How To Solve Microsoft's Rectangle Corners Interview Question

[@NotQuiteFirst]

So if P can be either inside or outside, does it represent Brexit?

[@vis9487]

The True Fizz LMAO get this comment to the top

[@TotalTimoTime]

I'd name the rectangle NP

[@jiggely_spears]

If P = Brexit then P stands for Pile o'shit

[@shrikrishnajugtawat4639]

The True Fizz that's a nice one😂😂

[@devendra_c]

The True Fizz Very witty..hahaha!!

[@shikhanshu]

learnt something new! my first instinct was to draw 3 circles and see where the intersect! but this equation is super cool

[@darealbeesechurger]

Thats pretty good of an answer too

[@the_shiny_skorupi1874]

That was my instinct too, but it doesn’t work as the rectangle isn’t to scale.

[@ccost]

i tried this in desmos, it does actually work@@the_shiny_skorupi1874 using trial and error, usable points are a = (0,0) b = (6.7, 0) c = (6.7, 10.025) d = (0, 10.025) P is around (-0.23, 11) thus the distance between D and P are 1.0017 (would be 1 if it werent due to human error and thus id round to 1)

[@tryhardofdoom7682]

As usual, I will be attempting to solve this in this comment box, so a) this might not be a full solution and b) this probably will be very long: The only way I can imagine handling circles right now is in a cartesian coordinate system, so I will use a classical 2d coordinate system with dimensions x and y. The four points A, B, C, D and P will be inscribed in it, and for the sake of simplicity, point A will be located at the origin (0|0). I will accordingly define points B, C, D and P, so I now have: A(0|0) B(xB|yB) C(xC|yC) D(xD|yD) P(xP|yP) (The x's and the y's are just variable names, I couldn't think of better ones) Since xB = xC and yC = yD and xA = xD and yA = yB, these five points now are: A(0|0) B(xB|0) C(xB|yC) D(0|yC) P(xP|yP) Now the distances to the three points A, B, C and D: The distance between two points can be expressed through the pythagorean theorem, which is d = sqrt( xDistance^2 + yDistance^2 ) For the distance between A and P, this means: dA = sqrt( xP^2 + yP^2 ) = 11 or 121 = xP^2 + yP^2 (1) For the distance between B and P, this means: dB = sqrt( (xB - xP)^2 + yP^2 ) = 13 or 169 = (xB - xP)^2 + yP^2 (2) For the distance between C and P, this means: dC = sqrt( (xB - xP)^2 + (yC - yP)^2 ) = 7 or 49 = (xB - xP)^2 + (yC - yP)^2 (3) For the distance between D and P, this means: dD = sqrt( xP^2 + (yC - yP)^2 ) or dD^2 = xP^2 + (yC - yP)^2 (4) Now follows a series of "ahh I see a duplicate occurence of this term" and "ohh what happens when I substitute them in". Hope it leads me somewhere :) (3) 49 = (xB - xP)^2 + (yC - yP)^2 | - (xB - xP)^2 (yC - yP)^2 = 49 - (xB - xP)^2 | (3') (4) dD^2 = xP^2 + (yC - yP)^2 | substituting in (3') dD^2 = xP^2 + 49 - (xB - xP)^2 | (4') (2) 169 = (xB - xP)^2 + yP^2 | - yP^2 (xB - xP)^2 = 169 - yP^2 | (2') (4') dD^2 = xP^2 + 49 - (xB - xP)^2 | substituting in (2') dD^2 = xP^2 + 49 - (169 - yP^2) dD^2 = xP^2 + yP^2 + 49 - 169 | substituting in (1) dD^2 = 121 + 49 - 169 dD^2 = 1 | sqrt() dD = + - 1 | -1 is irrelevant when talking about areas dD = 1 Well, I hope this is right.

[@tryhardofdoom7682]

Woo I was right! Maybe not the fanciest approach because I didn't know the formula. Atleast now I learned one more formula & how it's derived :) By the way, I have to admit this puzzle was really easy, compared to the other puzzles you uploaded in the past (those I am interested in, not the viral picture puzzles).

[@qwertyTRiG]

TryHard ofDoom MindYourDecisions has finally found middle gears! His stuff is usually ridiculously difficult or ridiculously easy. That said, I didn't attempt to solve this one myself: too tired right now.

[@Akronox]

In fact it is much more faster when you notice that (1)-(2)+(3) right members is equal to (4) right member so you avoid all substitution intermediate steps and you see directly where the formula from the video is coming.

[@tarmotaipale5704]

Actually, you made a little mistake. In equation (4) you forgot to square dD (it should be dD^2=xP^2+(yC-yP)^2). This way, you would have ended up taking a square root from the result you got, to get the correct result, however, your answer happened to be correct because square root 1 equals 1.

[@tryhardofdoom7682]

Thanks for pointing it out, I added the squares and a slight modification, hope everything is right now. Just a classic "copy-pasta without using my brain" mistake :D

[@omedakreyi5152]

i would draw circles if i were allowed to

[@supreetsahu1964]

Not sure how your method would work. We're not given any length of rectangle, so how would we start with the circle method. I tried to do it and can't find the approach.

[@davidb5205]

I did that. It works. Just center Point A at (0,0) and Point C at (a,b), then you have four circle equations and the equations work out nicely so that you don't even have to figure out x, y, a, or b.

[@supreetsahu1964]

+Colton Graham Dude, I was talking to omedakreyi. I have not attenpted solving the question purely by cartesian coordinate geometry.

