3 problems to find the shaded area

[@GourangaPL]

in problem #2 you can also notice that bottom left triangle in bigger square and the small triangle in bottom left of shaded region are similar in scale 3:1 so the bottom left horizontal line (from left to above point C) of length 40 is divided into 30 + 10 in the intersection point with shaded region border, therefore the horizontal line inside shaded region is 10+20+20 = 50 and shaded region is 2 triangles of base 50, one height of 60, one of 10

[@Pizzzaxz]

Nice observation

[@BalderOdinson]

second triangle is height 20, but yes! I was trying to figure out how to word this observation and you nailed it!

[@GourangaPL]

right it's height 20, my bad

[@marxcarton3858]

MindYourDecisions pls post more Olympiad problems, those are the interesting ones!!

[@theelk801]

in the second one I noticed that the triangle is a right triangle and from there it’s easy to solve but your solution is definitely more elegant

[@rubikscubedude4028]

For #3, you could focus on one side of the rectangle, since both sides are just mirrored versions of each other. In the end, we scale the half by 2. If you overlay the first and final states of the page, you see that the missing area is just a right triangle, with the base of 3 and a height of the combined length 3 (with of page) + √2 (length of 45° hypotenuse that gets folded on center crease). So, area of the triangle is 3/2 (3+3√2). Area of 1/2 of the rectangle is 3*8 Area of 1/2 of new shape is 3*8 - (9/2)(1+√2) or 3(8 - 3(1+√2)/2) Total area = 3*8*2 - 9(1+√2) or 39 - 9√2) Or, just look how you are subtracting 2 45° triangles and 2 3*(3+√3)/2 triangles, subtracting the 45° from that (net 0 45° triangles). Or, I'm just overanalyzing a simple math problem lol

[@compiling]

I had different solutions for problems 1 and 3. For 1, the square plus the external triangle (congruent to the one you constructed) equals the rhombus plus the area we want. For 3, everytime you fold the corner in you bisect the angle at the top. So we need to take the original rectangle, and subtract 2 triangles with sidelengths 3 * 3 cot(pi/8). cot(pi/8) is 1+sqrt(2) (looked it up), so we get 48 - 2(9 + 9sqrt(2))/2 = 39 - 9sqrt(2).

[@duminformatyka]

putting everything into a giant square makes problem 2 "in box thinking" 😁

[@hlee4248]

To be rigorous for problem #3, one should show that the length of the initial rectangle is indeed sufficient to form the final shape as shown. Your method would not work if the length is less than 3+3*sqrt(2).

[@shohamsen8986]

Problem 1: easy. Note that a^2=625 for the big square, and a*b=525 for the rhombus where b(=a sin\theta) is the vertical projected length of the other side. The shaded area can be calculated form a and \theta. Problem 2: The vertices lies on a grid point. This signals picks's theorem. Problem 3. The final shape is a triangle on top of a rectangle. What we need to determine is the height of the triangle. The Angle for the final fold is 22.5 degree. Since the angles get halved at each turn. Since the base is always 6, you can compute the height of the triangle.

[@peterkwan1448]

For the first problem. we can find out that DG is 20 using the area of the parallelogram, since DG is perpendicular to EF, so it is the height of AEFD. Then use the Pythagorus' Theorem on DGF, GF = 15. This means that the area of DGF = 15 * 20 / 2 = 150. The area of AEGD = The area of AEFD - The area of DGF = 500 - 150 = 350 The area of ABCGEA = The area of ABCD - The area of AEGD = 625 - 350 = 275

[@foryx_]

you can use basic trig to solve the third one as well. All given triangles are isosceles, because in order for the side to align with the centre line, the other line needs to be the same length (I'm not sure about the proof off this, but it makes intuitive sense as an origamist). The second pair of triangles are isosceles. the first pair of triangles are isosceles as well as right angles, so the non-right angles are 45 degrees, and the supplementary angles are 135 degrees. now we can use 1/2 ab sin C to find the area of the second pair, and subtract that from the total

[@Qermaq]

For #3, the first fold can be ignored. We need the sum of the area of an isosceles triangle (with base 6 and height 3root2 + 3) and the area of a rectangle (with base 6 and height 5 - 3root2). Seemed more direct to me.

