you it's a problem to children at age of 10 MONTHS, Not years. And they are not from earth, They are from ANDROMADA galaxy.
@@UnimatrixOne I know, but if the videos with Gougu in them get more engagement then they get more views, which is the opposite of what we need. When a video is getting poor viewer engagement the creator gets a notification about it, so he'll know we're unhappy and he won't benefit from the engagement we could create by disliking the video.
the last statement about it doesn't matter the side lengths the area will be the same, is what I started with, I set base and height lengths, found the slope of the lines, and then the intersection points, subtracted the areas of the same two triangles you used, and got 3/7. went through it all again with different base and height lengths to verify that I'd get the same answer, and I did.
I went with coordinate geometry (as usual for me). I defined D to be the origin and B to be (6,1) for ease of calculations (in reality, I flipped it horizontally, but it's the same thing as defining the positive x direction as going left of the origin). This gives me a rectangle of area 6, so I will later scale up my blue triangle by 10/6. The equation for BD is y=x/6. FG has the points (2,1) and (3,0) so the equation is y=-x+3. GE has the points (3,0) and (4,1) so the equation is y=x-3. Solving these equations for the intesection points gives I=(18/7,18/42) and H=(18/5,3/5). Drawing a vertical line at G (x=3) splits the blue area into two triangles with the same base of 1/2. The one with the side GH has a height of (18/5-3), while the one with the side IG has a height of (3-18/7). So the total area is (1/2)(1/2)(18/5-3)+(1/2)(1/2)(3-18/7) = (1/4)(18/5-3+3-18/7) = (1/4)(18/5-18/7) = (1/4)(126/35-190/35)= (1/4)(36/35)= 9/35. But that's for area ABCD = 6. So scaling it up 9/35*10/6= 3/7.
The interesting thing I found with my method is the way the two triangles added up, I could have simply said the area is (1/2)(vertical distance from G to HI)(horizontal distance between H and I). And I can do this even if I rotate the triangle. So basically there is a triangle area formula that can be used without using a side as the base.
You can also use similarity and find the EH/HG=2/3 and FI/IG=4/3 then you can easily find the area of triangle EFG then by using area of a triangle in sine form and figure out the area of triangle GHI
We'll use integration to determine the area. Let's set the rectangle to be the 1 square unit bounded by x = 1, y = 1, and the x and y axes. That will make it simpler to determine the equations describing the lines and to project the answer back onto an area 10 rectangle. The equation for BD is f(x) = -x+1. The equation for EG is g(x) = -6x+3. The equation for GF is h(x) = 6x-3. The intercept between g(x) and h(x) is x = 1/2, obviously, but the intercepts between f(x) and the two other formulas aren't as easy to determine. f(x) = g(x) 1-x = 3-6x 5x = 2 x = 2/5 f(x) = h(x) 1-x = 6x-3 7x = 4 x = 4/7 We'll split the integral in 2 at 1/2. A₁ = ∫(½-⅖) (-x+1) - (-6x+3) dx A₁ = -(x²/2)+x - (-3x²+3x) |(½-⅖) A₁ = (-1/8 + 1/2 + 3/4 - 3/2) - (-2/25 + 2/5 + 12/25 - 6/5) A₁ = -3/8 + 10/25 = 2/5 - 3/8 A₁ = 1/40 A₂ = ∫(4/7-½) (-x+1) - (6x-3) dx A₂ = -(x²/2)+x - (3x²-3x) |(4/7-½) A₂ = (-8/49 + 4/7 - 48/49 + 12/7) - (-1/8 + 1/2 - 3/4 + 3/2) A₂ = 56/49 - 9/8 = 8/7 - 9/8 A₂ = 1/56 A = 10(1/40+1/56) = 10(12/280) = 3/7 Edit: Wow, I was actually right. I haven't done integration in 30 years, and I don't think I ever want to do it again in a TH-cam comment...
Maybe I’m wrong, but I’m solving it now in my way back home in the subway. I got another answer. It’s (5/12). 1-The square of EFG= (10/6) 2- on HI lays the point of the middle of the diameter of the rectangle. Let’s call it O. 3-from O we draw a parallel line to BC... and so on... so I got the square of HIG = 5/12.
I'll say one thing for Presh's latest slogan: "Solving the world's problems one video at a time" is a heck of a lot better and catchier than "We can math the world a better place".
Any 10 years old student can draw the rectangle in checkered paper and easily find the proper size 70x35 that give integer intersections, thus solving the problem in no time. That what we kids would have done back in the last century, at least.
A quick approximate answer would be 5/12 or 10/24. This is since the triangle is 1/6 of the rectangle, and since the line crosses directly through the center it means the blue portion is approximately 1/4 of the triangle. Edit: This is a 1/84 difference from the correct answer, which I'm sure could be figured out...somehow. I do remember some questions in early math where they'd want us to figure out approximate percentages of shapes like this without actually using math.
If you have a maths package the quick hack for this is to extend the coordinates by appending a 3rd coordinate 1. E.g. A, the origin, is (0,0,1); E is (s/3,h,1) etc. These are homogeneous coordinates. Make "line coordinates" from cross products e.g. EG is the line E^G. Then line intersections are also cross products, H=BD^EG, I=BD^FG. Get back to Cartesian coordinates by normalising, H -> H/H(3). Make 2D vectors in the usual way, GH2=H-G, GI2=I-G. Leave in the 3rd coordinate, it's zero but this doesn't matter. Then triangle area is the 3-component of GI2^GH2/2. The result is (3/70)hs as shown..
My answer was wrong, but was so close! I got 10/24= 0.417, solution was 3/7= 0.429! I 1st proved that Area of triangle EFG= 1/6 of the total rectangle's area, so it's 10/6. Then I "proved" (wrongly) that our problem's area of triangle GHI is the same as the lower area of EFG if we sliced it horizontally at the half height. This gave the lower half triangle's area = 1/4 of EFG (since both its height and base were divided by 2). Final Area was 10/(6*4) I think in a school test my answer would still be wrong, but at least the teacher would have to think a bit to show Why it was wrong, lol!
I visit this channel for challenging questions but the whole comment section is filled with 12 year olds solving calculus and at this points it really degrades my self-esteem and my confidence.