[@khondakersadman8867]

I'm seeing 4 equations and 5 unknowns (x,y,L,W,R), so how can you solve the system? x^2 + y^2 = 121 (x-L)^2 + y^2 = 169 (x-L)^2 + (y-W)^2 = 49 x^2 + (y-W)^2 = R^2

[@danielpie.8926]

I drew circles too. I can't put it into math, but i am able to deliver a solution.

[@DennyMapleSyrup]

I got 1 by using complex numbers :P If you put p on the line AB, and connect P to C, you can use the pythagorean theorem with ∆PBC. You'll get BC=i√120. Draw a line from D to P and use pythagorean theorem on ∆DAP. 11^2 + (i√120)^2 = 121 - 120 = 1

[@SMOK3SCR3EN]

I don't know why I watch these videos I know nothing about math. They just pop up in my feed and I watch it

[@SNIPERINTX]

Whenever I think I'm smart, I watch another MindYourDecision video and humble myself!

[@8bit_pineapple]

My solution: Let A = (0,0) , B = (c₁, 0) , C = (c₁, c₂) , D = (0, c₂) and finally P = (x, y) We can write formulas for the distances AP, BP, CP, DP. But we'll just use squared distances because working out square roots is annoying. [1] AP² = x² + y² = 121 [2] BP² = (c₁ - x)² + y² = 169 [3] CP² = (c₁ - x)² + (c₂ - y)² = 49 [4] DP² = x² + (c₂ - y)² = ? I then noticed that equation [3] + [1] - [2] = [4] So, the squared distance DP² = CP² + AP² - BP² , which is equal to 49+121-169 = 1. So the distance is the square root of 1, which is still 1. QED.

[@gabrielbrockhaus974]

These are the kind of riddles I visit this channel for! Keep it up :)

[@jumpman8282]

Here's how I did: We have four points forming a rectangle, 𝐴𝐵𝐶𝐷. For example: 𝐴: (0, 0) 𝐵: (𝑎, 0) 𝐶: (𝑎, 𝑏) 𝐷: (0, 𝑏) We also have a point 𝑃: (𝑥, 𝑦), such that: |𝑃 - 𝐴| = 11 |𝑃 - 𝐵| = 13 |𝑃 - 𝐶| = 7 |𝑃 - 𝐷| = 𝑑 ≥ 0 This gives us the following system of equations: 𝑥² + 𝑦² = 11² (𝑥 - 𝑎)² + 𝑦² = 13² (𝑥 - 𝑎)² + (𝑦 - 𝑏)² = 7² 𝑥² + (𝑦 - 𝑏)² = 𝑑² ⇔ 𝑦² = 11² - 𝑥² (𝑥 - 𝑎)² = 13² - 𝑦² (𝑦 - 𝑏)² = 7² - (𝑥 - 𝑎)² 𝑑² = 𝑥² + (𝑦 - 𝑏)² Through a series of substitutions we get: 𝑑² = 𝑥² + 7² - 13² + 11² - 𝑥² = 1 ⇒ 𝑑 = 1

[@asklar]

Microsoft engineer here. Cool video, love the channel :) I was not aware of this theorem, so... Analytic geometry to the rescue. AP=11 BP=13 CP=7 Assume WLOG A=\left(0,0 ight) B=\left(a,0 ight) C=\left(a,b ight) D=\left(0,b ight) Let P=\left(x,y ight) so we have: \sqrt{x^{2}+y^{2}} =11 \sqrt{\left(a-x ight)^{2}+y^{2}} =13 \sqrt{\left(a-x ight)^{2}+\left(b-y ight)^{2}} =7 \sqrt{x^{2}+\left(b-y ight)^{2}} =DP Or conversely: x^{2}+y^{2} =121 \left(a-x ight)^{2}+y^{2} =169 \left(a-x ight)^{2}+\left(b-y ight)^{2} =49 x^{2}+\left(b-y ight)^{2} =DP^{2} Note then that DP^{2}=49-\left(a-x ight)^{2}+x^{2}=49-\left(169-y^{2} ight)+x^{2}=49-169+x^{2}+y^{2} So DP^{2}=49-169+121=1 therefore DP=1

[@10spen11]

Align the rectangle to the Cartesian plane with point A at the orgin. Let the width of the rectangle be represented by w, and its height by h. The coordinates of the points would then be A: (0, 0) B: (w, 0) C: (w, h) D: (0, h) P: (x, y) Using the distance formula, we can determine the following distances (squaring both sides): A to P: x^2 + y^2 = 121 B to P: (x-w)^2 + y^2 = 169 C to P: (x-w)^2 + (y-h)^2 = 49 D to P: x^2 + (y-h)^2 = ? We can then subtract (x-w)^2 + y^2 = 169 - ( x^2 + y^2 = 121) to get (x-w)^2 - x^2 = 48 We then can subtract (x-w)^2 + (y-h)^2 = 49 - ( (x-w)^2 - x^2 = 48) to get x^2 + (y-h)^2 = 1 This is the square of the distance from D to P as defined above. Therefore, the distance from D to P is 1.

[@nicholaslau3194]

I solved using distance formula. A (xa,ya) B (xc,ya) C (xc,yc) D (xa,yc) P (0,0) [Both sides squared] xa+ya=121 - (1) xc+ya=169 - (2) xc+yc=49 - (3) xa+yc=? - (4) (2)-(1) xc-xa=48 (3)-(4) xc-xa=49-? ? = 1 PD=sqrt(1)=1

[@shubhrajit2117]

I know such a theorem (derived from the pythagorean theorem) which was not taught at school: Let OD, OE, OF be the perpendiculars to the sides BC, CA and AB in ΔABC then BD square + CE square + EA square is equal to DC square + EA square + FB square.

