Can you solve this viral problem?

[@stevebailey805]

PT: FYI, at <a href="#" class="seekto" data-time="760">12:40</a> you say "minus A" when you mean "plus A".

[@MindYourDecisions]

Thanks for letting me know! During editing I usually find a few mistakes and clean them up, but I missed this one.

[@stephaneallimant8294]

Im ok with you

[@jasonmcclane8744]

Or on the pentagon after dividing each section into triangles going clockwise you would have b+c=4, c+d=8, d+e=7, e+a=5 then a+(b+c)+(d+e)=a+4+7=a+11 and b+(a+e)+(c+d)=b+5+8=b+13 so that a+b+c+d+e=(a+11=b+13) then subtract 11 from each side to get a=b+13-11=b+2 then subtract b from each side to get a-b=2

[@intheinternet8597]

@@MindYourDecisions a+d=c+d-b+c+a+b a+b+c+d=b+c+ a+d (28cm2)

[@vekyxeivo5595]

I want to say this too.

[@GD-mw1kd]

You gotta change the video cover to the colourful pentagon, otherwise it is looking like a repost right now.

[@HenrikMyrhaug]

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!

[@vincent412l7]

I didn't think he did reposts?

[@simpleperson98]

bigger question: do they read comments nowadays?

[@CheckmateSurvivor]

@@simpleperson98 I just did. God bless you.

[@timseytiger9280]

Yes, I wasn't aware of the old video, solved the question in the thumbnail and then watched the video, expecting to compare approaches. I'd have preferred to spend my time on the more interesting pentagon question.

[@verkuilb]

I was surprised @<a href="#" class="seekto" data-time="694">11:34</a>, 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.

[@travcollier]

Yep. My instinct was to just keep going around. And it works out just the same of course.

[@TurkishKS]

Same. I was coming to say this.

[@diniaadil6154]

B was feeling left out

[@miguelviau3163]

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.

[@mrgabreis]

@@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.

[@marcosreal11]

This is becoming one of my favorite channels.

[@CheckmateSurvivor]

Sorry to hear that.

[@randomgamer-te8op]

Glad to hear that :D

[@TurkishKS]

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

[@slayzet4293]

Does it only apply when lines start form the middle of the sides?

[@fireball54110]

@@slayzet4293 Yes, because otherwise the triangles (in the proof) wouldn't have the same areas.

[@iainrobinson4280]

That's intuitively correct too; if you move the point to even out both the / corner pair and the \ corner pair , you'll have it divided into four even squares

[@heldercomp]

That is true because cross corner areas are comprised of 4 triangles: 2 add up for 1/4 of the square, and 2 have a constant base and the sum of their heights is constant too.

[@donaldshockley4116]

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.

[@aaab6054]

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]

But why would the sum of opposite corner ones be equal?

[@skanderbeg152]

​@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.

[@maxhagenauer24]

@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.

[@aaab6054]

@@maxhagenauer24 This works due to the rotational symmetry and equal spacing of the edge midpoints in these regular polygons.

[@RobbieHatley]

Interesting. I solved the first problem (with the square) in a slightly different way, closer to how you solved the pentagon problem: a+b=16 => b=16-a b+c=20 => c=20-b=20-(16-a)=a+4 c+d=32 => d=32-c=32-(a+4)=28-a Hence a+d=a+(28-a)=28

[@Doktor_Vem]

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

[@DrZaius3141]

I did the square one differently. Connect all the midpoints to form a square on the inside. This square separates all the regions into an "outer" part that's the same for every region and an inner part. Those inner parts have the same base (1/sqrt(2) of the square's side). All the heights of the triangles are orthogonal to the base, meaning that diametrically opposed heights are parallel to each other and actually meet - meaning together their length is the side of the inner square. This means that the inner triangles of bottom left + top right has to be the same as bottom right + top left.

[@highelectricaltemperature]

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.

[@kevinruggles206]

a very little nit: at <a href="#" class="seekto" data-time="760">12:40</a> 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.

