Yeah, I was actually kinda annoyed to see this video looking like a repost, but decided to click it just in case. I am really glad it turned out to be about a completely different problem!
I was surprised @11:34, that you suddenly jumped to the “B” side of the problem, rather than continuing with the A side: if you had continued with the A side, the next green triangle and the first purple triangle would each have been 6-A, and the second purple triangle is then A-2. Leaving us with A-2=B, which rearranges to A-B=2. Neither method is necessarily better than the other-it just seemed odd to switch horses midstream.
A-2 from the second green triangle does not = B. it is equal to what ever value you would call that triangle. 8 = 6-A+2+A is what you would have so 8=8 which doesn't help you find B. It was just a fluke that you made A-2=B and it worked.
@@miguelviau3163 He didn't say that the second green triangle is equal to A - 2. He said the if you continue to go anticlockwise you would get to A - 2 = B. The second green triangle is equal to 8 - (2 + A), which is 6 - A. So the second purple triangle would be 4 - (6 - A), which is A - 2. Since the second purple triangle is equal to B, A - 2 = B, therefore A - B = 2.
I took a visual approach to the square problem and my initial guess was confirmed by the math approach in the video. I imagined a square with the point at the center so all 4 corner areas are equal. As you move the dot up or down, you add or remove equal areas above and below. Same things happen if you move left and right. Combine both movements and the opposite quadrants are linked / equal. So you end up with the same 16+32=20+? with 28 as an answer. But that only worked nicely with a square. It's still nice to know the more generalized method using the triangles.
Having done a similar visual approach from the thumbnail myself I think our method can be generalized to work for all regular even polygons, but it was nice to see the technique for odds demonstrated in the video.
@maxhagenauer24 im not sure what you meam by why - there's a proof of it in this video, both sets of opposite corners include areas (a+b+c+d), thus both sets must always equal each other, no matter where the vertex point is.
@skanderbeg152 What I mean is why the sum of the 2 corners is equal to the sum of the other 2 corners... Not sure what is confusing you about that. The video did not directly proof this but if it's If it's because there is an a + b + c + d is on both corners no matter where the vertex is so the corner areas are equal, I guess that makes sense but that still requires you to split them into triangles. So even this method still only makes sense if you split them into triangles I guess.
In the square problem, cross corner areas will always add to the same value, no matter where you put the point (even if you place it on one of the corners). 32 + 16 = 20 + x
After pausing the video on the completed pentagram, I determined that the values of A and B are either 3 and 1 or 4 and 2. Any other values would result in areas that are 0 or negative. Very interesting.
a very little nit: at 12:40 you say -a but should be +a (4-b plus 4+a) I think the -a came from prior step where A was minus. I love your content. Very interesting and educational.
What always bothers me in these kind of question is, when the sketch they provide is actually quite far from the real picture. I know the reasoning behind it, but it still bothers my inner Monk.
In the square, you can use the same triangles to prove that two opposite quadrilaterals sum to half the total area - since if you use the marked length as the base b for all four triangle area calculations, the sum of the heights can be factored out and identified as equal to 4b. (1/2)(b)(4b) = 2b^2, or half the total area.
These assignments are so unbelievably annoying to me because whenever I'm faced with a math problem, especially one with a visual representation like this one, I always feel the need to identify and put values on every single variable that I'm faced with, which is a sure-fire way to never be done with it lmao
I added the known areas, and figured that the unknown area had to be smaller than the largest, based on the relative sizes. I assumed it would be an even integer, and the sum of the known areas was 68, so 100 couldn't be the area of the entire square, as that would make the unknown area equal in size to the largest known area, which clearly wasn't the case. So I settled on 96, and subtracted the 68 to get the answer, 28.
