I feel that we have overcomplicated after generating 4 equations. Equation1 + Equation3 - (Equation2 + Equation4) will cancel out h1,h2,w1,w2, giving a single equation (25+196-x^2-100=0)
Correct. To also correct where P lies at the two endpoints: t = sqrt(75) corresponds to P lying on the left side of the rectangle, which has width sqrt(75) and height 16. t = 10 corresponds to P lying on the bottom of the rectangle, which has width 15 and height sqrt(96). Curiously, the maximum area of the rectangle is not achieved when it is a square (t = 9.5, w = h = about 13.4, A = about 179.7) but at about t = 9.417, with approximately w = 13.724, h = 13.116 and an area of exactly 180.
As others have mentioned, he got the formula (in blue) for the width of the rectangle wrong. The limits on t are [sqrt(75), 10]. When t is sqrt(75) the point P is on the left edge of the rectangle and the rectangle is 16 units high. When t is 10, the point P is on the bottom edge and the rectangle is 15 units wide. As t varies between these limits, the point P moves along an arc of radius 5 centered on the lower left corner of the rectangle (and, equally valid to state, along arcs of lengths 11, 14, and 10 centered on the other three corners of the rectangle).
Yes, it can obviously never land on the right nor upper side. The lower triangle is streched before the upper one is, and similarly the left one before the right one.
Si se equivoco en el ancho por eso salia ese 0 extraño e imposible y como √75 es un poco mas de 8.5 ya que 8.5*8.5=289/4=72.25 no son tantas opciones para el rectángulo como deberia ser
No need to use such a complex method...Use - 1/5² + 1/10² = 1/(h1)² and by some length changing by the pythagorean theorem you get the last equation as (√116)² + (√5)² = x² => x = 11.
14:55 I may be just a entertaining clown posting timestamps but I wanna give a shoutout to everyone double checking Michael videos and spotting mistakes when they occur. It helps the channel to grow too.
The Linear Algebra way is very cool in my opinion, but there is a way quicker solution. Consider A,B,C and D the projections of P onto the sides of the rectangle. Thus ABCD is quadrilateral with perpendicular diagonals, so AB^2 + CD^2 = BC^2 + DA^2. So x^2 + 10^2 = 5^2 + 14^2, and thus x=11.
Here is a nice stylish solution:😉 Using the British flag theorem we have: t^2+100= 25+196 So to = 11. And so we are done! (Note: just google the theorem, I am a bit lazy to type its statement here 😅)
Overly complex solution. Pythagoras can be used to form a pair of simultaneous equations. Subtracting one from the other gives: 96= X squared - 25, so X squared equals 121; therefore X=11. So much simpler than all the matrix stuff.
@@alanclarke4646Well if you are not familiar with matrices then I guess it may look confusing. My point is that matrix algebra is quite complex in compression but actually none of the "powerfull" stuff was used here. Michael did more or less the same thing as you did, just wrote it down differently. When you look at 6:51 then you should see for example equation "1*h1 + 0*h2 + 1*w1 + 0*w2 = 25" , and when he writes R2-R1 - > R2, it means "take equation 2 and subtract equation 1", then he manipulates then until he achieves at 0 = x^2 -121.
At time 4:10 subtract equation #1 from equation #2 and equation #4 from equation #3. The RHS of each of the results is the same and so you can equate the LHS so you get x^2 - 25 = 196 - 100. So, x^2 = 121 so x = 11.
My attempt. Let's draw heights of all four triangles from the point P. I denote lower triangle (5, 10, ___) as 1st, rightmost triangle (10, 14, | ) as 2nd, upper triangle (x, 14, ----) as 3rd and left triangle (x, 5, | ) as 4th. So, the height of 1st triangle h₁, and so on, and from this we can have four Pythagorean relations: h₁²+h₄²=5², h₁²+h₂²=10², h₂²+h₃²=14², h₃²+h₄²=x². It happens we can figure out x² from this system. Let's add (1)+(3) and subtract (2), so x²=5²+14²-10²=121=11². In other words, this means, that if x=11, we can have such a rectangle and there'll be *underdetermined* system for its side lengths which has infinitely many solutions. If x≠11, the system up there will have no solutions, i. e. no such rectangle could possibly exist.
Here is how I solved it: 1 - parallel shift of the line X to the right so it starts from the top right corner. 2 - parallel shift line with a length of 5 to the right so it starts from the bottom right corner. As a result we get a quadrilateral with sides 14, x, 5 and 10 listed clockwise. Let A and B be the split starting from the left part of the horizontal diagonal by the vertical one, then we got 2 equations: 196 - A^2 = X^2 - B^2 and 100 - A^2 = 25 - B^2. Subtracting second from first we get 96 = x^2 - 25, which yields x = 11
At first I wondered why you continued with the augmented matrix when you could already find the solution. Even from the beginning, you could form w_1^2 + w_2^2 + h_1^2 + h_2^2 two different ways. But given your explanation at the end, using the free variable t, gives a lot more insight! Glad you didn't just focus on finding the result, thereby ignoring the intricacies of the problem :)
yeah, if you add up equations one and three then you can substitute in x^2 for height 2 plus width 1, then you can substitute in 100 for height 1 plus width two and get x^2 + 100 = 221 so x = 11
So t moves from sqrt(75) to 10. During that, the ratio height/width shifts from sqrt(256/75) > 1 to sqrt(96/225) < 1. I.e. there is an intermediate value t' for which the rectangle is a square. I think t' = (7/2)*sqrt((1363 + 57*sqrt(23/17))/194). Maybe Michael wants to prove that?
