I figured it this way: PQ is the perpendicular bisector of AM, so its slope is 1/2 and it passes through (2.5,5). The x-coordinate of its intersection with x+y=10 is 12.5/3; multiply by √1.25 to get the length.
@RGP_Maths
Жыл бұрын
Yes! Same method. 😊
@bigbrainers990
Жыл бұрын
Yea me too
@joostvanrens
Жыл бұрын
I solved it like this: I paused the video to look at the problem, then unpaused it and 7 minutes later, like magic, the answer appeared on my screen. It's different from coordinate geometry but works almost as well.
@ccmplayer87
Жыл бұрын
The important things are AP = PM and AQ = QM
@DamienConcordel
Жыл бұрын
You could have also just remembered that the fold line will be perpendicular to AM and intersect it in the middle of AM. That gives you the equation of PQ and therefore the coordinates of both points
@ArabianShark
Жыл бұрын
I figured that ABC = 90º and that BCA = BAC = 45º, therefore MBP = 90º and PMQ = MCQ = 45º. Furthermore, because BC = 10 and M is the midpoint of BC, BM = MC = 5. Knowing that BP + PM = 10 and that CQ + QM = 10√2, I used the law of cosines applied to triangles PBM and MCQ to figure out that PM = 6.25 and MQ = (25√2)/6 (and also that BP = 3.75 and CQ = (35√2)/6, although that's not particularly useful). Applying the law of cosines again to triangle PMQ yields PQ = 4.658 (approx.).
@trnfncb11
Жыл бұрын
Much as I like setting up a coordinate system to solve geometry problems, in this case it is really much quicker to use basic geometry. Call a=PB, b=PM, c=QM, d=QC, x=PQ, L=AB=10. Having established that the original triangle is a half square, elementary geometry gives right away the five equations a^2+(L/2)^2=b^2, a+b=L, c^2=(L/2)^2+d^2-Ld cos(pi/4), c+d=L Sqrt(2), x^2=b^2+c^2-2bc cos(pi/4), which immediately yields b=5L/8, c=5 Sqrt(2)L/12 and x=5 Sqrt(5)L/24.
@user-ew1io5nd7t
Жыл бұрын
Designate the midpoint of AM as D. ⑴ Use the cosine law to calculate cosMAC and then tanMAC is obtained to be 1/3 (thus DQ = 1/3 AD) ⑵ Triangle APD is similar to triangle AMB. So AD must be twice the length of PD. ⑶ We all know the length of AD (1/2 AM), and that of PQ can be obtained by (1/3 + 1/2 = 5/6) AD (5/12 AM) ⑷ Problem solved without the construct of coordinate system!
@drnandkishorbagul3087
Жыл бұрын
COORDINATE GEOMETRY 1. Computer graphics. 2. Connects geometry and algebra. 3. Useful in advanced Mathematics. 4. Powerful problem solving technique. Useful features... Thanks👍
@Tiqerboy
Жыл бұрын
The diagrams that start your puzzle are confusing. It's pretty obvious that ABC is right angled 45° triangle, but PQ is NOT parallel to BC. Once I realize that much, I can begin to solve the problem. But if I used your diagrams as drawn to visualize the problem, i'd be making some incorrect assumptions pretty quick! **SPOILER ALERT** How I solved it ..... I found the value p as shown but then proceeded to find out ∆PBM is a 3-4-5 triangle. Therefore easy to calculate angle BPM. Calculate angle APQ (since angle APQ = angle QPM). You now have all the angles of ∆APQ. You know the length of AP, so use the law of sines to calculate PQ. After watching Presh's co-ordinate geometry solution, I don't think his method is any faster or better to be honest. It's only better if you aren't allowed to use trig or don't have access to a calculator. The nice thing about my solution, if you're ever asked to make a 3-4-5 triangle by starting with a square, and you're not allowed to use a ruler (just allowed to fold a square piece of paper), well now you know how, so if that problem is ever tossed your way, you know how to solve it!