On the square problem you can also add a coordinate system to solve it, say, put the origin at the down-left corner. Let l be the length of the side of the square, (x,y) be the point connected to the middles of each side. Just like it was done, trace lines from the (x,y) to the corners. The triangles will have have side l/2 and height x, y, l-x or l-y depending on the point, you can have 3 equations on the areas: (l/2)x/2 + (l/2)y/2=16 [I] (l/2)x/2 + (l/2)(l-y)/2=20 [II] (l/2)(l-x)/2 + (l/2)(l-y)/2=32 [III] You can solve it by adding [I] and [III] (l/2)x/2 + (l/2)y/2 + (l/2)(l-x)/2 + (l/2)(l-y)/2=16+32 (l/2)/2 (x + (l-x)) + (l/2)/2 (y+(l-y)) = 48 l²/4+l²/4=48 l²/2=48 l²=96 We want to find the area of the other region, we know that the area of the square as a whole is 96, so we know 16+20+32+?=96, solving for ? we get ?=28
I didn't do any of that! I just added the top right and bottom left (so 48) and knew that the top left and bottom right were also 48. Then subtracted the top left (-20) , and got 28 as my final answer
@@verkuilb It's the correct answer to the first question, solved in a simpler manner than what is shown in the video. Here the time stamp, since you obviously skipped this part of the video: 6:53
this question is a beginners question for the one's who have just learnt this concept of cevians and triangles.... in fact that square question is too old, like I did this when I had just learnt these concepts but I accept that that pentagon type question was very hard at the first glance if we didn't know this concept... But thanks a lot as I get too many questions to practice on this channel
Or, more algebraically: given brown(a+b)=5, blue(b+c)=7, green(c+d)=8, purple(d+e)=4. Let A=a and B=e. We solve for a-e, which we can do by (a+b)-(b+c)-(d+e)+(c+d) = (a-e). 5-7-4+8=2
If you speak Spanish you can find more problems like this in the book _"Geometria 10 - Areas de Regiones Poligonales y Circulares by Didy Ricra Osorio, editorial Cuzcano"_
Maybe another way to solve stuff like this is that if it works for an arbitrary point, then it must work for an easy point. Move the point around until you get some nice shapes that easily fit the numbers
Great problem. I solved the pentagon one slightly different. With a similar strategy as you used for the square, I came to the conclusion that: A + 4 (purple) + 7 (blue) = B + 8 (green) + 5 (orange) A + 11 = B + 13 13 - 11 = 2 so A = B = 2
Yes I can't solve the problem. That said, bro really only read the comments after 5 years... So much for making the community great... *salty in never getting noticed*
Just seemed intuitive that the upper left region plus the lower right region equal the upper right plus lower left regions. I wonder if I learned that somewhere or if it just seemed logical.
P be the interior point of this square. ABCD . Mid points of AB, BC, CD and DA be E, F, G and H. |∆APE| = |∆BPE| = a, say |∆BPF| = |∆CPF| = b say |∆CPG| = |∆DPG| = c, say |∆DPH = |∆APH| = d say Given a + d = x a + b = y b + c = z We need to evaluate c + d = b + c + d + a - ( a + b) = z + x - y
A regular pentagon ( all side equal) condition may not be needed. Since for four sided you have taken it square. It may work for rectangle (any quadrilateral)?
11:07 wouldnt you be unable to simplify this since 5 - A is a quanitity in parenthesis and the associative property doesnt apply to subtraction? ie. (10 - 5) - 2 = 3 but 10 - (5 - 2) = 7
Essentially, since subtracting a positive and adding a negative are the same thing, you are distributing a negative one through the parenthesis. i.e. 10 - (5 - 2) = 10 + -1(5 + -2) = 10 + -5 + 2 = 7 As rohit71090 shows in his reply, this applies the same to variables.
It becomes clear that, if the polygon has an odd number of sides, you can work out a difference, whereas if it's even sided, you can work out a sum. Another thing: we need a follow-up on the Turkish student. They must be less than a year from graduation right now.
@MindYourDecisions I used the first shown approach and solved the problem equally correct. A+B=? .......(1) B+C=4 ......(2) C+D=8 ......(3) D+E=7 ......(4) E+A=5 ......(5) By subtracting equation 5 from 4, we get D-A=2 ......(6) By subtracting equation 2 from 3, we get D-B=4, ......(7) Lastly, subtracting equation 6 from 7, we get -B-(-A)=2, which is similar to A-B=2.