I think at 14:26 you meant to say that t is bounded in [sqrt(75),10] instead of [0,10], because you got a mistake on saying that the total width is t+sqrt(t^2+75), and that it is instead t+sqrt(t^2-75), since you missed the negative sign for W1=sqrt(t^2-75).
For any three vectors u,v,w, it is easy to check that ||w-u-v||^2+||w||^2-||w-u||^2-||w-v||^2=2(u,v). If (u,v)=0, we deduce that x^2+10^2=5^2+14^2, so x=11. This also shows how to solve the same problem for any parallelogram. Also, the vectors need not even be in the same plane.
A whole other way to do it... Use the law of sines to create relationships among the sines and cosines of all the angles, and the lengths of the given line segments. (Cosines enter into it because each vertex angle is split into an angle and its complement, and the sine of the complement is the cosine of the angle.) Clear the denominators of all of the equations to get 4 equations that look like: 5 sin a = 10 cos b, etc. Square them all and then add pairs together that contain matching sin^2 + cos^2. Solve for x^2. Guess what? You get 121.
Good problem and very approachable. All you need to know is the Pythagorean Theorem and some introductory ideas involving using matrices to represent equations (i.e. you might learn this in a course that is a pre-req. for stats; something like Finite Mathematics or another class with very elementary ideas involving set theory, truth table logic, and so on). I think an average high school freshman could handle this if pointed in the right direction and given a bit of prep (and not just a brilliant math stand-out in an advanced competition).
Well that was cool. I appreciate you taking that a step further than just "solve for X" because, sure, I probably could have found X, but I probably would have missed the snazzy ramifications
I generalized it, and since I found fitting equalities, no need for "brut force" linear algebra/matrices: parallel lines at intersection. left=a, right=b, bottom=c, top=d. from intersection to bottom-left=p, to bottom-right=q, to top-right=r, to top-left=s. using Pythagorean theorem: 1) p²=a²+c² 2) q²=b²+c² 3) r²=b²+d² 4) s²=a²+d² from 1: a²=p²-c² from 3: d²=r²-b² into 4: s²=p²-c²+r²-b²=p²+r²-(b²+c²)=p²+r²-q² so: s=√(p²+r²-q²) for: p=5, q=10, and r=14, s=√(25+196-100)=√121=11 notice: sum of squares of "diagonals" are equal! p²+r²=q²+s²
While that may be true, I guarantee he's burning more interest in his videos as educational tools than he is inspiring others to take up the subject. Using the simpler relationships makes it approachable, and by introducing the answer and openly saying he's going to take a long route allows people to choose to watch the whole thing.
It would be interesting if someone made an animation of the parameterized diagram with t varying in [0,10], highlighting those diagonal segment are constant “rods” in a flexible rectangular structure. At first I thought he might try using Heron’s formula for the areas of the triangles from their side lengths, but you’d end up with a sum of four square roots which could be hard to deal with. 🤷♂️
No need for the linear algebra stuff, one can simply alternately add and subtract the equations you wrote to cancel all the variables on the LHS and get 0 = 25 - x^2 + 196 - 100 which means x = 11.
at 4;10: in those 4 equations we add first with rhird and substract forth. rezults the third equation, and that x^2=121. The Matrice just abstract this, but becomes lengthy.
With capital letters as the squares we have 25 - W1 = 100 - W2, which gives W1 = W2 - 75. Put that into the equation 196 - W2 = X - W1. You get X = 196 - 75 = 121.
Thanks for going over some of the very very basics of the linear algebra since It's been so long since I've done some of that stuff I've forgotten a lot of it. It comes back quickly though, cheers!
Thm: With rectangle OABC and any interior point P, then OP² + PB² = AP² + PC². Which I thought was rather well known. Anyway, it is then immediate that x² = 196 + 25 - 100. BTW the pastel colored chalk does not show up very well, imo. When I was using a blackboard I went to an artist supply store and got a package of VERY brightly colored chalk. The colors were unmistakable across a large lecture hall, and would do very well for your camera, I think.
I just used very simple pythag, splitting the rectangle into four rectangles, through P. And then solved via using simultaneous equations, finding it in terms of the sum of the squares of the top left segmant, square rooted
I doubt you'd get extra marks in the competition for solving in such a convoluted way, but it is interesting to see how various methods lead to the same answer. Also, thumbs up for another entertainingly crazy thumbnail.
If the problem isn't underconstrained, then we can put the point P on the bottom edge of the rectangle. Then on the right we have a 10-h-14 right triangle, so h²=14²-10²=96, and on the left we have a 5-h-x right triangle, so x²=h²+5²=121, so x=11.