[@MindYourDecisions]

Thanks for letting me know! I just misspoke when reading. I usually have to clean up a few sentences during editing but I missed this one.

[@marcelluswallace6240]

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.

[@bobh6728]

Never take a drawing as being accurate. There are problems where the given values are impossible, but an inaccurate drawing makes it seem possible. The solver will sometimes use something from the drawing (like a certain angle is 90° or two segments are equal length) and not be able to figure out why there is no solution.

[@johnnye87]

Because they're supposed to test your ability to do maths, not your ability to operate a ruler

[@russellblake9850]

intentionally ... but sometimes the problem is impossible (there was a "mind your decisions" on this).

[@awfuldynne]

I ended up using the same method for both problems: after the video gave the hint about connecting the corners to make triangles, I labeled the triangles, thought "this is a system of equations", and solved for the expression the question asked for, or an expression equal to it: a+d for the square, and A - B for the pentagon.

[@jasonmcclane8744]

Or on the pentagon after dividing each section into triangles going clockwise you would have b+c=4, c+d=8, d+e=7, e+a=5 then a+(b+c)+(d+e)=a+4+7=a+11 and b+(a+e)+(c+d)=b+5+8=b+13 so that a+b+c+d+e=(a+11=b+13) then subtract 11 from each side to get a=b+13-11=b+2 then subtract b from each side to get a-b=2

[@ilregulator]

<a href="#" class="seekto" data-time="409">6:49</a> when you add the top two equations you get a+b+c+d = 48 Now you can just substitute in b+c = 20 and get a+d+20 = 48 and then a+d = 28

[@matheusjahnke8643]

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

[@whoff59]

Looks more complicated than the solution in the video. But an alternative way, okay.

[@richardthomas3639]

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.

[@dashyz3293]

It would be highly interesting to see your work and method.

[@richardthomas3639]

@@dashyz3293 It is rather simple. If A is 5 or greater than the section that is 5 - A would be 0 or negative. Similarly, if A is 2 or less than b would be 0 or negative as it is B = A - 2. Therefore 2 < A < 5 and 0 < B < 3. I simply choose the integer values in that range given that all the other areas were integer values. I admit that any values in that range could be the answer.

[@RJ-pm2hp]

Is there any particular reason you restricted A and B to being whole numbers?

[@richardthomas3639]

@@RJ-pm2hp At the time I was only thinking in whole numbers as I was looking at the diagram. When I was replying to an earlier message I realized that there is actually a range of values as long as A is greater than 2 and less than 5.

[@jeremyfmoses]

Fun problem - I wouldn't have got either of them without your reminder about triangle area at the beginning.

[@whoff59]

at <a href="#" class="seekto" data-time="424">7:04</a> a+b+c+d = 48 b+c = 20 you can substract both equations to get a+ d = 48 - 20 = 28

[@flyingspirit3549]

Intriguing problem! I admit I had to listen through the explanation a couple of times before I fully understood it.

[@hippophile]

A very geometrical way of looking at it is to imagine sliding the nexus point and noting that this does not change the sum of diagonally opposite quadrant areas (this takes a little proving). If the nexus is in the middle, opposite quadrants are all the same. Therefore 16+32 = 20+ "?", and therefore ? = 16 + 32 - 20 = 28. The bonus is that you have a general result that the sum of diagonally opposite quadrants is half the area of the large square (which also follows from the algebra, or look at the diagram at <a href="#" class="seekto" data-time="459">7:39</a>: both diagonally opposite quadrants are a+b+c+d, so no actual calculations needed!

[@mytube001]

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.

[@muskyoxes]

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

[@Aryan-o4y7g]

I like the way u explain. I use to learn the theorems but now I understood it today.

[@hikari1690]

I use to know these. But I forgot. Lately I take a minute to calculate 23 * 3

[@TedHopp]

The method generalizes to any convex polygon. If the polygon has an even number of sides, you can figure out the area of one corner region from the others. If ut has an odd number of sides, you can only figure out the difference between the areas of the two pieces when the unknown corner region is divided by a line to the vertex.