@sharpmind2869
Жыл бұрын
I did this by cosine laws but i must say that coordinate geometry solution was very nice.
@krabkrabkrab
Жыл бұрын
Problem is simplified by using the fact that the slope of PQ is 1/2. Use this and slope of AC being -1 to find point Q coordinates.
@popupro
Жыл бұрын
You have to remember that you first have to prove that the slope of PQ is 1/2 by finding that the slope AM is -2
@Auroraborealis949
Жыл бұрын
@@popupro But isn't this obvious?
@popupro
Жыл бұрын
@@Auroraborealis949 Even the obvious things need to be pointed out sometimes, because it's not obvious to everyone
@ilialvov8142
Жыл бұрын
tanBAC = 1 (pi/4 angle) tanBAM = 1/2 (M is midpoint of BC) Hence, tanMAC = (1 - 1/2) / (1 + 1/2) = 1/3 PQ divides AM in half and is perpendicular, hence PQ = (tanMAC + tanBAM) * AM/2 AM = sqrt(10^2 + 5^2) by Pythagorean Substitute in expression above - and hence the answer 🙃
@harikatragadda
Жыл бұрын
If O is the intersection of AM and PQ, then ∆AOP is *Similar* to ∆ABM. Hence, PO = ½AO Drop a perpendicular from M on AC to meet at R, making an Isosceles Right ∆MRC. MR = RC = 5/√2 AR = AC - RC = 15/√2 ∆AOQ is *Similar* to ∆ARM. Hence, OQ = ⅓AO PQ= ½AO + ⅓AO = ⅚AO AO = ½5√5 PQ = 25√5/12
@josephatjose7524
Жыл бұрын
Excuse me, how do you MRC is isosceles ?
@harikatragadda
Жыл бұрын
@@josephatjose7524 ∆MRC is a Right Triangle with ∠C = 45°
@TheVocaloidNyan
Жыл бұрын
Ahh that's actually a pretty smart solution that doesn't use coordinates or trig. Was thinking how to do it without those two tools and the right isoceles made clicked for me.
@ReynaSingh
Жыл бұрын
Great channel. Keep it up
@ibrokhimbakirov2138
Жыл бұрын
I got a different answer, because the triangle is a 45-45-90 triangle, for the triangle to fold and still reach the midpoint means that the height of the smaller triangle is 1/2 the original, meaning that the height is 5, and because the small triangle and big triangle are similar, they are in a 1:2 ratio, meaning PQ is 5.
@ibrokhimbakirov2138
Жыл бұрын
never mind, i am wrong
@spiderjump
Жыл бұрын
Once we get coord of P, APMQ is a kite so PQ is perpendicular to AM , so we can get the equation to PQ and Q can be calculated by solving equations for PQ and AC.
@spiderjump
Жыл бұрын
@Verycoolgun post your solution in a video
@MathOrient
Жыл бұрын
Fascinating! This folded triangle puzzle presents an intriguing geometric challenge. I'm excited to see how the length of the crease PQ is determined in this unique scenario. The combination of mathematical problem-solving and visual thinking makes this video both engaging and intellectually stimulating. Thank you for sharing this intriguing puzzle with us!
@jaimeyomayuza6140
Жыл бұрын
Wonderful. I love it!
@akinamegu9896
Жыл бұрын
as usual , a good geometry exercise to turn our brains on , great approach mentor presh !
@mehulpunia6174
Жыл бұрын
I never thought that coordinate geometry can solve this very easily.I solved this problem using cosine rule and it takes a lot of time.
@goelchats
Жыл бұрын
I have also done it by cosine rule...
@hussaineh89
Жыл бұрын
Wouldn't QM be //el to AB? and PM //el to AC? and if P and Q were mid points of AB & AC, would vertex A folded - reach M at the midpoint of BC?