"I think it's pretty interesting to see what path this point travels along as we increase this t from 0 to 10." I tried to show that in this video (with another domain of t) ruclips.net/video/wp_pbEcw0H8/видео.html
i solved this using geometric algebra. let q, r, s, and t be unit vectors from the corners of the rectangle to P, where q is the bottom left one, so that line segment is represented by 5q, and so on (10r, 14s, xt). let e1 and e2 be an orthonormal basis for the rectangle, with e1 being a unit vector pointing right at the base and e2 being a unit vector pointing up. if you let w be the width of the rectangle, then we know: 5q-10r=we1 xt-14s=we1 5q-10r=xt-14s xt=5q+14s-10r now take the geometric algebra square of both sides, which is the magnitude of the vector squared: (xt)**2=x**2=321+140q•s-280r•s-100q•r now since the two sides are orthogonal: (5q-10r)•(10r-14s)=0 50q•r-70q•s-100+140r•s=0 -70q•s+140r•s+50q•r=100 140q•s-280r•s-100q•r=-200 but this is exactly the expression on the RHS of the x**2 equation: x**2=321-200=121 and since the magnitude of a vector can only be positive, x=11
Until the matrices came in, that was how I solved it. But then, I would add up equations 1 and 3 together, and 2 and 4 together. That would leave me with h1^2 + h2^2 + w1^2 + w2^2 = 221 h1^2 + h2^2 + w1^2 + w2^2 = x^2 + 100 x^2 + 100 = 221 x^2 = 121 x = 11
Another way of doing this: if there is only one answer, then we just need to find one rectangle that works. Take the rectangle with a width of 15 (5+10) so that the 5 and the 10 lines are perfectly horizontal and P lies on the bottom edge of the rectangle. It is then possible to select a height for the rectangle such that the the line from P to the upper right corner is 14. Using Pythagoras this height is sqrt(14^2-10^2) = sqrt(96). In this configuration x is the hypotenuse of a triangle with base 5 and height sqrt(96), using Pythagoras again, x=11.
@@bobh6728 That's not what we were asked. The question itself implied that whatever works for one rectangle will work for all (applicable) rectangles. It's a useful thing to pay attention to with problem questions like this, the fact that there is a singular numerical answer gives us loads of information to work with.
I love your presentations! A thought (likely equivalent to others below; I did not check all the comments): if you go to time 8:18, you will see that the only way the second and fourth equations can be consistent is if -75 = x^2 - 196 which gives the solution x =11 right away. Mind you, you likely knew that...I enjoyed your further exploration of the problem...
...beyond just getting the solution x = 11. While there was a typo at the end re: the domain of t, I find your presentations clear and elegant. Keep up the good work!
glad im not the only person who thought matrices was over complicating it. Just start algebra-ing the four Pythagorean equalities together and all the Ws and Hs cancel out.
This looks like an over complication to me; matrices do not make this algebra problem any simpler in my opinion. If the rectangle has dimensions w by h and the point is at (a,b) then we are given that a^2+b^2 = 5^2, (w-a)^2+b^2=10^2, (w-a)^2+(h-b)^2=14^2 and we are asked to find x where x^2=a^2+(w-b)^2 x^2=a^2+(w-b)^2= [add and subtract same terms] a^2 + (b^2-b^2) + ((w-a)^2-(w-a)^2) + (w-b)^2=. [rearrange] (a^2 + b^2) - ((b^2) + (w-a)^2) + ((w-a)^2 + (w-b)^2)= 5^2-10^2+14^2 = 121 so, x=11.
@@olau5478 Yes. Given a rectangle ABCD and a point P anywhere on the plane (not necessarily inside ABCD) then |AP|^2 + |CP|^2 = |BP|^2 + |DP|^2. You can prove this easily by dropping verticals from P to each side (extended as needed) and noting that each of AP, BP, CP, and DP is the hypotenuse of two right triangles. Just substitute using Pythagoras and both sides end up as the same expression (provided that you pick the correct right triangle for each substitution).
@@olau5478 Yep. But then Michael couldn't have filled a 15-minute video out of solving the problem. :) Seriously, I think this is probably the solution that the test designers had in mind, rather than the long detour into linear algebra.
Since the problem is high school level, the complex matrix can be avoided. The problem can be solved by considering 2 equations. That is, x^2 - a^2 = 14^2 - b^2…..….(1) and 5^2 -a^2 = 10^2 -b^2 …………..(2) Where a + b is the length of the rectangle and x is the required length. Solving the 2 equations by subtracting (2) from (1) gives x = 11 as the answer.
One of few Penn videos I'm able to solve the problem. Then I view the solution and suddenly realize x=11 is pretty dull and don't capture what's cool with the problem 😞
who else thought it was impossible/pointless to proceed with matrix operations on a system of 4 eqns with 5 unknowns?? i thought there had to be some neat geometry insight to solve for at least one of the variables in terms of another..
What *is* the path P takes through the rectangle as you vary t? (The rectangle itself will change dimensions, too, so perhaps some rescaling would be interesting as well.)
Let's Z - distance we want to found Imagine the long side of the rectangle is A then perpendicular from point P divides it for A-X and X parts. Let's denote two heights from point P to long sides as C and D, then from the Pythagorean theorem: C^2 = 14^2 - (A - X)^2 C^2 = Z^2 - X^2 and D^2 = 5^2 - X^2 D^2 = 10^2 - (A - X)^2 and system of equations: 14^2 - (A - X)^2 = Z^2 - X^2 25 - X^2 = 100 - (A - X)^2 substitute in first (A - X)^2 with (75 + X^2): 14^2 - (75 + X^2) = Z^2 - X^2 or Z^2 = 121 Z = 11 that's it
At 8:17 we already know x^2-196 = -75. That immediately gives the solution x=11 (because x=-11 is impossible). Why do all the extra steps then? EDIT: I should have watched the rest of the explanation, whoops ^^'
wait what logic was used to claim that a solution exists? Initially Michael said that if the row reduced form didn't have 0s in the last row, then this implied that no solution exists. So how do u disregard that case and show that the solution exists.
I have absolutely no idea why you went into matrices for what is a simple algebraic linear system of equations. EQ1 + (EQ3 - EQ4) = EQ2 = 121 = x^2 and its done.