[@mattiassollerman]

I can only solve those riddles "only geniuses can solve" and "93.2% get wrong" and "stumps Harvard math professors".

[@maxie4937]

I'm so proud for figuring it less than 2 minutes. Thank you Presh.

[@MrRyanroberson1]

we all need this in school. middle school, or as early as possible the image at 3:36 is what gave me everything

[@johnchessant3012]

Since the question didn't specify exactly where P or the rectangle's vertices were, I put P on CD so that PCB and PDA were right triangles with right angles at C and D, respectively. I solved for BC using Pythagoras's theorem on PCB (worked out to be sqrt(120)), and BC = AD, so using Pythagoras's theorem again on PDA, I was able to solve for PD = 1. However, the British flag theorem is nice to know!!

[@Magnasium038]

This is actually a good trick. If the answer is a constant which is same for all cases, then it is fine to take a single case to simplify the problem. One does not need to solve for the general case everytime; as long as you believe the question is correct.

[@saboorjalili3715]

Tony Math

[@nsnick199]

@John Chessant I did the same, and came to the realization it had to lie on CD, after looking at the angles PA and PB made in the corners of A and B, and doing some finagling with triangles.

[@nsnick199]

I started to write out my reasoning when I realized that I don't think the way I was thinking about it necessarily proves P must lie on CD, it just worked out nicely (the same way John got there). Whoops!

[@msscoventry]

Me too.

[@Etothe2iPi]

4:06 At this point it's much easier to use (1) AP^2 - y^2 = DP^2 - z^2 and (2) BP^2 - y^2 = CP^2 - z^2 . Simply subtract (1) - (2)

[@Jimpozcan]

My solution was pretty much the same except I used Cartesian coordinates. Put the origin at P; i.e. P = (0,0) Put the x-axis parallel to AB & the y-axis parallel to BC. Let B = (i,j) A = (x,j) C = (i,y) D = (x,y) (I used different letters: x & i for horizontal, y & j for vertical. Also they could be negative.) BP = 13 so, using Pythagoras' Theorem, i^2 + j^2 = 13^2 (eqn 1) Likewise x^2 + j^2 = 11^2 (eqn 2) i^2 + y^2 = 7^2 (eqn 3) Add equations 2 & 3 then subtract equation 1. x^2 + y^2 = 121 + 49 - 169 = 1 Take the square root to get the distance AD, i.e. 1.

[@molunderground716]

I see many solutions in the comments assuming without loss of generality that the point A is at (0,0) and the point C is at (a,b) with a rectangle with sides parallel to the axis. I started myself this way, but in my eyes a much easier solution is when you assume without loss of generality that the point P is at (0,0), the point A is at (x1,y1), and the point C is at (x2,y2). The equations become: x1^2+y1^2 = 121 (1) x2^2+y1^2 = 169 (2) x2^2+y2^2 = 49 (3) and when you simply do (1)-(2)+(3) you get what you want: x&^2+y2^2 = 1

[@paulkennedy8701]

At what point did you say that P lies on the plane? (I know that saying it could lie inside or outside the rectangle probably necessitates that, but I think it would be best to make it explicit.)

[@Robertlavigne1]

wow I was able to work this one out. I'm usually terrible at the geometry problems. I do think that these kind of interview question are the worst. So inapplicable to job performance. I doubt I would have worked it out under interview pressure, so it would be either I knew some obscure equation or I didn't. You may as well ask whats the mass of the statue of liberty.

[@axemenace6637]

Robertlavigne1 k I disagree. I already knew of the British Flag Theorem, but I forgot the exact formulation, so I just solved it as I would a regular geometry problem. the thing about this problem is that if you know the theorem, you're probably smart enough to know how to prove the theorem. It's a good problem because everyone knows Pythwgoras, so everyone can solve it, but it is still reasonably difficult (for the average person).

[@Robertlavigne1]

I am coming to agree with you. This problem may be a decent way to watch how a person attempts to solve a problem they haven't encountered before. I guess it depends on how much time the person is given to try and solve it in an interview setting. Also I went and tried to estimate the mass of the statue of liberty, and using some guesses on it's height, diameter, using the density of water, and a rough guess of what percent of it's volume is solid I was able to get a mass in grams within an order of magnitude. So even a stupid problem like that can still show an ability to work things out using rough heuristics.

[@palmomki]

@Maxim Enis I think you're falling for the fallacy that forgetting the details (precise statement and proof) of a theorem means being back to square one - in many cases, such as this one, the intuition behind a theorem, which is what remains at least in your subconscious when you have internalized and understood the concept of the theorem, is more than enough to make the difference between "I spent days thinking about this problem in vain and I think I would never be able to solve it" and "It took me five minutes to solve this problem, it seemed quite obvious".

[@axemenace6637]

palmomki What you are saying makes perfect sense. However, the first time I encountered the theorem was in problem very similar to this one, which I solved without much difficulty. The problem itself is rare (I only found it because I've spent solving math problems every day) so it is obscure enough to be used in an interview.

[@GopherAtl]

The point of this kind of interview question is not to see if you know the answer. Somebody who knew the British flag theorem and produced the answer in seconds would actually be a disappointment to the interviewer. The point of these kind of questions is to let the interviewer watch how you approach solving a problem. If they just wanted to test your knowledge, there are easier and more direct ways.