They way I solved it (a wee bit more complicated...) Let's call the upper left triangle "W", the shape above the blue triangle "X", the blue triangle itself "Y", and the triangle to the right of that "Z". (1) W = 4/9(Y+Z), since both the base and height of W are 2/3 times those of Y+Z. (2) W+X = 16/9(Z), since here the base and height of (W+X) are both 4/3 times those of Z. Subtract equation 1 from equation 2, and you get rid of W, leaving X = -4/9(Y) + 4/3(Z). We can get rid of X by adding Y to both sides. This is because we know that X+Y = 10/6 or 5/3, since it is a triangle with 1/3 the length of the rectangle, and you divide by 2 for triangles. 5/3 = 5/9(Y) + 4/3(Z). If you're still reading, you'll be interested to know that you can use other information to solve for Z. We know that the area of the upper right triangle formed by cutting the rectangle in half with a diagonal is 5. Let's also draw a line from F to D. This upper right triangle is now divided into 4 sections. The first two sections are W+X, which is 16/9(Z). The third section is (5/2)-Z. This is because triangle GFD is 10/4=5/2, using the same sort of logic that gave us the area of X+Y. Finally, the fourth section is 5/3 just like X+Y. 16/9(Z)+(5/2)-Z+(5/3)=5. You can solve for Z, then plug in Z into 5/3 = 5/9(Y) + 4/3(Z) to solve for Y, the blue triangle.
Got it, solved it more or less the way the Presh did it. HINT: Look for similar triangles and remember that the area of two similar triangles is in the same proportion as the square of the proportion of their sides (though he didn't use that fact directly) Doing this, you end up with some simple algebra and a lot of arithmetic involving fractions. Who in 5th grade learns algebra, though ?
And here I am watching that big triangle is 1/6 of the box, and blue is about 1/4 of the triangle (little over). So it is about 10/24 ~ 0.417. and result was 0.429. less than 3% wrong, close enough for approximated in 10 seconds.
I'm too tired right now to solve this one, but here's the steps I would take: 1) Set a coordinate system with origin on point A. Consider point D to be (L,0), then point B is (0, 10/L). 2) Determine the equation f(x) of the diagonal, considering it passes through points B and D. 3) Determine the equation for line g(x) that passes through (L/2,0) and (L/3,10/L) 4) Determine the equation for line h(x) that passes through points (L/2,0) and (2L/3,10/L) 5) Calculate point H as the intersection of f and g; and point I as the intersection of f and h. 6) We now have all the coordinates for the triangle. We can put them on a matrix and calculate the area.
Here's the easy way: define a = height of rectangle define b = width of rectangle Chop and glue the triangle to construct an isosceles traingle with: height = a/2 width = b/6 The area of this triangle thus becomes ab/24 = 10/24 So this is wrong but easy and close to the truth: 10/24 = 0.416667 3/7 = 0.4286 My guess is this is the solutions the teachers were looking for. Still hard for 10 year-olds IMO
For the 10 years old I think they (kids) should have drawn more similar, including one orizontal that divides the rectangle in half, and see that it can be divided in 24 equal parts, also they learn aproximation at this age, and the answer would be aproximativ 10/24, which is not far from the exact answer 3/7.
3:08 It is not REMARKABLE it is OBVIOUS. You are just doing a simple transformation. Imagine a plane Y vs X in which there is a closed curve. Integrate to get the area A insidethat curve, then transform Y--->sY and X--->X/s where s is a constant and integrate to get the area again. Since Ydx---> sYdx/s=Ydx you get the same area. This is the second time that I hear him say that kind of nonsense.
Presh, you might wanna start seeing kids as kids, not as someone with your level of intellect. At the age of 10, kids are barely learning shapes and the formulae for their areas. They do not have area algebra or geometry or any form of analytical tools such as variable substitution. This problem clearly begins from 9th or 10th class. You might wanna fact check the person who E-mailed you what he/she really meant. This is not a problem for a 10 yr old, unless he is Sheldon Cooper.
@@jackburns6403 i didn't understood what you are trying to explain ? I mean , how can a person comment earlier than the upload of a video . And if he has a time machine , then he must have got it patented .
well, the answer i got is 0.8 i.e 5/6. so, 25/3 + 5/3 = 10 therefore, 8.33+1.66 = 10(approx as it is 9.99) as the two trapeziums are equal, the triangle is isosceles, hence the diagonal divides it into 2 parts. so, 1.66/2 = 0.8(approx)
Are you kidding? 😆... Concepts of similar triangles in 4-5 th class student?!! I don't think that even international boards would be such hard... BTW great fan of yours and maths.... I Love ❤ ur videos 😊
me: thinking for a minute.... ok, i give up, lets use analytic geometry to solve this. but its really shameful that i didnt see the "smart" solution. i should have really noticed all the similar triangles.
So the blue area is about 4% of the overall area. This might be rephrased as an optical illusion because the area looks a whole lot bigger than that in the picture.
Using analytic geometry to find GH and GI as vectors, then finding A(∆GHI) = ½[cross-product GH x GI], I get A(∆GHI) = 3/7 This clearly isn't something the average fifth-grader (10-yo) would do, even in a fully competent educational system. Let's see what Presh has up his sleeve this time, that could be expected of such a student... Meanwhile, here was my initial, ballpark method of estimating the desired area. First, use the fact that ∆EFG has ⅓ the base of, and equal height as, the rectangle, to get A(∆EFG) = ⅙A(ABCD) = 5/3. Interestingly, if you then erroneously assume that the rectangle's diagonal, BD, cuts through the isosceles ∆EFG in a way that the blue ∆ has ¼ the area of ∆EFG, just as a horizontal cut midway up the rectangle would do, you get A(∆GHI) ≈ ¼(5/3) = 5/12 which is almost exactly right, being 35/36 of the true area, 3/7. Post-view: That's a lot of work, but each step is something that at least the top tier of those students could come up with. Bravo/brava to those who did solve it! Fred
if they are similar,that means that all you need to scale (and rotate) them. in first occurence ( 1:16 ) he knows one of sides of bigger one is half of S anc corresponding side of smaller one is third of S that gave him ratio for scaling them 3 / 2 to get value from smaller to larger, you multiply it by three and divide by two to ge value from larger to smaller, you multiply by two and divide by three he then stated that height of larger one is x from that follows that if you multiply x by 2/3 , you get height of smaller one so height h = x + 2/3x that translates to (3/3 + 2/3 = 5/3) h = 5/3 x so to get from x to h you multiply by 5 and divide by 3 if you want to reverse it,you multiply h by 3 and divide by 5 that got him x = 3/5 h no magic there :)
2001 AMC 12 problem 22 was nearly the same problem: artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_22
I'm always amazed at the challenging problems assigned to young students! Here are some more videos you might enjoy (list below):
A gifted 10 year old student in China solved this in 1 minute:
th-cam.com/video/OuJQaxZvlYs/w-d-xo.html
Can You Solve Homework For 10 Year Olds In China?