[@martijndekok]

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

[@jasonmcclane8744]

Or on the pentagon after dividing each section into triangles going clockwise you would have b+c=4, c+d=8, d+e=7, e+a=5 then a+(b+c)+(d+e)=a+4+7=a+11 and b+(a+e)+(c+d)=b+5+8=b+13 so that a+b+c+d+e=(a+11=b+13) then subtract 11 from each side to get a=b+13-11=b+2 then subtract b from each side to get a-b=2

[@russellblake9850]

1) for the first problem, extend to the general solution ... opposite quadrilaterals sum to the same, so 32+16 = 20+X, X = 28. 2) for the second problem I applied the same solution as the first (dividing quads into 2 equal triangles) then summing and subtracting you quickly get A-B = 2.

[@zdrastvutye]

<a href="#" class="seekto" data-time="200">3:20</a> there is a congruency theorem, so i have chosen la= in line 30 and later calculated a factor fg, read line 250: 10 print "mind your decisions-viral square december 2024":dim x(3,3),y(3,3),a(3) 20 f1=16:f3=32:f4=20:v1=f4/f3:v2=f4/f1:rem das verhaeltnis der flaechen, die gauss- 30 la=1:x(0,0)=0:y(0,0)=0:x(0,1)=la/2:y(0,1)=0:x(0,3)=0:y(0,3)=la/2:rem flaechenformel 40 x(1,0)=la:y(1,0)=0:x(1,1)=la:y(1,1)=la/2:x(1,3)=la/2:y(1,3)=0:rem anwenden *** 50 x(2,0)=la:y(2,0)=la:x(2,1)=la/2:y(2,1)=la:x(2,3)=la:y(2,3)=la/2 60 x(3,0)=0:y(3,0)=la:x(3,1)=0:y(3,1)=la/2:x(3,3)=la/2:y(3,3)=la 70 sw=1/23:ys=sw:goto 200 80 for au=0 to 3:x(au,2)=xs:y(au,2)=ys:next au:for au=0 to 3 90 a(au)=0:next au:for au=0 to 3:for bu=0 to 3:bp=bu+1:if bp=4 then bp=0 100 bm=bu-1:if bm=-1 then bm=3 110 da=x(au,bu)*(y(au,bp)-y(au,bm)):a(au)=a(au)+da:next bu:a(au)=abs(a(au))/2:next au 120 dg=a(2)*v1:dg=(dg-a(3))/la^2:return 130 xs=sw:gosub 80 140 dg1=dg:xs1=xs:xs=xs+sw:xs2=xs:gosub 80:if xs>la then return 150 if dg1*dg>0 then 140 160 xs=(xs1+xs2)/2:gosub 80:if dg1*dg>0 then xs1=xs else xs2=xs 170 if abs(dg)>1E-10 then 160 else return 180 xbu=x*mass:ybu=y*mass:return 190 gosub 130:df=a(0)*v2:df=(a(3)-df)/la^2:return 200 gosub 190:if xs>la then else 220 210 ys=ys+sw:goto 200 220 df1=df:ys1=ys:ys=ys+sw:gosub 190:ys2=ys:if df1*df>0 then 220 230 ys=(ys1+ys2)/2:gosub 190:if df1*df>0 then ys1=ys else ys2=ys 240 if abs(df)>1E-10 then 230 else print xs,ys 250 fg=(a(0)+a(2)+a(3))/(f1+f3+f4):print fg:av=(la^2-a(0)-a(2)-a(3))/fg 260 print"die verbleibende flaeche=";av:masx=1200/la:masy=850/la:vdu5 270 if masx 16 28 run in bbc basic sdl and hit ctrl tab to copy from the results window.you may add "@zoom%=1.4*@zoom%" at the start for fullscreen graphics.