You gotta change the video cover to the colourful pentagon, otherwise it is looking like a repost right now.
Yeah, I was actually kinda annoyed to see this video looking like a repost, but decided to click it just in case. I am really glad it turned out to be about a completely different problem!
I didn't think he did reposts?
bigger question: do they read comments nowadays?
I was surprised @11:34, that you suddenly jumped to the “B” side of the problem, rather than continuing with the A side: if you had continued with the A side, the next green triangle and the first purple triangle would each have been 6-A, and the second purple triangle is then A-2. Leaving us with A-2=B, which rearranges to A-B=2. Neither method is necessarily better than the other-it just seemed odd to switch horses midstream.
Yep. My instinct was to just keep going around. And it works out just the same of course.
Same. I was coming to say this.
B was feeling left out
A-2 from the second green triangle does not = B. it is equal to what ever value you would call that triangle. 8 = 6-A+2+A is what you would have so 8=8 which doesn't help you find B. It was just a fluke that you made A-2=B and it worked.
@@miguelviau3163 He didn't say that the second green triangle is equal to A - 2. He said the if you continue to go anticlockwise you would get to A - 2 = B. The second green triangle is equal to 8 - (2 + A), which is 6 - A. So the second purple triangle would be 4 - (6 - A), which is A - 2. Since the second purple triangle is equal to B, A - 2 = B, therefore A - B = 2.
This is becoming one of my favorite channels.
I took a visual approach to the square problem and my initial guess was confirmed by the math approach in the video. I imagined a square with the point at the center so all 4 corner areas are equal. As you move the dot up or down, you add or remove equal areas above and below. Same things happen if you move left and right. Combine both movements and the opposite quadrants are linked / equal. So you end up with the same 16+32=20+? with 28 as an answer. But that only worked nicely with a square. It's still nice to know the more generalized method using the triangles.
Having done a similar visual approach from the thumbnail myself I think our method can be generalized to work for all regular even polygons, but it was nice to see the technique for odds demonstrated in the video.
But why would the sum of opposite corner ones be equal?
@maxhagenauer24 im not sure what you meam by why - there's a proof of it in this video, both sets of opposite corners include areas (a+b+c+d), thus both sets must always equal each other, no matter where the vertex point is.
@skanderbeg152 What I mean is why the sum of the 2 corners is equal to the sum of the other 2 corners... Not sure what is confusing you about that. The video did not directly proof this but if it's If it's because there is an a + b + c + d is on both corners no matter where the vertex is so the corner areas are equal, I guess that makes sense but that still requires you to split them into triangles. So even this method still only makes sense if you split them into triangles I guess.
@@maxhagenauer24 This works due to the rotational symmetry and equal spacing of the edge midpoints in these regular polygons.
In the square problem, cross corner areas will always add to the same value, no matter where you put the point (even if you place it on one of the corners).
32 + 16 = 20 + x
Does it only apply when lines start form the middle of the sides?
@@slayzet4293 Yes, because otherwise the triangles (in the proof) wouldn't have the same areas.
After pausing the video on the completed pentagram, I determined that the values of A and B are either 3 and 1 or 4 and 2. Any other values would result in areas that are 0 or negative. Very interesting.
a very little nit: at 12:40 you say -a but should be +a (4-b plus 4+a) I think the -a came from prior step where A was minus.
I love your content. Very interesting and educational.
What always bothers me in these kind of question is, when the sketch they provide is actually quite far from the real picture. I know the reasoning behind it, but it still bothers my inner Monk.
In the square, you can use the same triangles to prove that two opposite quadrilaterals sum to half the total area - since if you use the marked length as the base b for all four triangle area calculations, the sum of the heights can be factored out and identified as equal to 4b. (1/2)(b)(4b) = 2b^2, or half the total area.