@ArabianShark
Жыл бұрын
To answer your first question, no. if QM were parallel to AB and M was still the midpoint of BC, that would mean that MQ would be equal to 5, because the angle MCQ is the same as the angle BCA, MQ = 5 and the triangles ABC and QMC are similar (only if QM were parallel to AB). However, MQ + QC = 10√2, and, if QM were parallel to AB, QC would have to be equal to 5√2 (because the triangles are similar). However, 5 + 5√2 does not equal 10√2. For a similar reason, to answer your second question, also no. If P and Q were midpoints of AB and AC respectively, then triangle APQ would be identical to triangle QMC, and both would be similar to triangle ABC, because they'd each share one angle with ABC and the sides that form it would be proportional (and exactly half as long). Therefore, if P and Q were midpoints of AB and AC respectively, PQ would be parallel to BC, and folding along that line would place A squarely on top of B. You can verify this yourself by cutting a square piece of paper down its diagonal, folding each side of the resulting triangle in half, then find PQ and fold along that line.
@hussaineh89
Жыл бұрын
@@ArabianShark thank you
@ArabianShark
Жыл бұрын
@@hussaineh89 My pleasure 😉
@terrylane1492
Жыл бұрын
If it's already in the coordinate plane, perp. bisectors through AM midpoint is much easier like several others have said.
@phalanxutsav6300
Жыл бұрын
we can eaisly use cosine rule if we want to do it with pure geometry and it works but i like ur approach also but it requires brute force
@Bob94390
Жыл бұрын
As an engineer, I am happy with a graphical solution. So I make a drawing according to the specifications, and then I measure how long the line is.
@ArabianShark
Жыл бұрын
As an engineer, I recognise that I can't make a drawing precise enough to draw any significant conclusions from measuring it.
@rohitmadashri7250
Жыл бұрын
Bravo! As a pest control engineer, I just spray chemicals in the area and count the number of dead bugs. That gives me total area without using my head.
@geoffreytrang8670
Жыл бұрын
Shapes not drawn to scale could easily fool you. In particular, despite what the shape not drawn to scale may suggest, the folding line PQ is not parallel to BC, and the bottom quadrilateral PQCB is not a trapezoid.
@HoSza1
Жыл бұрын
This one can be solved using only the law of cosines (even for the generalised case when folding to arbitrary point!) Then just substituting the side lengths ine gets the same answer.
@Vishw_1234
Жыл бұрын
"there are many ways to solve this problem" Meanwhile me being a bio student: idk any of them
@cdmcfall
Жыл бұрын
Using coordinate geometry, I would find the equations of lines AM (y = -2x + 10) and AC (y = -x + 10). Since the distance from PQ to A should equal the distance from PQ to M, then the point of intersection of PQ and AM will be at the midpoint of AM: (5/2, 10/2) or (5/2, 5). Now we can find the equation for PQ. It is perpendicular to AM, so its slope will be negative the inverse slope of AM, or 1/2, and it passes through point (5/2, 5). Using the point-slope format, we can find the equation for PQ, then reformat as slope-intercept: PQ: y - 5 = (1/2)(x - 5/2) => y = (1/2)x - 5/4 + 5 = (1/2)x + 15/4 Setting x to 0, the coordinates of P are (0, 15/4). The coordinates of Q can be found by solving the two linear functions representing PQ and AC: y = (1/2)x + 15/4 y = -x + 10 (1/2)x + 15/4 = -x + 10 (3/2)x = 25/4 x = 50/12 = 25/6 y = -25/6 + 20 = -25/6 + 20/6 = 35/6 So Q is at point (25/6, 35/6) and P is at (0, 15/4). We can then use the distance formula to find the distance between P and Q: PQ = 25√5/12
@cdmcfall
Жыл бұрын
Seems I wasn't the only one.