[@edwardhuang1502]

It happends that we can do it with no hands. Because of the particular configuration, if we move the intersection point in the middle leftward for some amount, such that the area of upper-left change by -x, then these changes follows: upper-right: +x lower-left: -x lower-right: +x Similarly, if the intersection point move downward for some amount such that upper-left area change by +y, then upper-right: +y lower-left: -y lower-right: -y By comparing the differences of areas in the problem, we found if the point was moved from the true center, then x=6, y=2, the upper-left area have had -x+y, and the lower-right area we care have had +x-y, so we have lower-right = upper-left +2x-2y = 20 +12 -4 = 28. # We can also try: lower-right = upper-right -2y = 32 -4 = 28. lower-right = lower-left +2x = 16 +12 = 28. By the way, I believe algebra is a decent way to go if nice alternatives do not show up.

[@thegemguy1334]

Divide and conquer. A concept, while ancient, is always efficient.

[@Rand0081]

It is more of a logic problem. If you draw a point in the center and move it around, area will be stolen from a quarter and given to the opposite one, so their sum is always 1/2 of the area. So 32+16 = 48, 48-20=28.

[@INFOWeatherMedia]

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

[@sievertemyt3701]

If you were to do that square problem, same way by separating midpoints of each line segment on the square, if you are to drag a point to any part of that square, the areas diagonal of each other must equal the diagonals of the other areas sum. (32 + 16 = 20 + x)

[@jannegrey]

So for people who say "brown" is "orange" - this depends on contrast to the background. There is a video about Brown not being a "color". At least when it comes to light (which is what monitors emit).

[@jrb6434]

oops - I think you misstated something (at ~<a href="#" class="seekto" data-time="761">12:41</a>). Sorry, I did this a lot in school, too, while following along, as I was here. You said "2 minus A" where I think you meant "2 plus A" setting equal to 8 for the green region. It still made sense, because the explanation was so good. Fun problem, interesting solve, and good review.

[@jrb6434]

...and sorry again, as I now see that someone else already pointed this out, AND you even pinned it. Didn't mean to pile on.

[@EnginAtik]

Draw an inner square of which corners are the midpoints of the bigger square. You will see that the inner square is divided into four triangles. You will notice that the sum of the areas of the opposite triangles are the same. You can prove this by drawing lines that pass through the point in the center and they are perpendicular to the sides of the inner square. The length of this line is the sum of the heights of the opposite triangles. Then the sum of the areas of the opposite triangles is the length of this line times the base over 2. The height is also equal to the side of the inner square. The areas in the corners of the big square are the same. Then it directly follows that 16+32=20+x => x=28

[@proboiz_50]

I learned so many things on this channel that my school will never teach me

[@rev0live752]

<a href="#" class="seekto" data-time="266">4:26</a> How do we KNOW? Like yeah, it looks like they are the same but 1. Drawings are not always accurate and 2. It could also just be off by a tiny bit, unless I'm missing something

[@bananalegend194]

pretty sure its the double lines on the sides

[@randomgamer-te8op]

it is stated in both problems "connected to the MIDPOINT" 1:26 and the one from the other video

[@Grizzly01-vr4pn]

The hash marks denote line segments of equal length in the diagram. Therefore, it is a given of the question.

[@jasonmcclane8744]

Or on the pentagon after dividing each section into triangles going clockwise you would have b+c=4, c+d=8, d+e=7, e+a=5 then a+(b+c)+(d+e)=a+4+7=a+11 and b+(a+e)+(c+d)=b+5+8=b+13 so that a+b+c+d+e=(a+11=b+13) then subtract 11 from each side to get a=b+13-11=b+2 then subtract b from each side to get a-b=2

[@Just1nBa1ley]

Proud of myself. Solved in 10 seconds for 28 just adding them up in diagnols.

[@scarletevans4474]

I started from purple and went (almost) all the way around, using one equation at the time, to eventually reach A=2+B 🙂

[@KramRemin]

I have a desire to see an animation of the "locus," i.e., the curve made by all the different points inside the square which have the property of giving those 3 specific values to the quadrilaterals.