th-cam.com/video/a7q3AJXeFuc/w-d-xo.html
Viral VERY HARD Problem For 11 Year Olds In China
th-cam.com/video/q81fjAYRDo0/w-d-xo.html
Adults Stumped By Geometry Problems For 10 Year Olds. Can You Solve Them?
th-cam.com/video/M8DrWjEo3IE/w-d-xo.html
Can You Solve The Sand Mixing Riddle? (Homework For 10 Year Olds Singapore)
th-cam.com/video/viK-Od_7hbQ/w-d-xo.html
HARD Math Problem A 13 Year-Old Solved 1 Second! 2017 MathCounts Final Question
th-cam.com/video/NzhXipwiz_E/w-d-xo.html
Can You Solve This Problem For 12 Year Olds In Singapore? 4 Overlapping Circles Puzzle
th-cam.com/video/fpdIdtYg4YE/w-d-xo.html
2x÷3y-1 if x=9 and y=2 ?
can you do this pls, its popular nowadays
Will you please solve this question??
A really difficult question.
In a triangle ABC, Angle A=84, B=78. Points D & E are taken on the sides AB and CB so that angle ACD=48 and CAE=63. Find angle CDE.
@@VipinSingh-mm1jk wrong question because angle c should be equal to 16 but you mentioned that angle acd is 48 and 48>16
....so it's wrong.....
Guys, if you love solving puzzles and riddles, do visit #PuzzleAdda
You can give any problem to 10 year olds, the key question is, did they solve it….
This problem was given to ten year olds? Ha, that's NOTHING, I gave this problem to a bunch of TWO year olds!
They didn't solve it either.
Math : Geometry problem
th-cam.com/video/4dkOeJXWDW4/w-d-xo.html
See it for fun
😂😂😂
Yes, the problem was given to 10-year-olds, and they are now 20 and can solve it! ;)
I'm 24 and still couldn't solve this.
th-cam.com/video/TCYJ4XYKR4g/w-d-xo.html
I am 15 and can't understand it.
This math is not for 10 y old kids 🤓
This was problem 22 of the 2001 AMC 12 as well:
artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_22
There's an old trick to this kind of problems in case you don't want to do algebra. The fact that the problem doesn't give you any specific lengths or ratios of the rectangle's sides implies that the answer (assumed to exist) must be the same regardless of those conditions. Therefore you can just give the sides whatever lengths and proceed to compute the area without any unknowns. For example you can make it a square with each side's length equal to 1. You can use particular lengths that make the computation way easier. After you've obtained a number which is the area of the triangle under your assumption of the side lengths, and if your rectangle's area is not 10, simply scale the triangle area up by the ratio of your rectangle area to 10. This kind of tricks are common in Chinese middle and high schools that exploit lazy problem making. For example you can solve some very general and difficult problems quickly by making a triangle equilateral or certain angles right angles. The reasoning is that if the ultimate answer is assumed to be true regardless of certain conditions, then the answer obtained by bending those conditions to give you additional information that is not offered by the problem must also be true.
So your answer is?
We had a ratio here however, h = 10/s
But of course it is not correct to *assume* that the result is independent from the size of the rectangle, just because the question does not mention it. Proving it is independent is part of the problem. However, I think it is obvious why, to a smart 10 years old.
Math problem
Geometry problem
th-cam.com/video/4dkOeJXWDW4/w-d-xo.html
One time see it
When I try to solve this question , I thought about assume it is a square . More information I got is an angle in the problem , but it is an unusually angle . This will be difficult to find the 3 sides with trigonometry . Then need to solve the area with Heron Formula . This sound very complicated . Use the similar triangle method is way easier.
maybe when i was 10 years old i was more inteligent, now i cannot solve it
Yes may be
th-cam.com/video/TCYJ4XYKR4g/w-d-xo.html
These are questions in competitive exams, I believe currently in some Asian countries. The questions aren't bad per se, if you are solving purely for the joy of it. But the comments section is fullied with smart alecs from Asia.
I wasn't even aware of the property of similar triangles. I still don't quite get how he got the height
Guys, if you love solving puzzles and riddles, do visit #PuzzleAdda
I'm 72. I have just solved it after an overnight "cogitate sit" and now about an hour of slowly working out the areas of all the shapes inside the rectangle. I used the same similar triangles but instead of using ratio of their heights ( definitely the more clever way to go ) I used the ratio of their areas. Rather long-winded. I was surprised the area stayed constant no matter the perimeter but assumed that had to be the case given what information we had been supplied to solve the problem. Actually pleased with myself to ot there. Now I can go die in peace.
Hey there, I guess we are similar too just like the triangles cause you did exactly the same way I did, It was toooooo lengthy but it got solved atleast.
yeah and where in the world are 10 year olds learning this?!
asia lol
@@icurras that is true, that is VERY VERY TRUE
10 martian years.
Not even in asia we were taught that in our 10 yrs old age.
I m from india from grade 9 i study calculus .not every one in asia only student preparing for jee and imo exam learn all thing bit early
Problem given to pregnant woman. Doctor asks the new born to solve the same problem.
Presh Talwalker: This problem was given to 10 min old baby.
Haha. Next on this channel: "This question was given to a student in its previous birth."
🤣😂🤣🤣🤣😂🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣
i thought this video was finally gonna have problems that I could actually solve...
Math : Geometry problem
th-cam.com/video/4dkOeJXWDW4/w-d-xo.html
See it for fun
Me and my 11 years old friends can solve this
I’m a high school math teacher and it took me a half hour to figure it out. 10 year olds that can solve this are exceptionally bright. I used similar triangles like he showed and then used Cavalierie’s principle to find areas of the triangles. The key for me was to subdivide the whole rectangle into smaller rectangles whose areas can be found based on the side length ratios.
Why is this wrong: We start with the area of 10 and then divide by:
3 to trisect the whole rectangle because we are focussing on the middle third of the rectangle
2 because we are only looking at half of that rectangle which is the triangle EFG
4 because the triangle we are looking for is a 4th of the big triangle.