@gborzi
Жыл бұрын
I did it this way. Compute p like you, determine the area A1 of triangle PBM=75/8; then determine [QC] using the equation (10*sqrt(2)-[QC])^2=[MC]^2+[QC]^2-2*[MC]*[QC]*cos(pi/4), since the angle QCM is 45 degrees, or pi/4. Solving gives [QC]=35*sqrt(2)/6. Next, compute the area of triangle QMC as A2=0.5*[QC]*5/sqrt(2)=175/12. Note that 5/sqrt(2) is the height of QMC w.r.t. base [QC]. Finally, compute the area of quadrilateral APMQ as A3=Area(ABC)-A1-A2=50-75/8-175/12=625/24. But A3=0.5*[PQ]*[AM]=[PQ]*5*sqrt(5)/2 => [PQ]=625*2/(24*5*sqrt(5))=25*sqrt(5)/12. A little bit of a mess, but it works.
@gborzi
Жыл бұрын
Actually there is a simpler, euclidean method: let O be the intersection between PQ and AM. Triangles ABM and AOP are similar, hence OP:AO=BM:AB, AO=AM/2=sqrt(AB^2+BM^2)/2=5*sqrt(5)/2. It follows that OP = AO*BM/AB=5*sqrt(5)/4. Now, let's determine OQ. Let R be the intersection point between AC and the perpendicular to AC passing through M. Clearly MR=CR=MC/sqrt(2)=5/sqrt(2). Triangles ARM and AOQ are similar, because they share angle MAC and are both rectangular. AR = AC-RC=AC-MR=15/sqrt(2). Then OQ:AO=MR:AR => OQ = MR*AO/AR=5*sqrt(5)/6. Finally, PQ=OP+OQ=5*sqrt(5)*(1/4+1/6)=25*sqrt(5)/12.
@_Dearex_
Жыл бұрын
Nice one, didn't think about coordinates at all! Took me way to long to figure out how to get all the angles but got it and got the correct answer :D
@AK2117official
Жыл бұрын
Everything went from upside from brain.
@_P_a_o_l_o_
Жыл бұрын
Does anyone know a purely geometric solution to this problem? Using similar triangles I only managed to compute the length of PO(calling ) the intersection between lines PQ and AM), but I have no clue as to how to compute OQ. Any suggestions?
@zy8900
Жыл бұрын
To calculate OQ, I calculate the length of AQ. Since AQ=QM, QC=AC-AQ, consider triangle QMC to relate QM with QC by using the cosine law。
@twwc960
Жыл бұрын
Yes. Call ∠PMB α and call ∠QMC β. Because ∠PMQ is the top corner folded over it is 45°. These three angles must add to a straight angle so α+β+45°=180°, so α+β=135°. Now, the angle at B is a right angle, so PBM is a right triangle. M is the midpoint of BC and BC is 10 units long, so BM=5. Let a be the length of BP. By the folding, we have that BP+PM=10, so PM=10-a. We can use the Pythagorian Theorem on that triangle to find that a=15/4 and PM=25/4. We thus have a 3-4-5 triangle and sinα=3/5 and cosα=4/5. Now, the angles in triangle QMC add up to 180° and ∠QCM=45°, ∠MQC+β=135°, so ∠MQC=α. Now, using the law of sines in triangle QMC, we have 5/sinα=QM/sin45° and recalling sinα=3/5, we get QM=25√(2)/6. Recalling PM=25/4 and ∠PMQ=45°, we use the law of cosines on triangle PQM to get PQ=25√(2)/12.
@jamessanchez3032
Жыл бұрын
First, since AP = PM, then BP = 10 - PM, so (10-PM)^2 + 5^2 = PM^2. AP = PM = 6.25. Now, call "S" the point on the left leg of the triangle that is the same height of Q. AS = SQ since it's a 45-45-90 triangle. Therefore, 2SQ^2 = AQ^2. Then, drop a line down from Q that hits the base of the triangle at a perpendicular angle. This line is 10-AS, or 10-SQ. The short base is 5 - SQ. And QM = AQ. So (5-SQ)^2 + (10-SQ)^2 = AQ^2. Factors out to 2SQ^2 - 30SQ +125 = AQ^2. Since we already found out that 2SQ^2 = AQ^2, we can do some cancellation. SQ = 25/6. PS^2 + SQ^2 = PQ^2. We have SQ, and PS is 6.25 - SQ, so we plug in the values for PS + SQ to get PQ.