These assignments are so unbelievably annoying to me because whenever I'm faced with a math problem, especially one with a visual representation like this one, I always feel the need to identify and put values on every single variable that I'm faced with, which is a sure-fire way to never be done with it lmao
I added the known areas, and figured that the unknown area had to be smaller than the largest, based on the relative sizes. I assumed it would be an even integer, and the sum of the known areas was 68, so 100 couldn't be the area of the entire square, as that would make the unknown area equal in size to the largest known area, which clearly wasn't the case. So I settled on 96, and subtracted the 68 to get the answer, 28.
On the square problem you can also add a coordinate system to solve it, say, put the origin at the down-left corner.
Let l be the length of the side of the square, (x,y) be the point connected to the middles of each side.
Just like it was done, trace lines from the (x,y) to the corners.
The triangles will have have side l/2 and height x, y, l-x or l-y depending on the point, you can have 3 equations on the areas:
(l/2)x/2 + (l/2)y/2=16 [I]
(l/2)x/2 + (l/2)(l-y)/2=20 [II]
(l/2)(l-x)/2 + (l/2)(l-y)/2=32 [III]
You can solve it by adding [I] and [III]
(l/2)x/2 + (l/2)y/2 + (l/2)(l-x)/2 + (l/2)(l-y)/2=16+32
(l/2)/2 (x + (l-x)) + (l/2)/2 (y+(l-y)) = 48
l²/4+l²/4=48
l²/2=48
l²=96
We want to find the area of the other region, we know that the area of the square as a whole is 96, so we know 16+20+32+?=96, solving for ? we get ?=28
Proud of myself. Solved in 10 seconds for 28 just adding them up in diagnols.
I took a wild guess and got A -B = 5,798. Not even close.
My guess is closer. 42
I like the way u explain. I use to learn the theorems but now I understood it today.
I use to know these. But I forgot. Lately I take a minute to calculate 23 * 3
Fun problem - I wouldn't have got either of them without your reminder about triangle area at the beginning.
Divide and conquer. A concept, while ancient, is always efficient.
An interesting viral question.
I didn't do any of that! I just added the top right and bottom left (so 48) and knew that the top left and bottom right were also 48. Then subtracted the top left (-20) , and got 28 as my final answer
Which isn’t the right answer…because it’s not the right problem. The new problem is presented @8:42.
@@verkuilb It's the correct answer to the first question, solved in a simpler manner than what is shown in the video. Here the time stamp, since you obviously skipped this part of the video: 6:53
god bless you presh !
Simple yet elegant
I learned so many things on this channel that my school will never teach me
I started from purple and went (almost) all the way around, using one equation at the time, to eventually reach A=2+B 🙂
Never saw such an easy Q in bharatiya entrance exam.
this question is a beginners question for the one's who have just learnt this concept of cevians and triangles.... in fact that square question is too old, like I did this when I had just learnt these concepts but I accept that that pentagon type question was very hard at the first glance if we didn't know this concept... But thanks a lot as I get too many questions to practice on this channel
Or, more algebraically: given brown(a+b)=5, blue(b+c)=7, green(c+d)=8, purple(d+e)=4. Let A=a and B=e. We solve for a-e, which we can do by (a+b)-(b+c)-(d+e)+(c+d) = (a-e). 5-7-4+8=2
If you speak Spanish you can find more problems like this in the book _"Geometria 10 - Areas de Regiones Poligonales y Circulares by Didy Ricra Osorio, editorial Cuzcano"_
That was a challenge.
Maybe another way to solve stuff like this is that if it works for an arbitrary point, then it must work for an easy point. Move the point around until you get some nice shapes that easily fit the numbers
Nice video brother!
Coordinate geometry is only a little messy but gives the answer.
Presh - Spends hours making an awesome video.
The Internet - That's Orange dude, not brown.
Eyvallah aga bak AYT çıkmış sorusu 😅❤🤕
Amazing indeed!
Great problem.
I solved the pentagon one slightly different.
With a similar strategy as you used for the square, I came to the conclusion that:
A + 4 (purple) + 7 (blue) = B + 8 (green) + 5 (orange)
A + 11 = B + 13
13 - 11 = 2 so A = B = 2
Fantastic!!!
Yes I can't solve the problem.
That said, bro really only read the comments after 5 years... So much for making the community great... *salty in never getting noticed*
Brilliant!