[@Alex-e9r8o]

<a href="#" class="seekto" data-time="489">8:09</a> so at this point d - c is equal to d - a. And that's how we can get a-b

[@braydentaylor4639]

<a href="#" class="seekto" data-time="247">4:07</a> You call that brown?

[@reubenmanzo2054]

Being colourblind, I wouldn't know.

[@akinamegu9896]

god bless you presh !

[@busogkid7715]

i did them all by just looking at the shapes. For square, just pair the shapes opposite to each other. the sum of each pairs must be equal (20 + a) = (32 +16). For pentagon, (A+7+4)=(B+5+8). It will save you a lot of time.

[@PoppySuzumi1223]

An interesting viral question.

[@davidzagorski9756]

Coordinate geometry is only a little messy but gives the answer.

[@josephbodindeboismortier7759]

Another solution is to notice that the sum of two non adjacent areas is half the area of the square. This gives 16+32=20+x. The demonstration starts with the square obtained by joining the four midpoints of the original square. Within this smaller square you have four triangles with an equal basis. The sum of the heights of 2 non adjacent triangles is constant and equal to the side of the smaller square. QED.

[@shashwatgupta4613]

Simple yet elegant

[@MacGyver5AF]

I am so glad I am no more in the age of need to resolve those geometric problems. I hate(d) geometry as a whole... especially "Create a triangle, if ..." I studied librarianship at Charles University in Prague, at Faculty of Arts. I have absolutely no math brain...

[@hippophile]

And for a hexagon with midpoints and known areas divided in the same fashion?? Then let the area of the unknown "sextant" be X. It can be written in terms of the other areas. Let the areas going clockwise be X, A, B, C, D and E. Then X = A - B +C - D +E. All even numbered polygons will have a formula rather like this...

[@jonorgames6596]

<a href="#" class="seekto" data-time="244">04:04</a> The area color is not brown, it is orange.

[@Violaetor]

I got stuck at <a href="#" class="seekto" data-time="258">4:18</a> trying to color calibrate my monitor to "yellow" and "brown".

[@vcvartak7111]

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)?

[@NHWY25]

Bro You Just Blow My Mind

[@FrancisZerbib]

The sum of opposite (diagonal) triangles in a rectangle are equal

[@OmOm-lu2rn]

How about we take side length of tge square as 'a' and the inner point as (x,y) and use area of polygon determinant formula to form three equations in a,x,and y and solve them

[@DChandra-g8q]

Never saw such an easy Q in bharatiya entrance exam.

[@遠傳五華]

Maybe we could let the length of the our most square as 2A and target area as Y. Then it's clear we could draw an inner square with side-length as sqrt(2)*A and know y+32+20+16=4*A^2, y=4*A^2-68 The value of A will be limited by minimum area,which is 16. A=5, Y=32 is the only solution.

[@pi-sx3mb]

I took a wild guess and got A -B = 5,798. Not even close.

[@hikari1690]

My guess is closer. 42

[@someblokeontheinternet]

⚠minor script error at <a href="#" class="seekto" data-time="761">12:41</a> : you incorrectly say "2-A", but on screen it is correct as "2+A"

[@MindYourDecisions]

Thanks for letting me know! I just misspoke when reading. I usually have to clean up a few sentences during editing but I missed this one.

[@markthompson2874]

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.

[@Ar4tic_Blizzard1626]

i... definitely did not spend like 4 hours trying to solve for A+B because i paused the video before seeing the question was A-B. this has been a fun experience regardless though

[@tj_roehre]

I just went on with this logic going from A counterclockwise to B and ended up with the same solution. This can be even done without writing it down.

[@michaelyyho]

This is not the most straightforward method. You just need to draw 6 lines in different ways. And you will get the relationship X+20=32+16. Then X=28. Most straight forward.

[@deusege7520]

<a href="#" class="seekto" data-time="246">4:06</a> brown?

[@MrGrumbleguts]

How are you proving that the triangles in the pentagon have the same perpendicular heights? because the lines from the midpoint aren't perpendicular to the base.