@user-gb4dc2fy2n
Жыл бұрын
perfect question
@FoxMcCloudV2
Жыл бұрын
I used geometry and the Law of Cosines to arrive at the solution, which is the same as what you got.
@kelvintowns5217
Жыл бұрын
Cute❤
@abulfadlahmadi
Жыл бұрын
suggestion: use CMU font
@santiagoarosam430
Жыл бұрын
(“T” es punto medio de AB ; “O” es punto medio de AM ; la horizontal por P corta a AC en “R” ; la vertical por Q corta a PR en “S”). Si AC =10√2 → ∆ABC es rectángulo y sus ángulos son 45º/90º/45º → Si después del pliegue A se superpone a M → PQ es perpendicular a AM y la corta en “O” → BM=5 → TO=5/2 → TP=(5/2)/2=5/4 → AP=AT+TP =5+(5/4)=25/4 =PR → En ∆PQR: PR=PS+SR=2b+b=25/4 ; QS=SR=b → PS= 25X2/4X3=50/12 → SR=QS=25X1/4X3=25/12 → PQ²=PS²+QS² =(50/12)²+(25/12)² =3125/144=5X625/144 → PQ=25√5/12 =4.65847 Problema muy interesante. Gracias y un saludo cordial.
@Tiqerboy
Жыл бұрын
I got the right answer by drawing the right diagram. I had to use the law of sines on ∆APQ to get exactly your answer. Misc notes: ∆ABC is 45° right angle. ∆PBM is a cute 3-4-5 triangle. It's easy to calculate the angles of a 3-4-5 triangle.
@ArabianShark
Жыл бұрын
@@Tiqerboy Nice catch; I missed that PBM is a 3-4-5 triangle (although I didn't need that particular piece of information).
@sirjohnson1626
Жыл бұрын
Good 👍 idea 💡
@kevinhughes8577
Жыл бұрын
The coordinate system is adding nothing to this proof but making it harder to understand.
@aadityajaiswal_1644
Жыл бұрын
I can understand your problem, but that's because u lack the basics..
@furno_2761
Жыл бұрын
What's a point of creating a image describing the triangle if it shows a completly different figure? Why include it? Is it there purely to artificially decrease the amount of people who solve the problem correctly? I don't see any other use for it.
@bobbytheferret6809
Жыл бұрын
Hi please do question “Determine all integers n > 1 such that every power of has an odd number of digits”
@Pythagorium-vh2cb
Жыл бұрын
The question is missing several things. One part of it says, "...every power of has..." Every power of what exactly?
@MathematikTricks
Жыл бұрын
Nice way 😊😊😊 I tried another way 😍
@mitachowdhury2423
Жыл бұрын
Good shot
@shadrana1
Жыл бұрын
At 1:50, Draw a line QR parallel to BC to intersect AB at R, Let PB=S, AP=PM= 10-S Consider triangle PBM, PM^2= S^2+BM^2 (10-S)^2= S^2+5^2 100+S^2-20S= S^2+25 20S=75 S = PB=15/4.....................................(1) AP=10-S=40/4-15/4=25/4..............(2) Consider triangle ABM, AB=10,AB=5, AM^2=AB^2+BM^2 (Pythagoras) AM^2=10^2+5^2=100+25=125 AM=sqrt(125)=sqrt(25*5)= 5 sqrt5............................(3) Triangle ABM is therefore a (1,2,sqrt5) triangle .........................(4) Consider triangle QRP, Let QR=t, Triangle QRP is similar to triangle ABM (angles are all equal), QR and RP are horizontal and vertical components of target line PQ, Comparing triangles QRP and ABM, QR=t,PQ= (sqrt5)*t/2....................................................................(5) RP= t/2..........................................................................................(6) AR= t ..............................................................(7) triangle ABC is a (45,45,90) degree triangle) AP=AR+RP 25/4= t+t/2=3t/2, t=(2/3)*(25/4)=50/12=25/6............................(8) PQ=(sqrt5)*t/2 from (5) PQ=(sqrt5*25/12) units and that is our answer. Just think,this puzzle is made easy because some of the triangles in the structure are similar. I thought I knew a lot about triangles and circles about 60 years ago but a brilliant physics student from Norway put me right. He ended up being a professor of theoretical physics at the university of Oslo.