It's just a rewriting of the system of equations. The numbers in the matrix are coefficients of the variables in the system, and the vector multiplying the matrix effectively tells which columns correspond what variable. The top element of the vector being the left-most column, and bottom being the right-most column. Say you have this system of equations: 12x + 3y + 7z = 134 5x + 11y + z = 93 4x + 9z = 73 An equivalent matrix equation would be: [12 3 7] [x] [134] [ 5 11 1] [y] = [ 93 ] [ 4 0 9] [z] [ 73 ]
@@YOM2_UB oh okay, I get it know. This kind of matrix system is new to me, I thought there's more than the usual setup for this kind of system of equation
There are two degenerate cases. One case is where the five unit edge is horizontal, and the other case is where the five unit edge is vertical. In the first case h sub 1 is 0, in the second case w sub 1 is 0. Both cases give the answer that the unknown length is 11. In fact if you describe h sub 1 and w sub 1 as the cosine and sine of the angle, then the trigonometry expressions disappeared during simplification and the answer is 11 regardless of the angle! ... Unless I've done something really stupid.
It looks sign of the t^2 in total width (in left drawing) is changed from minus to plus and range of the would be from 5*sqrt(3) to 10. At first it looks strange that it solves, while here is 4 equations with 5 unknowns (x, w1, w2, h1, h2). But for some reason they seem to be eliminating so that x can be solved. In this point of view it couldn't make sense. Maybe some "special case" where its allowed to be more variables than equations and one of the variables will solve perfectly and there would remain 4 equations with 4 unknowns that do not solve with exact solution (this parameter sweep will be solution for the w1, w2, h1 & h2). The equations are maybe not coupling the variables "efficiently enough" to give exact solution. I checked the image and it looks possible geometries, but I couldn't post the image here into these comment section. But I would set it w1=t=5*sin(theta), h1=sqrt(25-t^2)=5*cos(theta), w2=sqrt(75+t^2), h2=sqrt(121-t^2) and sweep 0
I think you are asking why isn’t there one solution for the four equations with four unknowns. The four equations are independent, leading to infinite solutions. You can see that by looking at the drawing and changing one of the segments. If you try to change h1, you either have to change h2 or you wouldn’t have a rectangle.
I have very short method Name rectangle PQRS and point inside rectangle O. Put P as origin in Cartesian plane, Q as (0,b), R as (a,b), S as (0,a), O as (c,d). Write distance formula for line segments OP, OQ, OR, OS as follows, c^2 + d^2 = 25 ....(1) c^2 + (b-d)^2 = t^2 ......(2) (a-c)^2 + ( b-d)^2 = 196 ......(3) (a-c)^2 + d^2 = 100 ......(4) Add 2 and 4 eq. and put value from 1 and 3 eq. We get t=11. comment if I have done something wrong!!
Stafford has gone downhill if this is an example.. did a long winded way to solve ( didn't need half of that equation) and got it wrong anyway..🤣🤣🤣 ime not impressed. Solved it in my head while waiting for the Internet to buffer..Oxford would laugh at this 10th grade problem.
The frist matrix is made by taking the coefficients from the system of equations. The vector multiplying the matrix tells which variables have what coefficients, with the top element corresponding to the left-most column and the bottom to the right-most. The second matrix is just a shorthand for the first equation. It adds the vector on the right-hand side of the equation to the matrix and hides the multiplying vector to make row operations quicker, as they effect both sides of the equation but not the multiplying vector.
I feel that we have overcomplicated after generating 4 equations. Equation1 + Equation3 - (Equation2 + Equation4) will cancel out h1,h2,w1,w2, giving a single equation (25+196-x^2-100=0)
yup, exactly my approach. (Although, this does not directly give us the family of possible rectangles, I suppose.)
That's what I did too! I think this is quite a general result called the British Flag Theorem (I guess it's because it looks like the Union Jack).
i agree, the british flag theorem would result exactly to this equation you gave.
Furthermore, I also did this approach
Yup I solved in 20 sec, without pen and paper: t=sqrt(196-100+25)=11
Small typo at the end, should have sqrt(t^2 - 75), which means t is on [sqrt(75), 10].
Correct.
And then the t must be from interval sqrt(75) up to 10.
@@ThePfilip Yes, Mr/Ms Optimality already said that.
How is someone supposed to make a typo with no computer keyboard?
Correct. To also correct where P lies at the two endpoints:
t = sqrt(75) corresponds to P lying on the left side of the rectangle, which has width sqrt(75) and height 16.
t = 10 corresponds to P lying on the bottom of the rectangle, which has width 15 and height sqrt(96).
Curiously, the maximum area of the rectangle is not achieved when it is a square (t = 9.5, w = h = about 13.4, A = about 179.7) but at about t = 9.417, with approximately w = 13.724, h = 13.116 and an area of exactly 180.
t = 0 makes w1 become complex number, which is invalid. The range of t is actually [sqrt(75),10].
The errors are the salt of the channel
@John Tse which side? I think you misread sqrt(196-t^2)
Another mistake is "t + sqrt(t² + 75)" in blue on board instead of "t + sqrt(t² - 75)", this should be the reason of t-range mistake
As others have mentioned, he got the formula (in blue) for the width of the rectangle wrong. The limits on t are [sqrt(75), 10]. When t is sqrt(75) the point P is on the left edge of the rectangle and the rectangle is 16 units high. When t is 10, the point P is on the bottom edge and the rectangle is 15 units wide. As t varies between these limits, the point P moves along an arc of radius 5 centered on the lower left corner of the rectangle (and, equally valid to state, along arcs of lengths 11, 14, and 10 centered on the other three corners of the rectangle).