So we end up with 10 devided by 24 which is close to this solution but not the same.
I somehow feel like the error is in the last step, but why? No matter how the line HI is drawn, it always goes through the center of the triangle and thus should devide the triangle in 1/4 and 3/4 even if it is not parallel to the bottom (or top in this case).
I came to the same conclusion. the only thing I presumed is that the line dividing the rectangle through AB en CD, cuts of a small triangle on te left and leaves a blanc on the right of the centerline of the small triangle, but those 2 parts are equally big (one in excess fills out the blanc. @MindYourDecisions where did we go wrong?
The last step is obviously wrong. Why do you think area ghi is 1/4th of efg. That doesn't make any sense. And btw, what do you mean by centre of triangle? Never heard of it before. Do you mean centroid ?
@@pankajjain8794 You are right and I can't really say what I was referring to. Since it was wrong I guess there is no right answer to that question ;-)
I Solved it by Coordinate Geometry... It was long but easy...
Same lol
I solved it by simply logic. Quick and easy.
Same
th-cam.com/video/TCYJ4XYKR4g/w-d-xo.html
Me too! :D
No Gogu = thumbs-up
Math : Geometry problem
th-cam.com/video/4dkOeJXWDW4/w-d-xo.html
See it for fun
I represented this on a co-ordinate plane. Recreated this diagram with the following equations:
X=5
Y=2
Y=2-2x/5
Y=12x/5-6
Y=6-12x/5
From there I prolly could have solved it using distance formula to get values I needed or something like that but I wanted to have some fun so I used integration by splitting the area in 2 pieces lol.
Anyways thought I'd share it its one of the few problems I've been able to solve on this channel 😅
Interesting solution.
Another solution is, at 2:19, we can use the area of the equilateral triangle is 1/6 of the area of the rectangle. As we computed the two edges of the target triangle are 3/5, 3/7 of the corresponding equilateral triangle edges, respectively, we get the area of the target triangle is 9/35, which is 3/70 of the rectangle, which is 3/7.
You always say that this Chanel is one of the bests in TH-cam, who told you so? It isn’t.
Interesting, the real answer of 3/7 is roughly 0.428. You can get very close to that by realizing that there are exactly 6 of the big triangles in the rectangle, and the area of the blue triangle is almost the same as the area if you rotate that diagonal (cc) on its center to be a horizontal line. Then you get a perfect fourth of that bigger triangle. So I estimated it to be approximately a quarter of a sixth, which is 5/12 which is 0.41666... and I can see visually that real answer is just a hair larger. So if this was a multiple choice answer, I'd be golden.
Did anyone else see this relationship?
I think 10-year-old kids are not given the problem cold like we are.
They have gone over several problems already with similar principles. They are taught how to solve them and the individual tools show up again and again. So the kid will see this and _know_ to start looking at all the triangles he can find (not just the ones mentioned in the puzzle) and sides that have known lengths.
They know to extend lines beyond the drawing because every pair of lines are parallel or intersect _somewhere_ . They know they can add more lines, anywhere they have defined points. They've seen examples before where the individual lengths are not known but a sum or product or difference is known, and can relate that combination to a geometric formula containing the same pattern.
I wish that was the case here and they would teach such things everywhere...sadly that's not the case
Math : Geometry problem
th-cam.com/video/4dkOeJXWDW4/w-d-xo.html
See it for fun.
My answer is close: 5/12 [ .41(6) ] compared to 3/7 [ .(42857) ].
This is what I did:
1. Assume S=10=2x5; draw 2 by 5 rectangle on grid.
2. Scale this rectangle by multiplying each side by 6 to draw the lines in new 12 by 30 rectangle with the grid lines inside.
3. Count whole [8] and half squares [>7] inside and divide this number by 36, according to the scale.
This method is primitive and more practical, because it requires less calculations. The downside is always precision.
I got the same answer by using symmetry / similar triangles.
I also got 5/12
... but as the exact value.
I don't understand how Presh Talwalkar got 3/7 😐
If you draw a horisontal line splitting the rectangel in two narrower rectangles of equal size, then you see a little triangle cut off from the triangle in question. And the little triangle is the exact mirror image of the one missing in order to construct a new triangle of the same area as the original one, but with a horisontal baseline instead of the slanted one in the original triangel. And then it is easy to see that the new triangle has a length equal to 1/6 of the horisontal sides of the rectangle (a) and a height equal to 1/2 of the vertical sides of the rectangle (b). So the triangle has area:
A = 1/2 × base × height
= 1/2 × (a)1/6 × (b)1/2
= (a × b) × 1/24
= 10/24
= 5/12
Ah, now I see ... The little cut off triangle is NOT a true mirror image of the one missing. It can't be, because everything in the big triangle shrinks the closer you get to the buttom of the rectangle.
Wrong method 😉
Why 2/3x or 4/3y? From which rule is it? Or how? 🤔
Help please.
Look at the labels along the top and bottom. The intersections at the top are s/3 and 2s/3, while the bottom is s/2. Think of the top (or bottom) edge as a ruler. How long is the base of each triangle?
I'm not 100% sure but i think i got it. So for the first one (to get 2/3x), we draw an imaginary line at the cross intersection. Since we know that the intersection is between the top s/3 and the bottom s/2, we set a value x to one of the height (in this vid x is on s/2). And if you flip the bottom triangle up to line up with the top smaller triangle, we can see/imagine that it is actually the same triangle but different size. That means we can think that the s/2 triangle is a certain amount larger than the s/3 triangle. Once we know this, we can find how to get s/2 to s/3 (so s/3 divided by s/2 is 2/3). So the s/3 triangle is 2/3 the size of the s/2 triangle. Therefore we get 2x/3. Same thing works for the next one.
@@JohnDlugosz the base is s/2 and s/3 or what do you mean? 🤔😊
thales theorem
@@eriklotus132 You know the base lengths of the two triangles. You know they are similar. So, the proportions of the bases is also the proportions of the heights. And they touch in the middle so the sum of the heights is y.
1:25 i would qualify that with a small proof: since the top and bottom lines are parallel, any line which intersects both will meet at the same angle on both. therefore we have that two of the angles, and therefore all three, are the same; this is what it means to be similar.
Just because something is given to solve doesn't mean it was solved
3/7
HG/HE = 1/2 / 1/3 = 3/2 , so HG/GE = 3/5.