[@YouFydes]

Good, but i can't proof that after each fold we have a rhombus. It seems so obvious.

[@verkuilb]

You could do that-once you establish that the height is indeed “3root2 + 3”. In Presh’s solution, that height is never determined. (The 3root2 is determined, but it’s never proven that the rest of the height is 3.). As long as you complete that step, then yes, your method would indeed be correct.

[@Qermaq]

@@verkuilbWell yeah, that's predicated on the trig functions of pi/8. Naturally I'd be more rigorous when required.

[@oliviervancantfort5327]

No, in the final airplane the top blue triangle height is 3. The base rectangle height is well 5-3root2. Proof that the triangle right is 3 can be obtained if you observe that the green triangle are all isocèles. This can be proven by their angles. In the last green triangles, top angle = pi/8, middle angle is 3pi/4, and as the sum must be pi, bottom angle is also pi/8.

[@Qermaq]

@@oliviervancantfort5327 It would depend on which triangle you were looking at. My approach basically divides the finished paper airplane into an isosceles triangle with apex angle pi/4 and a small narrow rectangle under that. You can make different triangles and get to the correct value in various ways.

[@lollol-tt3fx]

Thank you for creating such nice educational content.

[@Blosp]

for problem 2, use the shoelace formula to simplify what you did. it does the same thing but quicker

[@billogdaeggog9071]

I have a completely different solution for problem #2. If you look closely we'll see that the length of AB will be 100, as we know A to lets assume D on the L.H.S is 60 and DB (next to A point) will be, 60+20=80 And from Pythagorean theorem , we get AB = 100 Now we know that ACB is a right angle triangle, so, if we find AC and multiply it using the Base (b) × 1/2 we will get the area. So to find the height I created the diagonal CB of the rectangle just like you created in the video, and again with the Pythagorean theorem we got BC (base) = 20√5 Now again using the theorem we get AC = 40√5 And now, 1/2 × 40√5 × 20√5 = 2000 So the area of the triangle is 2000.

[@omaryassien7130]

quick question since in problem 1 side of the triangle FGD (due to FD being a side of the rhombus) and after constructing the perpendicular EH we had a height that is equal to GD therfore GF was 15 by pythag and then we could get area of triangle by GF*GD/2=150 and then subtracting that area from the rhombus which gives us 350 for area of EGDA which we then subtract from square ABCD giving us the area of figure ABCGEA

[@missirina6912]

1) ❤ My thought was about ABCD-AEFD area (625-500) + area of AIE (which is = to DGF = 150) = 125+150 = 275 ❤ 3) WOW, how easy it is! 😂 I thought about the angle of that point top which is 45° in the finish finale... 😄 The 2-nd was way too easy! When you think of the vectors, and about those distances between the points of a segment and the origine of the Cartesian Reper, you are quickly used to see those segments you was talking about that made the whole big square... 👍

[@quigonkenny]