Yes, it can obviously never land on the right nor upper side. The lower triangle is streched before the upper one is, and similarly the left one before the right one.
Si se equivoco en el ancho por eso salia ese 0 extraño e imposible y como √75 es un poco mas de 8.5 ya que 8.5*8.5=289/4=72.25 no son tantas opciones para el rectángulo como deberia ser
No need to use such a complex method...Use -
1/5² + 1/10² = 1/(h1)² and by some length changing by the pythagorean theorem you get the last equation as
(√116)² + (√5)² = x²
=> x = 11.
14:55 I may be just a entertaining clown posting timestamps but I wanna give a shoutout to everyone double checking Michael videos and spotting mistakes when they occur. It helps the channel to grow too.
*give
@@a_llama Indeed, fixed. Thanks
Yup, as many have pointed out in the other comments:
w1 + w2 should have been t + sqrt(t² - 75), NOT t + sqrt(t² + 75)
Another Michael's "glitches" 😄
Hey now.
You're OUR entertaining clown posting timestamps.
The Linear Algebra way is very cool in my opinion, but there is a way quicker solution. Consider A,B,C and D the projections of P onto the sides of the rectangle. Thus ABCD is quadrilateral with perpendicular diagonals, so AB^2 + CD^2 = BC^2 + DA^2. So x^2 + 10^2 = 5^2 + 14^2, and thus x=11.
Here is a nice stylish solution:😉
Using the British flag theorem we have:
t^2+100= 25+196
So to = 11.
And so we are done!
(Note: just google the theorem, I am a bit lazy to type its statement here 😅)
@13:40 - the width should be t + sqrt(t^2 - 75), not t + sqrt(t^2 + 75). This means that t is in the interval [ 5 sqrt(3), 10 ], not [ 0, 10 ].
Why don't we just notice at 8:20 that x² - 196 = -75 and then solve for x?
You can find x very quickly without all of this, but he wanted to show the parameterization with t at the end
Yeah like when( h_1)²+(w_1)² = 25 we know it is a Pythagoras triple and h and w have to be (3,4) or( 4,3)
@@adityaekbote8498 Not really, they don't have to be integer at all.
Just looking through the comments to see whether this had already been spotted and yes it has 👍🏻
It is all for family (of solution rectangles)
Overly complex solution. Pythagoras can be used to form a pair of simultaneous equations. Subtracting one from the other gives: 96= X squared - 25, so X squared equals 121; therefore X=11. So much simpler than all the matrix stuff.
adding/subtracting and scaling equations is exactly the same as row operations, it's more common to do it this way because the notation is clearer.
@@MrMeztar looked a whole lot more complicated to me.
@@alanclarke4646Well if you are not familiar with matrices then I guess it may look confusing. My point is that matrix algebra is quite complex in compression but actually none of the "powerfull" stuff was used here. Michael did more or less the same thing as you did, just wrote it down differently. When you look at 6:51 then you should see for example equation "1*h1 + 0*h2 + 1*w1 + 0*w2 = 25" , and when he writes R2-R1 - > R2, it means "take equation 2 and subtract equation 1", then he manipulates then until he achieves at 0 = x^2 -121.
@@MrMeztar Ask yourself whether a typical student aged 16 would find simultaneous equations or row reduction easier to understand.
@@hb1338 row reduction once they understand the notation
At time 4:10 subtract equation #1 from equation #2 and equation #4 from equation #3. The RHS of each of the results is the same and so you can equate the LHS so you get x^2 - 25 = 196 - 100. So, x^2 = 121 so x = 11.
My attempt.
Let's draw heights of all four triangles from the point P. I denote lower triangle (5, 10, ___) as 1st, rightmost triangle (10, 14, | ) as 2nd, upper triangle (x, 14, ----) as 3rd and left triangle (x, 5, | ) as 4th. So, the height of 1st triangle h₁, and so on, and from this we can have four Pythagorean relations:
h₁²+h₄²=5²,
h₁²+h₂²=10²,
h₂²+h₃²=14²,
h₃²+h₄²=x².
It happens we can figure out x² from this system. Let's add (1)+(3) and subtract (2), so x²=5²+14²-10²=121=11². In other words, this means, that if x=11, we can have such a rectangle and there'll be *underdetermined* system for its side lengths which has infinitely many solutions. If x≠11, the system up there will have no solutions, i. e. no such rectangle could possibly exist.
Here is how I solved it: 1 - parallel shift of the line X to the right so it starts from the top right corner. 2 - parallel shift line with a length of 5 to the right so it starts from the bottom right corner. As a result we get a quadrilateral with sides 14, x, 5 and 10 listed clockwise. Let A and B be the split starting from the left part of the horizontal diagonal by the vertical one, then we got 2 equations: 196 - A^2 = X^2 - B^2 and 100 - A^2 = 25 - B^2. Subtracting second from first we get 96 = x^2 - 25, which yields x = 11
At first I wondered why you continued with the augmented matrix when you could already find the solution. Even from the beginning, you could form w_1^2 + w_2^2 + h_1^2 + h_2^2 two different ways. But given your explanation at the end, using the free variable t, gives a lot more insight! Glad you didn't just focus on finding the result, thereby ignoring the intricacies of the problem :)
Well this can be simplified just by relating the equations formed by using pythagoras theorem. But still, a great video!
yeah, if you add up equations one and three then you can substitute in x^2 for height 2 plus width 1, then you can substitute in 100 for height 1 plus width two and get x^2 + 100 = 221 so x = 11
Exactly.