[@ehllie]

I solved this problem using analytic geometry. Since ABCD is a rectangle with length l and width w, it's vertices can be described as: A(a1, a2), B(a1, a2+l), C(a1+w, a2+l), D(a1+w, a2) and point P as P(x, y) for a1, a2, w, l, x, y in the real number set. We can use the formula for a distance between 2 given points K(k1,k2), L(l1,l2) which is |KL| = sqrt((k1-l1)^2 + (k2-l2)^2) Then, |DP| = sqrt((a1+w-x)^2 + (a2-y)^2), which means we only really need to find the value of (a1+w-x)^2 + (a2-y)^2 We can find it from equation system for the 3 given distances: |AP| = 11 |CP| = 7 |BP| = 13 sqrt((a1-x)^2 + (a2-y)^2) = 11 sqrt((a1+w-x)^2 + (a2+l-y)^2) = 7 sqrt((a1-x)^2 + (a2+l-y)^2) = 13 We sqare each equation and get: (a1-x)^2 + (a2-y)^2 = 121 (a1+w-x)^2 + (a2+l-y)^2 = 49 (a1-x)^2 + (a2+l-y)^2 = 169 We calculate the value of (a1-x)^2 and (a2+l-y)^2 from the first 2 equations and substitute them in the 3rd: (a1-x)^2 = 121 - (a2-y)^2 (a2+l-y)^2 = 49 - (a1+w-x)^2 121 - (a2-y)^2 + 49 - (a1+w-x)^2 = 169 From the third equation we get the expression: (a2-y)^2 + (a1+w-x)^2 = 1 Which means that |DP| = sqrt((a2-y)^2 + (a1+w-x)^2) = sqrt(1) = 1 This is a much more convoluted and less elegant approach to the presented problem, but I generally prefer solving geometry problems the analytical way, since I have an easier time and feel more confident in my results when looking at equations than when doing geometry the traditional way :3

[@atharvas4399]

Hey Presh, can this theorem be generalized to a n-sided polygon (rather than a 4 sided)? Can you make a video about it?

[@WolfRose11]

Possibly, but once you have 3 distances to other points (in a 2D space) that point is fixed. Any other shape higher than 4 points or other than a rectangle/square would have to include the angles at the vertices in addition.

[@crazym108]

I put A,B,C,D into Cartesian coordinates: A(0,0) B(x,0) C(x,y) D(0,y). Then I used the distance formula to an arbitrary Px,Py from each of A,B,C,D equal to the given distances. You can't solve for x and y directly, but you can use substitution of the whole equations to show that 13^2 -7^2 = 11^2 -1^2. This is another (more complicated) way to prove the same theorem

[@jessstuart7495]

You don't need to memorize a special identity to solve this. Assume point p has cordinates Px, and Py. Point A is at (-L,0), Point B is at (0,0), Point C is at (0,W), and Point D is at (-L,W). Now write some equations of circles. 1. (Px)^2+(Py)^2=169 (Circle around B) 2. (Px+L)^2 + (Py)^2 = 121 (Circle around A) 3. (Px)^2 + (Py-W)^2 = 49 (Circle around C) The circle around point D is (Px+L)^2 + (Py-W)^2 = (Distance from D)^2 Add equations 2. and 3., then subtract equation 1 to leave only the (Px+L)^2 and (Py-W)^2 terms. So you get (Px+L)^2 + (Py-W)^2 = 121 + 49 -169 = 1. Sqrt(1) = 1, so the distance from D is 1.

[@davidb5205]

My solution: Four equations for four circles with Point A centered at (0,0) [Triangulation?] Circle A: x^2 + y^2 = 11^2 Circle B: (x-a)^2 + y^2 = 13^2 Circle C: (x-a)^2 + (y-b)^2 = 7^2 Circle D: x^2 + (y-b)^2 = distance^2 You don't even have to solve individually for x, y, a, or b. Just replace values. Answer = 1

[@vishalmishra3046]

Using *co-ordinate geometry* , the points are A (0,0) B (a,0) C (a,b) D (0,b) and point P (c,d). Note that distance^2 = deltaX^2 + deltaY^2 of the 2 co-ordinates. Therefore - 11^2 = AP^2 = c^2 + d^2 13^2 = BP^2 = (c-a)^2 + d^2 7^2 = CP^2 = (c-a)^2 + (d-b)^2 X^2 = DP^2 = c^2 + (d-b)^2 Hence, X^2 + 13^2 = 11^2 + 7^2 = 121 + 49 = 170 => X^2 = 170 - 169 = 1 => X = 1 (ANS).

[@fartinmartin5]

Alternative method, you can convert the vertices to Cartesian coordinates. A =(0,0) B =(B,0) D =(0,D) C =(B,D) P =(P1,P2) |DP| =d Then you can get the distance to P from each vertex by using Pythagorean theorem. Then solve the system for d, by expanding the formulas and using substitution. ( I also factored the terms like B^2 -2(B)(P1) = B(B-2(P1)) to make substitution easier)

[@8bit_pineapple]

Try to resist the urge to expand brackets, you should be able to find the answer once you get: AP² = P1² + P2² = 121 BP² = (B - P1)² + P2² = 169 CP² = (B - P1)² + (D - P2²) = 49 DP² = P1² + (D - P2²) = ? If it's hard for you to treat terms like (B - P1)² as a single variable in your head, do the substitutions on paper e.g. let w = (B - P1)² and z = (D - P2)².

[@villaholland]

Kept seeing a 3d rectangle. Once it's in your brains.... once you go 3d there's no going escapee

[@davidsoto8734]

F**k now I'm in

[@lordbubax3929]

Shit

[@salmakassimi5365]

Robb V. ikrrr

[@DizzyBagel]

3d rectangle? Rectangular prism?