IG/IF = 1/2 / 2/3 = 3/4........ so IG/GF = 3/7
Area of EGF = 1/6 of Area of ABCD, so area of GHI = 10 * 1/6 * 3/5 * 3/7 = 10 * 9/35 * 1/6 = 3/70 * 10 = 3/7
This is a straightforward task and general solution is available for any division of AD and BC
Presh, my friend the day I stumbled upon your channel was a golden day. I am in love with your channel. Eventhough I cannot solve these I enjoy math Thank You Bro
I was the person who was given this question at age ten and I solved it😀 after 10 years
You keep impressing me with your presentation skills. 👍👍👍
Please dont stop making such nice videos
I drew an aux line that bisected AB across the length of the rectangle. This created 2 triangles, one containing point H, the other point I. These two triangles would be symmetric. One triangle could be folded into the other to give an isosceles triangle with apex G and its base at the aux line drawn previously. If you imagine the rectangle folded at the aux line you now have a rectangle of area 5. There are now 12 triangles in that half so the area of each one is 5/12. My assumption that the two triangles created by my aux line are identical must not be correct (I did not rigorously prove it). Oh well. Seemed like a neat way to avoid algebra!
Hey I really like your videos and they really help me out but I was thinking if you could make the background black as my phone's screen really makes my eyes strain
A sketch of a co-geom. sol'n:
[1] Align the whole figure w/ the Cartesian plane s.t.
~ the corner of the rectangle (called R) where its side marked as bisected meets its marked diagonal is at (0,0); &
~ the 2 sides meeting at the said corner lie on the +ve axes.
[2] Let s (resp. 6t) be the length of R's side on +ve x-axis (resp. +ve y-axis), so its endpoints are, ordered anticlockwise, at (0,0), (s,0), (s,6t) & (0,6t).
[3] As area of R = 10, 6st = 10 => st = 5/3.
[4] Coordinates of the vertex of the blue △ that is on the y-axis is G(0,3t), while those of the "2 trisection points" on the side of R marked as trisected are F(s,2t) & E(s,4t).
[5] By slope-intercept form, eqn. of the line containing R's marked diagonal is L₁: y = (6t/s) x, whereas eqns. of the 2 lines, 1 of which containing GE & the other GF, are, respectively, L₂: y = (t/s) x + 3t & L₃: y = -(t/s) x + 3t.
[6] L₁ & L₂ intersect at (3s/5,18t/5), while L₁ & L₃ intersect at (3s/7,18t/7).
[7] By the "Shoelace Formula",
area of the blue △
| 0 3s/7 3s/5 0 |
= (1/2) | |
| 3t 18t/7 18t/5 3t |
= 9st/35 = 9(5/3)/35 = 3/7.
So, in my opinion, there is a much easier way to solve this. (Using the letters added to the diagram for simplicity.)
1. Let x and y be the vertical and horizontal side lengths of the rectangle respectively. Construct the second diagonal of the rectangle from A to C and call the points where it intersects the triangle GFE I' and H'. (where the segment I I' is parallel to AD and H-H')
2. Note: The triangle GIH is equal to the two triangles I I' G and I' I H. Note: both of these triangles have a base I' I.
3. Therefore we can write the formula to find the area as: len(I' I) * height(I I' G)/2 + len(I' I) * height(I' I H)/2 which simplifies to len(I-I')/2 * (height (I' I H) + height ( I I' G)). Where height is measured in the y-direction (vertically).
4. to find (height (I' I H) + height ( I I' G)) which is just the length of the vertical line from AD to H, hereby called h. Look at the similar triangles GDH and BEH we get the following relationship h/(x/2) = (y - h) /(x/3).
4a. this simplifies to yx/2 - hx/2 = hx/3 --> y/2 = h/3 + h/2 --> y = 2h/3 + h --> y = (5/3)h --> h = (3/5)y.
5. to find len(I'I) consider the horizontal distance from AB to I', herby called w, and note that it is the same as the horizontal distance from CD to I. Therefore, the len(I-I') = x - 2w. To find w, we look at the similar triangles GDI and BIF to get the following relationship w/(x/2) = (x - w) / (2x/3).
5a. this simplifies to 2wx/3 = (x^2)/2 - wx/2 --> 2w/3 = x/2 - w/2 --> (2/3 + 1/2)w = x/2 --> (7/3)w = x --> w = (3/7)x
5b. Plugging that into our formula len(I-I') = x - 2w --> len(I' I) = x - 2((3/7))x --> len(I' I) = x/7
6. Finally, plugging that into our equation in (3.) we get (x/7)/2 * (3/5)y --> xy *(1/14 * 3/5) --> (because xy = 10) 10 * 1/14 * 3/5 = 1/7 * 3 = 3/7.
Question
If the problem doesn't say anything about the rectangle but the area, can I assume a priori it not depends on the shape and, without loss of generatility, solve for a square?
I used a similar, but more convoluted method: basically, calculate just one height, then the area of triangle BFI, then write a set of equations for these five triangles (the shaded one, and the two pairs of similar ones). At the end, I got the same answer.
Solved it!!! In 15 mins......using similarity and coordinate geometry.......however this was a good sum ....... thanks for uploading such nice probs..... 🙂
I solved this using 3 line equations, and using line-line intersections to solve for the two upper vertices of the shaded triangle. Rect height is h, and width is 6s. Origin is put at bottom tri vertex:
L1: y = -hx/s
L2: y = hx/s
L3: y = hx/6s + h/2
p1 = [ -3s/5 , 3h/5 ] intPt L1&L3
p2 = [ 6s/14 , 6h/14 ] intPt L2&L3
The tri area can be solved using the vector cross product:
A_tri = 0.5 * || p1 X p2 || = 0.5* || p1.x*p2.y - p1.y*p2.x||
= 0.5 * || -18hs/70 -18hs/70 || = 18hs/70.
As a fraction of the rect area: A_tri_fraction = (18hs/70) / 6hs = 3/70.
If the rect has area 10, the the small triangle has area 3/7.
Thank you for helping the students by giving challenging problems
I hope that someday you’ll even make videos on Physics and Chemistry problems also
Math : Geometry problem
th-cam.com/video/4dkOeJXWDW4/w-d-xo.html
See it for fun.
This is essentially an affine geometry problem, as it relies entirely on properties that are invariant under affine transformations: straightness of lines, parallelism, ratios of lengths along parallel lines, and ratios of areas. So ABCD could be an arbitrary parallelogram and it would still work.