@rohitmadashri7250
Жыл бұрын
How old are you sir? Who is this brilliant professor?
@janeluooo
Жыл бұрын
I used trigonometry and get the same answer
@klh6729
Жыл бұрын
I used Pythagoras and law of sines to get the answer.
@blue_wool_ZL
Жыл бұрын
I want to share my problem i had on my math competition (i got first place out of all first graders in highschool), the problem is as follows: "If you would tie a rope around the equator of earty and add 10 meters to and hang it in air without loosing length of it on knobs, would you be able to go under it without leaning and why?" I did a strategy were i proved that you need 2*pi centemeters more of the rope to tie it around any sphere with leaving 1 cm between the sphere and the rope if you could hang the rope, then i calculated how much more rope i would need to be at my height and compared it to the 10 meters and my question is what is another way to do it?
@kohlsnofl5110
Жыл бұрын
Dunno to me it seems the radius of the circle formed by the rope increases from (2pi*R)/2pi to (2pi*R+10)/2pi so thats an increase of 10/2pi. 10/2pi is about 1.6, and since the units are in meters, we get a 1.6 meter increase. So I would have to lean under since I am more than 1.6 meters high. Did I do something wrong? Edit: R is the radius of the earth, which I could just google but is really irrelevant for the problem. The radius of the earth has its endpoint at the surface of the earth, aka where it touches the rope. All the increase in the radius is sticking out of the earth's surface, meaning the amount of space between the rope now increased in length and the earth is the exact amount the radius was increased, in this case 1.6 meters.
@blue_wool_ZL
Жыл бұрын
@@kohlsnofl5110 it falls apart with the 2πr+10/2π its supposed to be 2π(r+10)/2π because if we would calculate like you said it would turn out to be r+(5/π) and if you would go with the other version it turns into r+10 if im doing my calculations correctly
@kohlsnofl5110
Жыл бұрын
@@blue_wool_ZL Well with the formula 2π(r+10) you are increasing the radius itself by 10, are you not? In this case we just have a 10 meter increase of the rope's height above the ground so of course you would be able to walk under it without leaning. The way I understood the question the rope itself increased in length for 10 meters, so the circumference increased for 10 meters in length and thus the radius is now (2πR+10)/2π which is R+5/π, like you wrote. And 5/π is 1.6 (meters), so thats how I got my answer
@joostvanrens
Жыл бұрын
Another wat is to tie a rope around the earth, add 10 meters to it, suspend it in the air and see if you can walk under it without leaning. The experimental physicist approach. The reason why is: you are short enough to do it or too tall to do it, depending on the result.
@kohlsnofl5110
Жыл бұрын
I guess, but you'd have to try it in some sort of simulation for now
@alien0369
Жыл бұрын
Can we use mid point theorem??
@Dinesh-IIT-Bombay
Жыл бұрын
Please also share pure geometric solution
@HoSza1
Жыл бұрын
The figure in the beginning is exceptionally inaccurate.
@The_Cali_Dude_88
9 ай бұрын
Almost magic is highly subjective 😅
@12_xu
Жыл бұрын
The first image is very, very confusing
@brunogrieco5146
Жыл бұрын
You just used standard Linear Algebra. Don't know what "Coordinate Geometry" means. Also, this is not Computational Geometry used in computer graphics. I started working on this and I sort of gave up due to all the needed calculations. Was waiting for a simpler, geometrical solution.
@dfailsthemost
Жыл бұрын
Now I feel like I cheated doing it my way..