[@AMalas]

DizzyBagel as in pyramid with rectangular base

[@jimlocke9320]

Solving by construction: Draw 3 concentric circles, centered at point P, with radii 7, 11, and 13 units. Draw a horizontal line from P to the right through the circle with radius 7. Label the intersection point C. Draw a vertical line down from point C until it intersects the circle with radius 13. Label the intersection point B. Draw a horizontal line left until it intersects the circle with radius 11. Label the point A. Draw a vertical line up to the line from P to C. Label the intersection point D. Measure the distance from D to P. For more accuracy, use coordinate geometry. Let P be the origin (0, 0). Then C has coordinates (7, 0). From there, compute the coordinates of B, A, and D. D will have coordinates (1, 0), placing it 1 unit away from P. You can, alternatively, place A at (0, -11). Then, draw a horizontal line right until it intersects the circle with radius 13. Label that point B. Then, vertical until it intersects the circle with radius 7 (y coordinate is negative). Label the point C. Then horizontal to the y axis and label that point D, which should have coordinates (0, -1). The absolute value of the y coordinate is the distance from P. The problem statement implies that there is only one solution. We can prove that. There are, of course, an infinite number of rectangles, each with a point A between (0, -11) and the first location of point A (inclusive). Solve for the general case and all will locate point D 1 unit away from P.

[@emboar15]

An easier way to visualize this is by putting P on DC. Then you have two right triangles, which allows you to solve for CB, which gives you AD, which lets you solve for DP

[@davidchung1697]

If you let P at (x, y) and rectangle ABCD to have coordinates, (0, 0), (0, x1), (x1, y1), and (0, y1), you can use the distance formulas to obtain 3 equations from each of A, B, and C to P. You can obtain the distance from P to D using those 3 equations. Its quite easy, actually.

[@MrEdwinsantiago]

Another solution is to use the equation for the distance between 2 points. We can place the rectangle in the xy plane with A being in (0,0). We assign a lenght l and a widht w, so B is in (l,0), D is in (0,w) and C is in (l,w). We can now assign a coordinate to the point P, let's say that P is in (x,y). Now we can use that (AP)^2=x^2+y^2=11^2, (BP)^2=(x-l)^2+y^2=13^2 and that (CP)^2=(x-l)^2+(y-w)^2=7^2. With this equations alone we can conclude that x^2+(y-a)^2=1, being this the expression for (DP)^2.

[@briankeener9749]

hey Mr.talwaker I was wondering if you could do a video on game theory for the following game the game consists of 5 rounds in each round the teacher rolls a die if it's a 2,3,4,5,or 6 the students get those points but, if it's a 1 they lose all points for that round they can stop and claim there points whenever they want or they can keep going until the teacher roles a 1 what is the best strategy to win this game? I'm not sure but I think you should stop at 16 every time a video would be greatly appreciated thank you.

[@stromboli183]

If at some round you have gathered p points, the expected value for stopping and claiming the points is obviously p. The expected value of going for another roll is ⅙×0 (lose all) + ⅙×(p+2) (roll a 2) + ⅙×(p+3) (roll a 3) etc which is 5/6×(p+4). So it would make sense to stop when p ≥ 5/6×(p+4) which results in p≥20.

[@ZigZag5056]

You can solve this by considering the distances as being taken from the left side, right side, top or bottom of the rectangle. Where A is bottom left, B is bottom right, C is top right and D is top left: Let L = distance from left of rectangle, R = distance from right, T = distance from top and B = distance from bottom B^2 + L^2 = 121 B^2 + R^2 = 169 T^2 + R^2 = 49 DP^2 = T^2 + L^2 R^2 = 169 - B^2 L^2 = 121 - B^2 L^2 = R^2 + 48 As T^2 + R^2 = 49, T^2 + L^2 + 48 = 49 Therefore T^2 + L^2 = 1 So DP^2 = 1, DP = 1

[@myuu22]

I would not have been able to solve that problem myself. That's a very cool theorem, and I'm disappointed that I didn't learn about it in school.

[@drkInxgud]

an more straight forward method is to set A as origin in X,Y coordinate. set dummy variables for the coordinates of all points. write equation of circles centred A, B and C with the respective distance to the points. solve them simultaneously to find the coordinates of point P. then u can find the distance from P to D via Pythagorean theorem.

[@codebeard]

I thought it would be difficult, but it turned out to be easy. Let c be the length DP. I just set the coordinates: A = (0,0), B = (a,0), C = (a,b), D = (0,b) Then using Pythagoras' theorem and some expansion we get: c² = x² + y² + b² - 2by 121 = x² + y² 169 = x² + y² + a² - 2ax 49 = x² + y² + a² + b² - 2ax - 2by Note that we can substitute much of the last equation with c²: 49 = c² + a² - 2ax Next, we can substitute x² + y² in the third equation with 121 to get: 48 = a² - 2ax And from there, c² = 1. Since c is positive, c = 1. Done.

[@ObjectsInMotion]

If you're curious, this is what the rectangle looks like: ibb.co/ffpnQk

[@theoldfinalchapters8319]

If this is based on an interview question, you completely missed the point. Like you said, that formula isn't taught in schools. The point of the question is to see if you can figure out DP *without* that formula.

[@PaladinswordSaurfang]

I understood why the formula holds instantly. It's derived from the Pythagorean theorem. Lines that join opposite corners of a rectangle are of equal length, so if these lines are hypotenuses, then the hypotenuses are equal, thus the squares of the hypotenuses are equal, thus the sums of squares of adjacent sides of one triangle are equal to the sums of squares of adjacent sides of the other.