Presh said "I was told the question was given to 10-year olds". He did not say that it truly was given to 10-year olds. He merely was told so, whether the claim is true or not is a different matter.
Keep up the good work Presh!
what software do u use to create these awesome animations ?
Since the solution of the problem does not involve the area until the very end, do I understand right that if we generalize, the shaded triangle will be 3/70 of the rectangle anyways?
A quick evaluation gives "a bit more than 10/24".
Not bad, as 20/48 is really very close to 21/49...!
Plz explain your quick evaluation
@@pratham_4355 I repeated the V to the left and to the right of itself. Then I draw the horizontal median of the rectangle. Then I used this median to mirror the 3 V. Now, we have 24 identical triangles (collate the half triangles found at the left and right edges of the rectangle). Obviously, each one of these triangles has an area of 10/24. The diagonal of the rectangle cuts the two center triangles in the middle of their horizontal common edge. So, the resulting bow tie shape is made of two identical triangles and we can "cut-paste" one side to other. Doing so, the area is still 10/4. But to obtain the blue triangle, we need to add a tiny part from the neighbor triangle. This is how I evaluated that the area of he blue triangle is "a bit more than 10/24". Hopefully, it's much easier to draw than to describe with words...
I think that the area of the green or the first triangle at 2:20 is wrong. The base of the triangle should not be s/2. Please give an explanation to the s/2 expression used for the base.
*I like your solution but I like mine better:* Since we know the shaded region has the same area regardless of the proportion of the side lengths, we can allow the height to be 1 and the base to be 10. Knowing this, we can treat the four corners of ABCD as points A(0,0), B(0,1), C(10,1), D(10,0). Since the bottom side is bisected, the point G is at (5,0). Since the top side is trisected, the points E and F are (10/3,1) and (20/3,1). With these points we can then find equations of GE, GF, and BD to find the coordinates of points H(4,3/5) and I(40/7,3/7). Now we can set up vectors GI= and IH=. Plugging these vectors into a matrix, finding the determinant, and dividing by two gives us 3/7.
Presh Talwalkar.
You didn't understand him well.
He told you it's a problem to children at age of 10 MONTHS,
Not years.
And they are not from earth,
They are from ANDROMADA galaxy.
Lol
👌👌😅😅
Math : Geometry problem
th-cam.com/video/4dkOeJXWDW4/w-d-xo.html
See it for fun.
And is it always then 3/70s of the total area, regardless of total area value?
MindYourDecisions: *Post a Video
Gougu Theorem: Allow me to introduce myself..
It wasn't in this one though. Whenever I hear it I just immediately close the video without leaving a like/dislike/comment
Gougu = thumbs down!
@@UnimatrixOne don't even do that, TH-cam still counts a dislike as "viewer engagement" and thus is more likely to promote the video.
@@swingardium706 It´s for (against) Presh not for YT.
@@UnimatrixOne I know, but if the videos with Gougu in them get more engagement then they get more views, which is the opposite of what we need. When a video is getting poor viewer engagement the creator gets a notification about it, so he'll know we're unhappy and he won't benefit from the engagement we could create by disliking the video.
the last statement about it doesn't matter the side lengths the area will be the same, is what I started with, I set base and height lengths, found the slope of the lines, and then the intersection points, subtracted the areas of the same two triangles you used, and got 3/7. went through it all again with different base and height lengths to verify that I'd get the same answer, and I did.
I went with coordinate geometry (as usual for me). I defined D to be the origin and B to be (6,1) for ease of calculations (in reality, I flipped it horizontally, but it's the same thing as defining the positive x direction as going left of the origin). This gives me a rectangle of area 6, so I will later scale up my blue triangle by 10/6. The equation for BD is y=x/6. FG has the points (2,1) and (3,0) so the equation is y=-x+3. GE has the points (3,0) and (4,1) so the equation is y=x-3. Solving these equations for the intesection points gives I=(18/7,18/42) and H=(18/5,3/5). Drawing a vertical line at G (x=3) splits the blue area into two triangles with the same base of 1/2. The one with the side GH has a height of (18/5-3), while the one with the side IG has a height of (3-18/7). So the total area is (1/2)(1/2)(18/5-3)+(1/2)(1/2)(3-18/7) = (1/4)(18/5-3+3-18/7) = (1/4)(18/5-18/7) = (1/4)(126/35-190/35)= (1/4)(36/35)= 9/35. But that's for area ABCD = 6. So scaling it up 9/35*10/6= 3/7.
The interesting thing I found with my method is the way the two triangles added up, I could have simply said the area is (1/2)(vertical distance from G to HI)(horizontal distance between H and I). And I can do this even if I rotate the triangle. So basically there is a triangle area formula that can be used without using a side as the base.
The only 10 year-old that can solve this was Terance Tao.
hehehe
How do you know?
Yessssssss agreed.
What does ”all will be well if you use your mind for your decisions, and mind only your decisions” mean?
You can also use similarity and find the EH/HG=2/3 and FI/IG=4/3 then you can easily find the area of triangle EFG then by using area of a triangle in sine form and figure out the area of triangle GHI
We'll use integration to determine the area. Let's set the rectangle to be the 1 square unit bounded by x = 1, y = 1, and the x and y axes. That will make it simpler to determine the equations describing the lines and to project the answer back onto an area 10 rectangle.
The equation for BD is f(x) = -x+1. The equation for EG is g(x) = -6x+3. The equation for GF is h(x) = 6x-3. The intercept between g(x) and h(x) is x = 1/2, obviously, but the intercepts between f(x) and the two other formulas aren't as easy to determine.
f(x) = g(x)
1-x = 3-6x
5x = 2
x = 2/5
f(x) = h(x)
1-x = 6x-3
7x = 4
x = 4/7
We'll split the integral in 2 at 1/2.
A₁ = ∫(½-⅖) (-x+1) - (-6x+3) dx
A₁ = -(x²/2)+x - (-3x²+3x) |(½-⅖)
A₁ = (-1/8 + 1/2 + 3/4 - 3/2) - (-2/25 + 2/5 + 12/25 - 6/5)
A₁ = -3/8 + 10/25 = 2/5 - 3/8
A₁ = 1/40
A₂ = ∫(4/7-½) (-x+1) - (6x-3) dx
A₂ = -(x²/2)+x - (3x²-3x) |(4/7-½)
A₂ = (-8/49 + 4/7 - 48/49 + 12/7) - (-1/8 + 1/2 - 3/4 + 3/2)
A₂ = 56/49 - 9/8 = 8/7 - 9/8
A₂ = 1/56
A = 10(1/40+1/56) = 10(12/280) = 3/7
Edit: Wow, I was actually right. I haven't done integration in 30 years, and I don't think I ever want to do it again in a TH-cam comment...