Problem 1: Extend FE so that it intersects and is perpendicular to AB at H. The area of a rhombus is base × height, so the area of BCGH is going to be 625-500 or 125 m². Now we need to find the length of EH and AH so we can calculate the area of the triangle ∆AHE which is taking up the remainder of the red area. As suggested in the previous portion, the height of the rhombus, AH, is equal to the ratio of the areas of the two polygons times the length of a side of the square, which is √625 = 25 m. AH = (500/625)25 = 500/25 = 20 m a² + b² = c² a² + 20² = 25² a² = 625 - 400 = 225 a = √225 = 15 m A = bh/2 = 20(15)/2 = 150 m² Total area = 125 + 150 = 275 m² Problem 2: Extend the lines C and B are on so as to construct an 80 u it per side square. We'll call it ADEF, read counterclockwise so that C is on DE and B is on EF. By observation, the area we are looking for is this new square minus the triangles ∆ADC, ∆CEB, and ∆BFA. This is surprisingly simple. For simplicity's sake, the base in each of these triangle area calculations will be the long leg. A = s² - bh/2 - bh/2 - bh/2 A = 80² - 80(40)/2 - 40(20)/2 - 80(60)/2 A = 6400 - 1600 - 400 - 2400 A = 2000 unit² Problem 3: As each fold is effectively an angle bisector, two folds on each side mean that the final angle of each of the "wings" of the figure is 90°/4 = 22.5°. The final figure is effectively two congruent but flipped right triangles each with narrow angle of 22.5°, base of 3, and height of 8-x sitting on top of a narrow rectangle of (horizontal) width 6 and height x. it would be fairly simple to get the value of 8-x from the cosine of 22.5°, but we're not doing it that way. Let's note the initial width is 6. The "point" of the paper airplane will be O. On the first fold, the upper right corner (let's call it A) folds in to a position W/2 (6/2 = 3) units from O. This forms a new corner B which is 3 down and 3 to the right of O. c² = a² + b² = (3)² + (3)² = 18 c = √18 = 3√2 Note that the corner B is a 135° (180-45) angle. As our next fold will be creating a 22.5° angle, thst makes the area we will be folding over an isoceles triangle, and thus the distance from O to B and from B to the new corner of C are the same. As one line is straight down and the other is at a 45° angle, vertical distance from O is 8-x = 3√2+3. 8 - x = 3 + 3√2 x = 8 - 3 - 3√2 = 5 - 3√2 A = 2(bh/2) + wh A = 2(3)(8-x)/2 + 6x A = 3(3+3√2) + 6(5-3√2) A = 9 + 9√2 + 30 - 18√2 A = 39 - 9√2 = 3(13-3√2) ≈ 26.27 unit²

[@pros_mega]

In the third problem you can use the formula "A=1/2ab*sin alpha" to calculate the area of the second pair of triangles. This way it's faster

[@knutolavbjrgaas1069]

On problem 3# I just kept track of the angle of the right triangles formed by the folding, 45° -> 22.5°. then did tan22.5=3/x where x would be the height of the equilateral triangle in the final shape. Then i did (6*3)/(2*tan22.5) for the area of that triangle, then added 6*(8-3/tan22.5) for the remaining rectangle at the bottom. Got the same answer.

[@mirfayozmirgiyosov9594]

In the first problem, I subtracted the area of AEGD trapezium from the area of the square. Sabcgea = Sabcd - Saegd= 625 - 350 = 275

[@EmanuelsWorkbench]

For the first one, I found it easier to calculate area of the square - area of the rhombus + area of the external triangle...

[@deerh2o]

Problem 1, I took the square and subtracted out triangle AEH and rectangle EGDH -- a little easier than the construction given in the video.

[@JLvatron]

I solved them! Granted the 2nd 1 was made easier, once you said think outside the box, the simple solution clicked!

[@mrinaldas5714]

Solved all 3 problems in first attempt (I'm still worthy)

[@Drawliphant]

problem 2 triangle is already a right triangle. find two leg lengths and solve sqrt(80^2+40^2)*sqrt(40^2+20^2)/2

[@Qaptyl]

im thinking you could also use picks theorem, since 20 is divisible by 60, form a lattice with 20x20 squares. 4 lattice points fall on the edge of the shape, and there are 4 points you can find inside it. then do the formula (4/2-1)+4 = 1+4 = 5. then since the squares in the lattice are all 20x20, multiply by 400. 5x400=2000 which did end up being the final answer.

[@stuartmcconnachie]

8:42 You should also check that the height of the rectangle (8) is sufficient to make the shape in the third diagram ;) i.e. Perhaps the second set of “wings” of the aeroplane fold down to below the bottom edge, and it is simply not possible to make the construction as shown. You should never assume that the problem diagrams are drawn in proportion or to scale.

[@bladdnun3016]

I would expect the correct topology.

[@BalderOdinson]

@@bladdnun3016 and you could be wrong. Even in a real world situation: what if the picture is accurate, but the given measurements were listed incorrectly?