[@alexanderhoward122]

Hm, interesting question. So P is the closest to C, furthest from B and closer to A than to B. In order to be further to B it must be placed on the left side of the line, that divides the rectangle in half. It means that the distance AB or DC is less than 7*2=14.

[@McBobX]

Chasle might help here: A, B, C, D 4 points from a plan (or space): We have: vect(AD) = vect(AB) + vect(BC) + vect(CD) So ||vect(AD)|| = ||vect(AB)|| + ||vect(BC)|| + ||vect(CD)||. I don't know but will watch...

[@nordicexile7378]

I solved by setting P at the origin and assigning coordinates to the points: A at (x1, y2) B at (x2, y2) C at (x2, y1) D at (x1, y1) This is possible by (WLOG) having the rectangle rotated parallel to the coordinate grids. Then used the equation of the circle: x^2 + y^2 = r^2. This gave (x1)^2 + (y2)^2 = 121, (x2)^2 + (y2)^2 = 169, (x2)^2 + (y1)^2 = 49, (x1)^2 + (y1)^2 = ? Trying to rearrange the first three equations into a way to derive the fourth gave Adding the first equation to the third gave (x1)^2 + (y2)^2 + (x2)^2 + (y1)^2 = 121 + 49 ((x1)^2 + (y1)^2) + ((x2)^2 + (y2)^2) = 170 [rearranging/combining terms] ((x1)^2 + (y1)^2) + 169 = 170 [substituting from equation 2] ((x1)^2 + (y1)^2) = 1 = 1^2 which is the desired result. Interesting that this is not explicitly using the Pythagorean Theorem, even through that is essentially what the equation of the circle is doing. :-)

[@QuackersForMath]

A quick way of thinking about it: (read full for quick solution) P must lie on the circles: (x-Ax)^2 + (y-Ay)^2 = |AP|^2 (x-Bx)^2 + (y-By)^2 = |BP|^2 (x-Cx)^2 + (y-Cy)^2 = |CP|^2 (x-Dx)^2 + (y-Dy)^2 = |DP|^2 And some function f must exist such that f(AP, CP) = f(BP, DP) since you can reflect the rectangle about two lines through the center. Therefore: |AP|^2 + |CP|^2 = |BP|^2 + |DP|^2 : where f(a, b) = a^2 + b^2 Therefore: |DP| = |sqrt(|AP|^2 + |CP|^2 - |BP|^2)| = |sqrt(11^2 + 7^2 - 13^2)| = |sqrt(1)| = 1

[@maxtaylor3531]

Random fact. The flag is not actually called the Union Jack. It's called the union flag. It's only referred to as the Union Jack when it's flown at sea. Specifically at the bow of a warship.

[@easy_s3351]

Funny thing is that point P is exactly on AD at a height of 11. AB=4√3 and AD=12 so the square is 4√3*12. I used circles at points A, B and C with a respective radius of 11, 13 and 7 to figure it all out but it's too much work to type out here. Which is when I realized there had to be another way, I won't be forgetting about AP^2+CP^2=BP^2+DP^2.

[@necrolord1920]

Hmm it seems like there could be multiple solutions. I got a solution by assuming point P was along the segment from D to C. Based on that assumption, the segment from B to C was equal to sqrt(13^2-7^2) Since the image was a rectangle, that means the segment from A to D was also sqrt(13^2-7^2) Then, using the right triangle there, the distance from D to P would be the sqrt(11^2-(sqrt(13^2-7^2))^2) Which equals sqrt(11^2 - 13^2 + 7^2) = sqrt(1) = 1 Drawing the whole rectangle in that situation, I see no conflicts, therefore this is a possible solution, but I don't know if it is the only solution.

[@davidbuffkin6531]

It's actually called the british flag theorem because the proof involves dropping altitudes from p to the four sides, presenting a shape similar to the british flag.

[@trueriver1950]

David Buffkin correct - especially if P is at the centre. Unfortunately if you put P at the center you dont get a US flag ;)

[@LughSummerson]

The name of the riddle gives away the solution! But surely it's so named because it looks like an off-centre Union flag when you draw the horizontal and vertical lines through _P_, not when you overlay the squares.

[@MrZakrencony]

I figured it out, but i would still kinda want you to go through the math behind the end part of equation or at least write all of the steps using Pythagoras theorem to get to that phase

[@Minecraftster148790]

MrZakrencony he pretty much did

[@FRIMION1]

I solved it in another method: Step 1: Locate P - It should be to the above right side next to point D ( This can be proved by analyzing its proximity To A and B , then to B and C) Step 2: Using Pythagorean theorems ( Use right triangles with hypotenuses having sides with known length, and project them to the right triangle with DP as its hypotenuse) Answer is the same: DP = 1

[@pepegasadge2977]

I just did what you did at when you proved it but I used the lengths of the verticies intead. So I came up with three equations where a,b and c is the respective lenghts: a^2+b^2=121 c^2+b^2=169 c^2+d^2=49 Then I just shuffled it around a bit and put it into a^2+d^2=r^2. Then I got 170-b^2-c^2=r^2. And from that 170-169=r^2 r=1.

[@almrls]

I just figured I would place point P on one of the edges of the rectangles, and solve the geometry problem from there. It couldn't possible be placed on edge AB because triangle BCP would not be possible, and similarly, it couldn't be placed on edge BC because triangle ABP would not be possible either. So I placed point P on edge CD, forming a right triangle BCP, allowing the triangle ABP to be possible. Edge BC is solved to be of length (13^2 - 7^2)^(1/2) = 120^(1/2). Then, another right triangle is formed between ADP. Since point P has the same y positon value as point D, the length of AD is equal to length BC. So length DP = (11^2 - BC^2)^(1/2) = (121 - 120)^(1/2) = 1.