Maybe I’m wrong, but I’m solving it now in my way back home in the subway. I got another answer. It’s (5/12).
1-The square of EFG= (10/6)
2- on HI lays the point of the middle of the diameter of the rectangle. Let’s call it O.
3-from O we draw a parallel line to BC... and so on... so I got the square of HIG = 5/12.
The midpoint between H and I is not the center of the rectangle.
I used similar triangles . Tedious but straight foward.
Exact same method as shown.
I love your channel ,from mauritania ❤❤❤
First person I ever see on the internet from mauritania! Greetings from Mexico!
I'll say one thing for Presh's latest slogan: "Solving the world's problems one video at a time" is a heck of a lot better and catchier than "We can math the world a better place".
The most surprising thing is the optical illusion. I couldn't believe one could squeeze 23 < 70/3 of these blue triangles into the rectangle.
I am so happy I solved it with same method😀😀😀
amazing video as usual! you give me inspiration to make my own videos :)
Solving geometry problems with min.given is a real challenge especially learning the insights which make it look easy👍👍in this case similarity &ratio
Any 10 years old student can draw the rectangle in checkered paper and easily find the proper size 70x35 that give integer intersections, thus solving the problem in no time. That what we kids would have done back in the last century, at least.
A quick approximate answer would be 5/12 or 10/24. This is since the triangle is 1/6 of the rectangle, and since the line crosses directly through the center it means the blue portion is approximately 1/4 of the triangle.
Edit: This is a 1/84 difference from the correct answer, which I'm sure could be figured out...somehow.
I do remember some questions in early math where they'd want us to figure out approximate percentages of shapes like this without actually using math.
If you have a maths package the quick hack for this is to extend the coordinates by appending a 3rd coordinate 1. E.g. A, the origin, is (0,0,1); E is (s/3,h,1) etc. These are homogeneous coordinates. Make "line coordinates" from cross products e.g. EG is the line E^G. Then line intersections are also cross products, H=BD^EG, I=BD^FG. Get back to Cartesian coordinates by normalising, H -> H/H(3). Make 2D vectors in the usual way, GH2=H-G, GI2=I-G. Leave in the 3rd coordinate, it's zero but this doesn't matter. Then triangle area is the 3-component of GI2^GH2/2. The result is (3/70)hs as shown..
My answer was wrong, but was so close! I got 10/24= 0.417, solution was 3/7= 0.429!
I 1st proved that Area of triangle EFG= 1/6 of the total rectangle's area, so it's 10/6. Then I "proved" (wrongly) that our problem's area of triangle GHI is the same as the lower area of EFG if we sliced it horizontally at the half height.
This gave the lower half triangle's area = 1/4 of EFG (since both its height and base were divided by 2). Final Area was 10/(6*4)
I think in a school test my answer would still be wrong, but at least the teacher would have to think a bit to show Why it was wrong, lol!
I took this again, and knew my previous method was wrong, but I couldn't solve it.
Your solution is so clever cleaver!
thanks! this was a very good problem. i need to watch it until i get it. wish you a great night!
I visit this channel for challenging questions but the whole comment section is filled with 12 year olds solving calculus and at this points it really degrades my self-esteem and my confidence.
Everyone's talking about the Trump-Biden debate, and I'm here like a weirdo with my tablet solving math! 😂
believe me... you are the normal one.
Such a clever flex bro
They way I solved it (a wee bit more complicated...)
Let's call the upper left triangle "W", the shape above the blue triangle "X", the blue triangle itself "Y", and the triangle to the right of that "Z". (1) W = 4/9(Y+Z), since both the base and height of W are 2/3 times those of Y+Z. (2) W+X = 16/9(Z), since here the base and height of (W+X) are both 4/3 times those of Z. Subtract equation 1 from equation 2, and you get rid of W, leaving X = -4/9(Y) + 4/3(Z). We can get rid of X by adding Y to both sides. This is because we know that X+Y = 10/6 or 5/3, since it is a triangle with 1/3 the length of the rectangle, and you divide by 2 for triangles. 5/3 = 5/9(Y) + 4/3(Z).
If you're still reading, you'll be interested to know that you can use other information to solve for Z. We know that the area of the upper right triangle formed by cutting the rectangle in half with a diagonal is 5. Let's also draw a line from F to D. This upper right triangle is now divided into 4 sections. The first two sections are W+X, which is 16/9(Z). The third section is (5/2)-Z. This is because triangle GFD is 10/4=5/2, using the same sort of logic that gave us the area of X+Y. Finally, the fourth section is 5/3 just like X+Y. 16/9(Z)+(5/2)-Z+(5/3)=5. You can solve for Z, then plug in Z into 5/3 = 5/9(Y) + 4/3(Z) to solve for Y, the blue triangle.
This is one of those problems that, the presented solution implies, what I call "supernatural intuition."
Got it, solved it more or less the way the Presh did it.
HINT: Look for similar triangles and remember that the area of two similar triangles is in the same proportion as the square of the proportion of their sides (though he didn't use that fact directly)
Doing this, you end up with some simple algebra and a lot of arithmetic involving fractions. Who in 5th grade learns algebra, though ?
Would be really nice to have a second channel where Presh dissects most popular wrong solutions like here with 5/12
This channel was quite original a few years ago ... Now it regurgitates the same old tired stuff... maybe time(for me) to move on...
Math : Geometry problem
th-cam.com/video/4dkOeJXWDW4/w-d-xo.html
See it for fun.
In India this type of problem is given to 13-15 year olds..
Yes
You said write someone is saying for 2 years old
maybe in olmpyiad but not in school
@@krishan9739 well i solved in schools too ...
@@rajeshsinghtomar1024 right*
And here I am watching that big triangle is 1/6 of the box, and blue is about 1/4 of the triangle (little over). So it is about 10/24 ~ 0.417. and result was 0.429. less than 3% wrong, close enough for approximated in 10 seconds.
I posted that solution too!