As a high school student i have no idea why he did this problem in matrix and i used ‘’gougu’’ theory to able solve it
@@dalan1999 Not sure, but I think matrixes can be used to solve system of equations. (Not sure, also a high school student)
So t moves from sqrt(75) to 10. During that, the ratio height/width shifts from sqrt(256/75) > 1 to sqrt(96/225) < 1. I.e. there is an intermediate value t' for which the rectangle is a square. I think t' = (7/2)*sqrt((1363 + 57*sqrt(23/17))/194). Maybe Michael wants to prove that?
My thoughts exactly (for the square).
I think at 14:26 you meant to say that t is bounded in [sqrt(75),10] instead of [0,10], because you got a mistake on saying that the total width is t+sqrt(t^2+75), and that it is instead t+sqrt(t^2-75), since you missed the negative sign for W1=sqrt(t^2-75).
I literally solved this in 15 seconds by using the British flag theorem x^2=14^2+5^2-10^2=11^2
Much love as always :)
For any three vectors u,v,w, it is easy to check that ||w-u-v||^2+||w||^2-||w-u||^2-||w-v||^2=2(u,v). If (u,v)=0, we deduce that x^2+10^2=5^2+14^2, so x=11. This also shows how to solve the same problem for any parallelogram. Also, the vectors need not even be in the same plane.
Appreciate the methods used, but simple substitution of equations works pretty easily. generated 100-25+w1^2+x^2-w1^2=196-->x^2=121-->x=11
x² + z² = y² + w² where x y z w are in circular order. xz and yw are opposite to each other
A whole other way to do it... Use the law of sines to create relationships among the sines and cosines of all the angles, and the lengths of the given line segments. (Cosines enter into it because each vertex angle is split into an angle and its complement, and the sine of the complement is the cosine of the angle.) Clear the denominators of all of the equations to get 4 equations that look like: 5 sin a = 10 cos b, etc. Square them all and then add pairs together that contain matching sin^2 + cos^2. Solve for x^2. Guess what? You get 121.
Good problem and very approachable. All you need to know is the Pythagorean Theorem and some introductory ideas involving using matrices to represent equations (i.e. you might learn this in a course that is a pre-req. for stats; something like Finite Mathematics or another class with very elementary ideas involving set theory, truth table logic, and so on).
I think an average high school freshman could handle this if pointed in the right direction and given a bit of prep (and not just a brilliant math stand-out in an advanced competition).
Well that was cool. I appreciate you taking that a step further than just "solve for X" because, sure, I probably could have found X, but I probably would have missed the snazzy ramifications
I generalized it, and since I found fitting equalities, no need for "brut force" linear algebra/matrices:
parallel lines at intersection.
left=a, right=b, bottom=c, top=d.
from intersection to bottom-left=p, to bottom-right=q, to top-right=r, to top-left=s.
using Pythagorean theorem:
1) p²=a²+c²
2) q²=b²+c²
3) r²=b²+d²
4) s²=a²+d²
from 1: a²=p²-c²
from 3: d²=r²-b²
into 4: s²=p²-c²+r²-b²=p²+r²-(b²+c²)=p²+r²-q²
so: s=√(p²+r²-q²)
for: p=5, q=10, and r=14, s=√(25+196-100)=√121=11
notice: sum of squares of "diagonals" are equal!
p²+r²=q²+s²
Ohh my god, the comment section is filled with school math peeps. He is not trying to solve a problem, he is giving perspective into math....
While that may be true, I guarantee he's burning more interest in his videos as educational tools than he is inspiring others to take up the subject. Using the simpler relationships makes it approachable, and by introducing the answer and openly saying he's going to take a long route allows people to choose to watch the whole thing.
It would be interesting if someone made an animation of the parameterized diagram with t varying in [0,10], highlighting those diagonal segment are constant “rods” in a flexible rectangular structure.
At first I thought he might try using Heron’s formula for the areas of the triangles from their side lengths, but you’d end up with a sum of four square roots which could be hard to deal with. 🤷♂️
I tried here: ruclips.net/video/Zgeelf-ejJY/видео.html
@@wernergamper6200 Thanks! Is it possible to keep the lower left corner at (0,0) so it is easier to see P move?
@@michaelact Since the lower left "radius" always is 5, P moves on a quarter circle.
ruclips.net/video/ghYQfNHj-xI/видео.html
@@wernergamper6200 Neat, thanks. 🙂
No need for the linear algebra stuff, one can simply alternately add and subtract the equations you wrote to cancel all the variables on the LHS and get 0 = 25 - x^2 + 196 - 100 which means x = 11.
at 4;10: in those 4 equations we add first with rhird and substract forth. rezults the third equation, and that x^2=121. The Matrice just abstract this, but becomes lengthy.
What a fantastic problem!
I love Geometry and Linear Algebra, so this was great.
Thank you, professor
With capital letters as the squares we have 25 - W1 = 100 - W2, which gives W1 = W2 - 75. Put that into the equation 196 - W2 = X - W1. You get X = 196 - 75 = 121.
Thanks for going over some of the very very basics of the linear algebra since It's been so long since I've done some of that stuff I've forgotten a lot of it. It comes back quickly though, cheers!
Thm: With rectangle OABC and any interior point P, then OP² + PB² = AP² + PC². Which I thought was rather well known. Anyway, it is then immediate that x² = 196 + 25 - 100. BTW the pastel colored chalk does not show up very well, imo. When I was using a blackboard I went to an artist supply store and got a package of VERY brightly colored chalk. The colors were unmistakable across a large lecture hall, and would do very well for your camera, I think.