[@r0ntuber]

How come I can't get it to work if I make the value of C as 8 (for instance)? Using that formula I get a value of 256 (that is to say 16 squared)? That doesn't seem right. AP = 11, BP = 13, CP = 8. 11squared + 8squared = 13squared + DPsquared. 185 - 169 = DPsquared. DP = 256. I don't see how such a shape is possible.

[@AScheeser]

Not sure how you got from DP²=16 to DP=256 Surely if DP²=16 then DP=4

[@r0ntuber]

That would be the square root. I was doing 16 squared which is 256 (unless I misunderstood what he was doing the video).

[@AScheeser]

r0ntuber Since the square is equal to 16, you need to take the square root in order to get from DP^2 to DP.

[@r0ntuber]

Ok, thank you. I was thinking backward.

[@r0ntuber]

Perhaps if he had used an example that hadn't ended in 1 = DP squared it would have been more obvious to me. Thanks.

[@kujmous]

What's more is that any regular polygon with an even number of sides (parallel opposite sides) can be segmented into rectangles, and this sum of squares theorem extrapolates to all opposite corner pairs. For giggles, I will informally call this The Stop Sign Theorem.

[@PvblivsAelivs]

That looks like a complicated method. I just slapped the rectangle on a coordinate system with P at the origin. It turned out to be easier than I expected.

[@J7Handle]

I figured it out, but didn't know that theorem. I just assumed that if you were so sure there was only one answer, that I wold just come up with the easiest rectangle solution to the problem. I stuck P on the line DC and then suddenly there was a simple trigonometry problem. cosines and sines.

[@l0lfuuuuuuuuuuuuuuuu]

I solved it with Pythagoras without doing the intermediate equation I labeled AE and DF as d1 and labeled FC and EB as d2, EP as h1 and FP as h2 and labeled DP as x so i got: 121 = d1^2 + h1^2 169 = d2^2 + h1^2 49 = d2^2 + h2^2 x^2 = d1^2 + h2^2 subtracting the second equation from the first gives: d1^2 = d2^2 -48 and then you put the values into the second equation: x^2 = d2^2 - 48 +h2^2 and then you use the third equation and get that: x^2 = 1 x = 1

[@topschnarf8430]

Could you draw circles of the specified radii around the listed points and find the intersection?

[@3seven5seven1nine9]

I tried that and got a very not-circle, so I waited for him to say that D was one unit away from P. If anything, it proves that the theorem doesn't work, because there's not a single 90 degree angle in this "rectangle" imgur.com/a/AtOJn

[@WolfRose11]

I did the Pythagorean theorem idea but instead of making 4 new variables along each direction, I assumed that point A was at (0, 0) on a 2d plane and used (x, y) for the location of P with the height of the rectangle as h and width as w. AP^2 = (x-0)^2 + (y-0)^2 = (x)^2 + (y)^2 BP^2 = (x-w)^2 + (y)^2 CP^2 = (x-w)^2 + (y-h)^2 DP^2 = (x)^2 + (y-h)^2 With each one solved for a variable in the next equation so: (y)^2= AP^2 - (x)^2 (x-w)^2 = BP^2 - (y)^2 (y-h)^2 = CP^2 - (x-w)^2 DP^2 = (x)^2 + (y-h)^2 Where I plugged in the distances starting with the first one and then plugged in that answer to the next equation. At the end you get DP^2 = (x)^2 + 1 - (x)^2 = 1

[@maikopskoy]

yaaay i got this one, I rarely get anything from this channel lol

[@alexeddie2479]

how world this work as the side of the rectangle approaches 0 or Infinity or is there a rule governing the size such as we are only considering rectangles that allows to work

[@quintopia]

Isn't this just a special case of the power-of-a-point theorem?

[@xBieux]

Can you solve this with vectors? Like, you can say BP = BC + CP, and because we're talking about a rectangle, that must mean DP = AP - BC...Seems plausible if we get AP right

[@ABhaim]

@MindYourDecisions this is strange: i approached the challenge using analytic geometry. i used vertex a as (0,0) and played with the radii as was given. i DID get 1 as the remaining radius, bud drawing it all in a square, it didn't work as only 2 circles were intersected

[@McBobX]

Yes, I figured it out, nice proof of the equation.

[@Sid-ix5qr]

That's really a good video! I just learnt some new thing... Something unthinkable..!

[@henkgaas9446]

when p can be anywhere, P can be on the DC-line we have CP=7 BP=13 so BC =10.9545... AP= 11 AD =10.9545... and you will find that DP=1

[@IorPerry]

Didn't the demostration be easer using law of cosines? because inside a rectangle the diagonals have same length

[@DocEJ]

i started drawing circles and estimated that P should be around DC, if you work it out for P on DC you get one single rectangle (8 by 2*root(30)) and with CP =7 you get DP =1. I started with an estimate but ended up with one single (correct) awnser...did i do good?

[@FasAntick]

i lile the flag theorum better but can you solve this with the sin/cosine rule?

[@Ratcat234]

I had nosebleeds for this hard mathematics problem this might genius will slove

[@danik0011]

why just not write circles: center A radius 11 center B radius 13 center C redius 7 find P on intersections of circles and easure distcane P to D?

[@mariamksovreli3243]

Good question, I solved almost the same way, just used only one variable x (FC=x), and represented other distances in terms of x.