PT: FYI, at 12:40 you say "minus A" when you mean "plus A".
"Plus 2 minus A" oops
He mean at 12:40
8:09 so at this point d - c is equal to d - a. And that's how we can get a-b
This was great.
This is my guess before watching the video 30 cm². May not be exact but definitely in that range 0:14
⚠minor script error at 12:41 : you incorrectly say "2-A", but on screen it is correct as "2+A"
Just seemed intuitive that the upper left region plus the lower right region equal the upper right plus lower left regions. I wonder if I learned that somewhere or if it just seemed logical.
The sum of opposite (diagonal) triangles in a rectangle are equal
From Türkiye 💗
Yeah I didn't expect people from Turkey to watch this channel I'm from Turkey too
(a+b)+(c+d)-(b+c)
P be the interior point of this square. ABCD . Mid points of AB, BC, CD and DA be E, F, G and H.
|∆APE| = |∆BPE| = a, say
|∆BPF| = |∆CPF| = b say
|∆CPG| = |∆DPG| = c, say
|∆DPH = |∆APH| = d say
Given a + d = x
a + b = y
b + c = z
We need to evaluate
c + d = b + c + d + a - ( a + b)
= z + x - y
I got stuck at 4:18 trying to color calibrate my monitor to "yellow" and "brown".
A regular pentagon ( all side equal) condition may not be needed. Since for four sided you have taken it square. It may work for rectangle (any quadrilateral)?
11:07 wouldnt you be unable to simplify this since 5 - A is a quanitity in parenthesis and the associative property doesnt apply to subtraction?
ie. (10 - 5) - 2 = 3 but 10 - (5 - 2) = 7
If you try to simplify 10 - (5 - 2) = 10 - 5 + 2 = 7.
Similarly, 7 - (5 - A) = 7 - 5 + A = 2 + A.
Was this your confusion ?
Essentially, since subtracting a positive and adding a negative are the same thing, you are distributing a negative one through the parenthesis.
i.e.
10 - (5 - 2)
= 10 + -1(5 + -2)
= 10 + -5 + 2
= 7
As rohit71090 shows in his reply, this applies the same to variables.
@@rohit71090 ah i see, thank you
Also from India
i got the area 13 cm². The result is obtained from the area of the nearest square,
the area obtained in question is 13 cm²
Okay here we go again A-B is maybe 1
wth happened at 5:23
You're back😂
Oh, once you break the segments of the square into those triangles, it becomes a lot like the recent weighing problems, doesn't it?
Прикольно вийшло!
I totally forgot that the line to the midpoint divides a triangle into 2 equal areas
20+12=32. Therefore 16+12=28. I got lucky.
Hi from India.
Me who guessed the area of the blue region as 28 because all the numbers were multiples of 4 and 24 seemed to small cause of 20😎
Why solve tor A-B ? Shouldn't it be solved for A+B ?
22 cm² idk, i havent take geometry yet
4:07 You call that brown?
Being colourblind, I wouldn't know.
How can you have such an easy question in a univercity entrance exam?
Why you call that orange, brown?!
Brown is just orange but darker.
Yes, brown and orange have the same hue. Technology Connections has a video about it.
did you just call orange brown 4:05
hermoso problema
28
28?
Wrong problem. 28 is the solution to the OLD “reposted” problem, not the new problem discussed in this video.
It becomes clear that, if the polygon has an odd number of sides, you can work out a difference, whereas if it's even sided, you can work out a sum. Another thing: we need a follow-up on the Turkish student. They must be less than a year from graduation right now.
Hi😊
Hi
@MindYourDecisions I used the first shown approach and solved the problem equally correct.
A+B=? .......(1)
B+C=4 ......(2)
C+D=8 ......(3)
D+E=7 ......(4)
E+A=5 ......(5)
By subtracting equation 5 from 4, we get D-A=2 ......(6)
By subtracting equation 2 from 3, we get D-B=4, ......(7)
Lastly, subtracting equation 6 from 7, we get -B-(-A)=2, which is similar to A-B=2.