I'm too tired right now to solve this one, but here's the steps I would take:
1) Set a coordinate system with origin on point A. Consider point D to be (L,0), then point B is (0, 10/L).
2) Determine the equation f(x) of the diagonal, considering it passes through points B and D.
3) Determine the equation for line g(x) that passes through (L/2,0) and (L/3,10/L)
4) Determine the equation for line h(x) that passes through points (L/2,0) and (2L/3,10/L)
5) Calculate point H as the intersection of f and g; and point I as the intersection of f and h.
6) We now have all the coordinates for the triangle. We can put them on a matrix and calculate the area.
That’s exactly what I did. It’s a very time consuming approach though.
Math : Geometry problem
th-cam.com/video/4dkOeJXWDW4/w-d-xo.html
See it for fun.
What software do you use?
Here's the easy way:
define a = height of rectangle
define b = width of rectangle
Chop and glue the triangle to construct an isosceles traingle with:
height = a/2
width = b/6
The area of this triangle thus becomes ab/24 = 10/24
So this is wrong but easy and close to the truth:
10/24 = 0.416667
3/7 = 0.4286
My guess is this is the solutions the teachers were looking for. Still hard for 10 year-olds IMO
For the 10 years old I think they (kids) should have drawn more similar, including one orizontal that divides the rectangle in half, and see that it can be divided in 24 equal parts, also they learn aproximation at this age, and the answer would be aproximativ 10/24, which is not far from the exact answer 3/7.
3:08 It is not REMARKABLE it is OBVIOUS. You are just doing a simple transformation.
Imagine a plane Y vs X in which there is a closed curve. Integrate to get the area A insidethat curve, then transform Y--->sY and X--->X/s where s is a constant and integrate to get the area again.
Since Ydx---> sYdx/s=Ydx you get the same area.
This is the second time that I hear him say that kind of nonsense.
Presh, you might wanna start seeing kids as kids, not as someone with your level of intellect. At the age of 10, kids are barely learning shapes and the formulae for their areas. They do not have area algebra or geometry or any form of analytical tools such as variable substitution. This problem clearly begins from 9th or 10th class. You might wanna fact check the person who E-mailed you what he/she really meant. This is not a problem for a 10 yr old, unless he is Sheldon Cooper.
bruh, there is regular 11 y/o doing this (IDK about 10 y/o though, it isn't that hard to solve though)
I am in class 10th. ....Still I don't get this😆
10 years olds? My eye....
Please tell how you comment before the upload
Dude forget geometry this man has a time machine
@@jackburns6403 i didn't understood what you are trying to explain ? I mean , how can a person comment earlier than the upload of a video . And if he has a time machine , then he must have got it patented .
@@allgenre2332 patreon I think
I thought Gougu Theorem is back here in this video😂
Math : Geometry problem
th-cam.com/video/4dkOeJXWDW4/w-d-xo.html
See it for fun.
Ah yes, the similar triangles, I still don't know how to use them
well, the answer i got is 0.8 i.e 5/6. so, 25/3 + 5/3 = 10
therefore, 8.33+1.66 = 10(approx as it is 9.99)
as the two trapeziums are equal, the triangle is isosceles,
hence the diagonal divides it into 2 parts.
so, 1.66/2 = 0.8(approx)
...clear as mud. So much for that math degree I have.
Are you kidding? 😆... Concepts of similar triangles in 4-5 th class student?!! I don't think that even international boards would be such hard... BTW great fan of yours and maths.... I Love ❤ ur videos 😊
"what's remarkable about the answer"..... is that it didn't require the use of the Gougu theorem at all
I don't presh sir ,why u don't make videos on ramanujan's master theorem and some high level mathematics.
me: thinking for a minute.... ok, i give up, lets use analytic geometry to solve this.
but its really shameful that i didnt see the "smart" solution. i should have really noticed all the similar triangles.
So the blue area is about 4% of the overall area. This might be rephrased as an optical illusion because the area looks a whole lot bigger than that in the picture.
Using analytic geometry to find GH and GI as vectors, then finding A(∆GHI) = ½[cross-product GH x GI], I get
A(∆GHI) = 3/7
This clearly isn't something the average fifth-grader (10-yo) would do, even in a fully competent educational system.
Let's see what Presh has up his sleeve this time, that could be expected of such a student...
Meanwhile, here was my initial, ballpark method of estimating the desired area.
First, use the fact that ∆EFG has ⅓ the base of, and equal height as, the rectangle, to get
A(∆EFG) = ⅙A(ABCD) = 5/3.
Interestingly, if you then erroneously assume that the rectangle's diagonal, BD, cuts through the isosceles ∆EFG in a way that the blue ∆ has ¼ the area of ∆EFG, just as a horizontal cut midway up the rectangle would do, you get
A(∆GHI) ≈ ¼(5/3) = 5/12
which is almost exactly right, being 35/36 of the true area, 3/7.
Post-view: That's a lot of work, but each step is something that at least the top tier of those students could come up with.
Bravo/brava to those who did solve it!
Fred
make video on why √x. sinx can not be integrated
Challenging problem given to newborns:
Prove riemann hypothesis
Thanks for the video!!
Which country's 10yr old are taught similar triangles and mid school geometry and application of algebra in geometry?
Everyone don't worry, he said just the problem was GIVEN to 10ys... No proof that they could solve.
Yes , you are right
Because in OLYMPIAD gifted students are selected
@@shivansh668 But I am 10y....... I solved
Math : Geometry problem
th-cam.com/video/4dkOeJXWDW4/w-d-xo.html
See it for fun
Another good problem
👍👍
Excellent problem.
But i dont understand how you got the values for the hights of the triangles just because they are similar
if they are similar,that means that all you need to scale (and rotate) them.
in first occurence ( 1:16 ) he knows one of sides of bigger one is half of S
anc corresponding side of smaller one is third of S
that gave him ratio for scaling them 3 / 2
to get value from smaller to larger, you multiply it by three and divide by two
to ge value from larger to smaller, you multiply by two and divide by three
he then stated that height of larger one is x
from that follows that if you multiply x by 2/3 , you get height of smaller one
so height h = x + 2/3x
that translates to (3/3 + 2/3 = 5/3) h = 5/3 x
so to get from x to h you multiply by 5 and divide by 3
if you want to reverse it,you multiply h by 3 and divide by 5
that got him
x = 3/5 h
no magic there :)
Excelente, obrigado por compartilhar.