[@jgoemat]

5? Going from one corner to any point and back to the opposite corner should be the same for a rectangle. A->P->C is 18, so B->P->D should be 18. Since B->P is 13, P->D should be 5. It needs to be left of center because A is closer than B.

[@jgoemat]

Dang, needed to square them...

[@deerh2o]

I maneuvered P to be on the rectangle. It fit on CD and thus I could derive the appropriate lengths.

[@MrFrak0207]

I'm literally shocked about myself that I did not already know that formula but came up with the right solution just by thinking it through that was kinda creepy

[@adirmugrabi]

i solved it in a very long and complicated way. using coordinates. yours is way better. A(0,0), B(AB,0), C(AB,BC), D(0,BC) P(x,y) then calculate all the distances and plug in the remaining one. since we only care about distance, i can move it to (0,0) and rotate it. works but horrible.

[@adirmugrabi]

1. x^2+y^2 = 11^2 2. (x-ab)^2 + y^2 = 13^2 3. (x-ab)^2 + (y-bc)^2 = 7^2 4. x^2 + (y-bc)^2 = DP^2 1. x^2 + y^2 = 11^2 2. x^2 - 2ab + ab^2 + y^2 - 2bc^2 + bc^2 = 13^2 3. x^2 - 2ab + ab^2 + y^2 = 17^2 4. x^2 + y^2 - 2bc^2 + bc^2 = DP^2 subtract line 2 from line 3 - 2bc^2 + bc^2 = 7^2 - 13^2 add line 1 to it and we get line 4 x^2 + y^2 - 2bc^2 + bc^2 = 11^2 + 7^2 - 13^2 = 1 DP^2 = 1 => DP =1

[@HD-fy2wu]

This also works even when P is not on the same plane as rectangle ABCD.

[@ivan1793]

Every time they talk about this interview questions I wonder: are these oral interviews in which you are supposed to just use your head and answer in a couple of seconds, or are you given pen, paper and a couple of minutes?

[@seanocansey2956]

this one was really interesting! I liked this one....

[@jayant9151]

Ummm i was thinking if it 3d geometry, is it possible to get answer from 3d view

[@Joffrerap]

the distance PD is one?

[@DRD363]

why cant we place p outside the rectangle such that this forms the triangle APB? using this triangle to find DC and then Pythagoras to get PD???

[@Minecraftster148790]

N D he said multiple times you could place it outside the rectangle

[@DRD363]

I know. my comment is not debating that.

[@DRD363]

never mind. now I know why APB cannot be a right triangle. it gives negative distance.

[@xenontesla122]

N D That's exactly what I did. It works, but it doesn't prove that DP is a constant for every rectangle with AP=11 BP=13 CP=7.

[@Kenlauderdale123]

I was thinking of the pythagorean theorem at first, but i dont know the rest lol.. thanks for the explanation

[@chattyw87]

Is there a solution for P inside? I think P has to be outside.

[@daniellassander]

questions based on prior knowledge to gauge intelligence is really stupid.

[@axemenace6637]

Daniel Lassander It's not based on prior knowledge. Nobody memorizes the "British Flag Theorem" because it's a silly theorem with a silly proof. the way you would solve this problem is by dropping perpendicualrs from point P, which is the way I solved it.

[@prakhardwivedi3649]

Maxim Enis or you could use coordinate geometry

[@malcolmbryant]

Maxim Enis -- Daniel Lassander is right. The answer AS GIVEN IN THE VIDEO assumes knowledge of an equation. Which is NO test of intelligence but of knowledge.

[@BingoHS]

melancholiac Are you saying that the answer is a bad test of intelligence, or that the question is?

[@Supermario0727]

This is easy. Just triangulate the position using a compass. All you need to know is the length of one unit.

[@JianJiaHe]

How about if P is on the 3rd dimension?

[@xDprinDx]

anything to the ^2 can be drawn with a square

[@johnboucher3795]

If P can be inside or outside why can it not be on CD? If it is, then doing Pythagoras twice with the 7 and 13 gives you two root 30 and then for 11 and 2 root 30 gives DP equals 1. Don't think outside the box or even inside: think ON the box.

[@TomKaren94]

The placement of P in the diagrams seems impossible given the distances to A and B.

[@Osguima3]

There is an easier way to explain why this equation works: Consider triangles APC and BPD Applying pythagoras, we get: AP^2 + CP^2 = AC^2 BP^2 + DP^2 = BD^2 Notice how AC and BD are the two diagonals of the rectangle? So, AC = BD, AC^2 = BD^2 AP^2 + CP^2 = BP^2 + DP^2 Done!

[@cheezemziez]

Oscar Guillen Mateu APC and BPD are not necessarily right angled triangles

[@Osguima3]

You are right! I wonder why does it still lead to the same answer, though

[@Rararawr]

I drew some circles in paint and lined them up and managed to get 1.11. I'm quite proud of my rough estimate

[@paulpeterson4216]

The people that thought those red and blue squares look like the Union Jack must have been a little bleary from too much Bombay Gin.

[@GaborRevesz_kittenhuffer]

Presh, why not just let A = (0,0), B = (h,0), C = (h,v), D= (0,v), and P = (x,y)? Then, regardless where P is, we have AP² + CP² = (x²+y²) + ((x-h)²+(y-v)²) = ((x-h)²+y²) + (x²+(y-v)²) = BP² + DP² and we're done.

[@magedelimage]

Yeah, equation's great. But, I would like to see the graphic of the solution ! I can't figure out where could be positioned P with those 4 numbers...

[@krisztian76]

Inside the rectangle. AB is shorter than BC, not like in the picture. So P is almost next to D.