@KitKat-xt4ti
Жыл бұрын
Hi, i'm a middle shcool student, who will be starting high school soon, and I'm wondering where/how I could learn the advanced math that you use in your videos, thank you.
@diskritis2076
Жыл бұрын
I just graduated high school with perfect math scores. I assume you are in grade 9 so maybe you can start with some advanced geometry and if you dont find it interesting, start learning basics of calculus. There are unlimited sources on youtube itself but you can go for higher grade books in your curriculum. Personally I think you can start learning from grade 11 books
@sergniko
Жыл бұрын
Why using 'm' for slope instead of 'k'? :)
@charlesboys9674
Жыл бұрын
4.658 🤨
@francois8422
Жыл бұрын
unnecessarily complicated one thing a non-scale drawing, another distorting the data of the problem creating a further problem for understanding them might as well not submit any drawing it does not make sense
@E--ViranshAgrawal
Жыл бұрын
Hi
@petertran5343
Жыл бұрын
I believe that your calculation of value of p is not correct . the value of p per my calculation is 25/4
@santiagoarosam430
Жыл бұрын
La solución que se propone en el vídeo me parece complicada, pero sus resultados son correctos. Por semejanza de triángulos es más sencillo y se obtiene el mismo resultado. AP=(20/4)+[(5/2)/2]=(20/4)+(5/4)=25/4 → p=PB=AB-AP=(40/4) - (25/4)=15/4 =75/20. Un saludo.
@ArabianShark
Жыл бұрын
How did you arrive at that value?
@santiagoarosam430
Жыл бұрын
@@ArabianShark Si dibujas el esquema con los puntos que yo indico en mi solución, lo verás con claridad; no obstante procuraré resumirlo →T es punto medio de AB y O es punto medio de PQ → AP=AT+TP → AT=10/2=20/4 → TO=BM/2=5/2→TP=TO/2=(5/2)/2=5/4 → AP=20/4 + 5/4 =25/4 → Si p=PB=AB-AP =(10*4/4)-(25/4)=15/4=15*5/4*5=75/20. Un saludo
@ArabianShark
Жыл бұрын
@@santiagoarosam430 I was asking @Peter Tran, but thank you, anyway.
@santiagoarosam430
Жыл бұрын
@@ArabianShark Perdón por mi error. Gracias
@d.christianrathjens7209
Жыл бұрын
Very misleading sketch. PQ looks parallel to BC.
@thegoldengood4725
Жыл бұрын
Z
@theremoteman4504
Жыл бұрын
Pin me for Pythagoras 👌
@LeaderTerachad
Жыл бұрын
Ohh sir how you managed make brilliant son(Pythagoras) you are awesome!!
@JamesWylde
Жыл бұрын
It still kills Presh to say Pythagorean Theorum or Pythagoras LOL
@kunalkapoor6588
Жыл бұрын
So early idk wut to say-
@smiling_buddha
Жыл бұрын
... say something I'm giving up on you
@adventureboy444
Жыл бұрын
just watch the video...?
@Ddntitmattrwhtuthnk
3 ай бұрын
Mind your own decisions.
@ericherde1
Жыл бұрын
7:25 How can this be one of the best communities on KZitem when the comment section bullied the host into no longer using his prefered name for the distance formula?
@harikatragadda
Жыл бұрын
This is the best community of bullies.
@JamesWylde
Жыл бұрын
Because he's doing great things with making math approachable, understandable, and less intimidating. But his insistence on refusing to use the most common term for a very common tool that is used across a lot of disciplines completely detracts from what he is trying to do.
@ericherde1
Жыл бұрын
@@JamesWylde Do you have a citation for Pythagorean being the most common term? If the term he was using is the common term for it in India, then it's probably the most common term throughout the English-speaking world. However, I don't know if most people in India call it the Gougu Theorem, the Pythagorean Theorem, or something else. Do you know, or are you just assuming that the name we use for it is the most common name?
@taylormanning2709
Жыл бұрын
I’m really tired of the thumbnail not being representative of the problem 😡 Quit it
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