[@dannyroosenboom3640]

the square and the rhombus was easy to solve. Nice round numbers so simple math in the head (by heart)

[@egyptian20091]

I tried to solve problem 1 for hours and then I gave up all because I thought that ADFE is a parallelogram not a rhombus and I didn't realize that it is a rhombus except when I saw the solution 😐 And also there is a much simpler way to solve problem 1 which is by getting the area of the triangle GFD then subtract it from the area of the rhombus then subtract the solution from the area of the square

[@georgestoica9744]

Hello I love tour videos and I was wondering if you could make a video about grahams number or tetration or pentation and their real world uses

[@Ditt_ty]

For 1st and 3rd... I found the answer quickly and they are 275u² and 39-9r2 u²

[@bxyhxyh]

For, third problem, area is never changed. Just because you folded paper doesn't change paper area. So it's never changed.

[@tonypapas9854]

Fun take and video - thanks.

[@عليالعبدالله-ق3ب]

Do you have this in PDF 📝??

[@iamadooddood4331]

Surely I can't be the only one who noticed that ABC is a right triangle. AB² = 80² + 60² = 10000 AC² = 80² + 40² = 8000 BC² = 40² + 20² = 2000

[@hyperboloidofonesheet1036]

The area of the paper doesn't change unless you actually cut it, so its still 48.

[@zecuse]

The question is about the area of the shape formed by the boundary of the folded paper's edges, not the paper itself despite being folded over itself and technically still being the same area if we considered the folded geometry. Only in 3D space does the paper not lose area when it's folded. Projecting into a 2D space reduces its area if it's not perfectly parallel.

[@verkuilb]

Listen to the problem again. At 1:58, he specifically states, “what is the area OF THIS SHAPE?” Not of the paper.

[@craftynoah452]

In problem one, if you split the rhombus into two equilateral triangles, each should have an area of 250. If you then split the right most triangle down the middle, it should give two triangles of area 125. now that you have split the rhombus into sections that overlap the square and that don't, subtract the 375 area that DOES overlap from the area of the square 625, to find the wrong answer of 250. why does this not work?

[@craftynoah452]

TURNS OUT, a rhombus can't be split into two equilateral triangles, it's been a while since I was in Preschool, I don't know my shapes lol

[@DevPaswan-b4d]

as a 10th grader i solved all of 3 but it takes much time as i am doing these questions with basic concepts

[@pierreabbat6157]

In problem #2, the triangle is a right triangle with both legs having a factor of √5. That's how I figured it out.

[@natashok4346]

Better, if Presh presents one hard problem of geometry, than a lot of, but easy. 🐶

[@Storybookie]

Its a lot easier to solve the 2nd problem using similarity

[@krzysztofmazurkiewicz5270]

First was easy, second was tricky as you need to "see" the big square to make it easy, the third one was a bit evil and wit this one i just watched the answer as i was not sure i had the right idea. And proposed resolution was way easier so either id get a wrong answer or i was going the wrong way ;)

[@j100j]

I didn't get the first one because I am not too good at english and the geometry vocabulary is new to me so I did not know what a rhombus was.

[@GaurangAgrawal2]

My Answers 1. 275 sq. units 2. 2000 sq. units 3. (i) 48 sq. units (ii) 39 sq. units (iii) 39 - 6√3 sq. units

[@marxcarton3858]

iii) is wrong it is supposed to be 39-9√3

[@vitalibahdanovich8072]

Too easy 😮

[@froge101]

find the shaded 136 area

[@xuanhuy8341]

275,2000

[@Nini_not_lucy]

Love thiss

[@mjolnir3309]

First. The area stays the same in 3 dimensions. Second, they are a couple folds short.

[@Sohail55447]

I provide online tutorial mathematics statistics per calculus calculus 1 and 2 physics biology and chemistry

[@mickelsantiagoquispenamuch4961]

25^2 + 1 like is here

[@Sohail55447]

I provide online tutorial mathematics statistics per calculus calculus 1 and 2 physics biology and chemistry

[@malkiz3750]

in 2nd i used Heron's formula xdd

[@abhinavgupta2511]

Evil maths

[@Sohail55447]

I provide online tutorial mathematics statistics per calculus calculus 1 and 2 physics biology and chemistry