I just used very simple pythag, splitting the rectangle into four rectangles, through P. And then solved via using simultaneous equations, finding it in terms of the sum of the squares of the top left segmant, square rooted
I doubt you'd get extra marks in the competition for solving in such a convoluted way, but it is interesting to see how various methods lead to the same answer.
Also, thumbs up for another entertainingly crazy thumbnail.
This...
If the problem isn't underconstrained, then we can put the point P on the bottom edge of the rectangle. Then on the right we have a 10-h-14 right triangle, so h²=14²-10²=96, and on the left we have a 5-h-x right triangle, so x²=h²+5²=121, so x=11.
"I think it's pretty interesting to see what path this point travels along as we increase this t from 0 to 10." I tried to show that in this video (with another domain of t) ruclips.net/video/wp_pbEcw0H8/видео.html
i solved this using geometric algebra. let q, r, s, and t be unit vectors from the corners of the rectangle to P, where q is the bottom left one, so that line segment is represented by 5q, and so on (10r, 14s, xt). let e1 and e2 be an orthonormal basis for the rectangle, with e1 being a unit vector pointing right at the base and e2 being a unit vector pointing up. if you let w be the width of the rectangle, then we know:
5q-10r=we1
xt-14s=we1
5q-10r=xt-14s
xt=5q+14s-10r
now take the geometric algebra square of both sides, which is the magnitude of the vector squared:
(xt)**2=x**2=321+140q•s-280r•s-100q•r
now since the two sides are orthogonal:
(5q-10r)•(10r-14s)=0
50q•r-70q•s-100+140r•s=0
-70q•s+140r•s+50q•r=100
140q•s-280r•s-100q•r=-200
but this is exactly the expression on the RHS of the x**2 equation:
x**2=321-200=121
and since the magnitude of a vector can only be positive, x=11
Theorem belongs to Mustafa Yağcı
'' height theorem''
x^2 - 5^2 = 14^2 - 10^2
used for similar questions
It follows quickly from the British flag theorem. 10^2+x^2=14^2+5^2, so x^2=121 and x=11.
Until the matrices came in, that was how I solved it. But then, I would add up equations 1 and 3 together, and 2 and 4 together.
That would leave me with
h1^2 + h2^2 + w1^2 + w2^2 = 221
h1^2 + h2^2 + w1^2 + w2^2 = x^2 + 100
x^2 + 100 = 221
x^2 = 121
x = 11
Another way of doing this: if there is only one answer, then we just need to find one rectangle that works. Take the rectangle with a width of 15 (5+10) so that the 5 and the 10 lines are perfectly horizontal and P lies on the bottom edge of the rectangle. It is then possible to select a height for the rectangle such that the the line from P to the upper right corner is 14. Using Pythagoras this height is sqrt(14^2-10^2) = sqrt(96). In this configuration x is the hypotenuse of a triangle with base 5 and height sqrt(96), using Pythagoras again, x=11.
But does that work for ALL rectangles.
@@bobh6728 That's not what we were asked. The question itself implied that whatever works for one rectangle will work for all (applicable) rectangles. It's a useful thing to pay attention to with problem questions like this, the fact that there is a singular numerical answer gives us loads of information to work with.
I love your presentations! A thought (likely equivalent to others below; I did not check all the comments): if you go to time 8:18, you will see that the only way the second and fourth equations can be consistent is if -75 = x^2 - 196 which gives the solution x =11 right away. Mind you, you likely knew that...I enjoyed your further exploration of the problem...
...beyond just getting the solution x = 11. While there was a typo at the end re: the domain of t, I find your presentations clear and elegant. Keep up the good work!
glad im not the only person who thought matrices was over complicating it. Just start algebra-ing the four Pythagorean equalities together and all the Ws and Hs cancel out.
This looks like an over complication to me; matrices do not make this algebra problem any simpler in my opinion.
If the rectangle has dimensions w by h and the point is at (a,b) then we are given that
a^2+b^2 = 5^2, (w-a)^2+b^2=10^2, (w-a)^2+(h-b)^2=14^2
and we are asked to find x where x^2=a^2+(w-b)^2
x^2=a^2+(w-b)^2= [add and subtract same terms]
a^2 + (b^2-b^2) + ((w-a)^2-(w-a)^2) + (w-b)^2=. [rearrange]
(a^2 + b^2) - ((b^2) + (w-a)^2) + ((w-a)^2 + (w-b)^2)= 5^2-10^2+14^2 = 121
so, x=11.
Matrices are overkill here. The sum of the squares of the respective diagonals thru P are equal: so 25+196=100+x^2. Simplify for x^2 = 121, or x=11.
You surely can use strategies from linear algebra, but also can use a fact, that in the rectangle x^2+10^2=14^2+5^2 ))
is that true?
@@olau5478 Yes. Given a rectangle ABCD and a point P anywhere on the plane (not necessarily inside ABCD) then |AP|^2 + |CP|^2 = |BP|^2 + |DP|^2. You can prove this easily by dropping verticals from P to each side (extended as needed) and noting that each of AP, BP, CP, and DP is the hypotenuse of two right triangles. Just substitute using Pythagoras and both sides end up as the same expression (provided that you pick the correct right triangle for each substitution).
@@TedHopp wow, i didnt know that. certainly seems useful in this problem
@@olau5478 Yep. But then Michael couldn't have filled a 15-minute video out of solving the problem. :) Seriously, I think this is probably the solution that the test designers had in mind, rather than the long detour into linear algebra.