[@saetainlatin]

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"_

[@BigFellaDj]

<a href="#" class="seekto" data-time="667">11:07</a> 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

[@rohit71090]

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 ?

[@SgtSupaman]

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.

[@BigFellaDj]

@@rohit71090 ah i see, thank you

[@Macchia_a]

could anyone explain me why the top equations are equal to the bottom? <a href="#" class="seekto" data-time="410">6:50</a>

[@futurepath]

Nice video brother!

[@kredwol2103]

Elegant way of solving this. You can solve it with brute force algebra on the rectangles and triangles and you'll find that the sides have length 4 * Sqrt(6) and the coordinates of the unknown point are at {Sqrt(6),5*Sqrt(2/3)}

[@uwelinzbauer3973]

I did it like this. Much of work. Watching the video showed me the short cut. Now my memory only has to keep it.

[@American_Language]

wth happened at <a href="#" class="seekto" data-time="323">5:23</a>

[@JohnJones-w7i]

Dear Presh, are you sure about how we are to interpret the 40 Questions and 7.6 average score? I think that Turks would use a 10 point scale. So 7.6 / 10 (out of 40 questions) does not look too bad.

[@mr.nelson7511]

So, can you tell me what the area of A+B is for the pentagon?

[@jonygamingengusa1660]

X

[@ayunda.alicia]

Where is the original video?

[@hikari1690]

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*

[@playmotion.trailer]

the image is not about the problem

[@kirkanos771]

I find the demonstration overcomplicated. The drawing shows it is a square with equal edges. The crossing not being at the center of the square is a red herring. 20 + X = 32 + 16. X = 28. square area is 96. Edges are 4.898989 and sides are 9.797979

[@munkustrap79]

brown??? Seriously?? That color is called orange

[@Meshamu]

Oh, once you break the segments of the square into those triangles, it becomes a lot like the recent weighing problems, doesn't it?

[@MightyKK006]

Why solve tor A-B ? Shouldn't it be solved for A+B ?

[@Iomhar]

(a+b)+(c+d)-(b+c)

[@Bokery1000]

Beautiful!

[@kaushalagrawal6258]

You either have faith in your approach, or you jump to B @<a href="#" class="seekto" data-time="690">11:30</a>

[@prajwalshrestha9818]

i can't unsee the midpoint indicators looking like a smiling emoji..

[@Donkamrad93]

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

[@EmilD1986]

How can you have such an easy question in a univercity entrance exam?

[@satrajitghosh8162]

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

[@Nikioko]

16 cm² + 32 cm² − 20 cm² = 28 cm²

[@evanrosman9226]

That was a challenge.

[@MarcioHuser]

Amazing indeed!

[@elliotlesser4373]

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]

Which isn’t the right answer…because it’s not the right problem. The new problem is presented @8:42.

[@lilliematthews7922]

​​@@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

[@sladjanavelickovic-stankov289]

Yes, my solution is 28cm2.Itis not that hard.

[@margaretcorfield9891]

Exactly what I did

[@spareyourbrain]

Correct me if I am wrong. If the area of the blue is 28 then then total area is 16+20+32+28 = 96 cm2. The total area is 96 which implied that 96 is a perfect square. It does not make sense.

[@winstongsd4273]

This was great.

[@diniaadil6154]

I totally forgot that the line to the midpoint divides a triangle into 2 equal areas

[@tombristowe846]

For once, with your videos, I got there first; with logic, not algebra. With the information given at the 6min 39 sec point, clearly C must be 4 more than B. For it to add up to 20 C must be 12 and B must be 8. Therefore A must also be 8. C and D equals 32 so D is 20. A and D is 28. QED

[@venkatk1968]

Brilliant!

[@physicslover909]

Eyvallah aga bak AYT çıkmış sorusu 😅❤🤕

[@bjarnieinarsson3472]

Why didn't you mark the mid-point ahead of the question? That's crucial information before try to solve. To rely on vision is a joke!