Since the problem is high school level, the complex matrix can be avoided. The problem can be solved by considering 2 equations. That is, x^2 - a^2 = 14^2 - b^2…..….(1) and 5^2 -a^2 = 10^2 -b^2 …………..(2) Where a + b is the length of the rectangle and x is the required length. Solving the 2 equations by subtracting (2) from (1) gives x = 11 as the answer.
One of few Penn videos I'm able to solve the problem. Then I view the solution and suddenly realize x=11 is pretty dull and don't capture what's cool with the problem 😞
who else thought it was impossible/pointless to proceed with matrix operations on a system of 4 eqns with 5 unknowns??
i thought there had to be some neat geometry insight to solve for at least one of the variables in terms of another..
There is, but if it was a square, or if there was some dependency between the rectangle's sides, for ex.
2(h₁ + h₂) = w₁ + w₂
What *is* the path P takes through the rectangle as you vary t? (The rectangle itself will change dimensions, too, so perhaps some rescaling would be interesting as well.)
ruclips.net/video/ghYQfNHj-xI/видео.html a quarter circle
@@wernergamper6200 Awesome! Thanks for the great animation!
t can’t be 0 because then t^2-75 would be negative in the square root
That was great.can you please solve some questions of IOI
Very cool problem.
well, this is a nice classical problem in Geometry called British Flag Theorem
Let's Z - distance we want to found
Imagine the long side of the rectangle is A
then perpendicular from point P divides it for A-X and X parts.
Let's denote two heights from point P to long sides as C and D, then from the Pythagorean theorem:
C^2 = 14^2 - (A - X)^2
C^2 = Z^2 - X^2
and
D^2 = 5^2 - X^2
D^2 = 10^2 - (A - X)^2
and system of equations:
14^2 - (A - X)^2 = Z^2 - X^2
25 - X^2 = 100 - (A - X)^2
substitute in first (A - X)^2 with (75 + X^2):
14^2 - (75 + X^2) = Z^2 - X^2
or Z^2 = 121
Z = 11
that's it
British Flag Theorem
At 8:17 we already know x^2-196 = -75. That immediately gives the solution x=11 (because x=-11 is impossible). Why do all the extra steps then?
EDIT: I should have watched the rest of the explanation, whoops ^^'
where can i buy that chalkboard
wait what logic was used to claim that a solution exists? Initially Michael said that if the row reduced form didn't have 0s in the last row, then this implied that no solution exists. So how do u disregard that case and show that the solution exists.
I have absolutely no idea why you went into matrices for what is a simple algebraic linear system of equations. EQ1 + (EQ3 - EQ4) = EQ2 = 121 = x^2 and its done.
This is might be the most ridiculous way to solve that problem...
Determine the real numbers of a and b such that a^2-4b and b^2-4a are both perfect squares? Watch now Michael Penn!!! 😁😁👍👍
How did you do the starting Matrix? 4:18
I know this can be done with Pythagoras, but this looks like Ceva’s theorem
Just after 8:10 (I think!) we have, x^2-196 = -75 and then we could just solve for x easily?? No?
How is the first matrix calculated? I can't figure it out
It's just a rewriting of the system of equations. The numbers in the matrix are coefficients of the variables in the system, and the vector multiplying the matrix effectively tells which columns correspond what variable. The top element of the vector being the left-most column, and bottom being the right-most column.
Say you have this system of equations:
12x + 3y + 7z = 134
5x + 11y + z = 93
4x + 9z = 73
An equivalent matrix equation would be:
[12 3 7] [x] [134]
[ 5 11 1] [y] = [ 93 ]
[ 4 0 9] [z] [ 73 ]
@@YOM2_UB oh okay, I get it know. This kind of matrix system is new to me, I thought there's more than the usual setup for this kind of system of equation
There are two degenerate cases. One case is where the five unit edge is horizontal, and the other case is where the five unit edge is vertical. In the first case h sub 1 is 0, in the second case w sub 1 is 0. Both cases give the answer that the unknown length is 11. In fact if you describe h sub 1 and w sub 1 as the cosine and sine of the angle, then the trigonometry expressions disappeared during simplification and the answer is 11 regardless of the angle! ... Unless I've done something really stupid.
It's just straight up British Flag Theorem.
Good problem
Bizdede 1966 üss de var bu soru. Demekki çalışırken dünyayı takip etmek gerekliymiş
It will be x^2= 14^2+5^2 -10^2 = 121
i.e x= 11
Using linear combination would be easier
This is the sickest proof of the British flag Theorem.
Solution is 11 by the British Flag Theorem
Why orthogonal over perpendicular?
Why so complicated?
British flag theorem has joined the chat 💀
We can use british flag theorem
It looks sign of the t^2 in total width (in left drawing) is changed from minus to plus and range of the would be from 5*sqrt(3) to 10.
At first it looks strange that it solves, while here is 4 equations with 5 unknowns (x, w1, w2, h1, h2). But for some reason they seem to be eliminating so that x can be solved. In this point of view it couldn't make sense. Maybe some "special case" where its allowed to be more variables than equations and one of the variables will solve perfectly and there would remain 4 equations with 4 unknowns that do not solve with exact solution (this parameter sweep will be solution for the w1, w2, h1 & h2). The equations are maybe not coupling the variables "efficiently enough" to give exact solution.
I checked the image and it looks possible geometries, but I couldn't post the image here into these comment section.
But I would set it w1=t=5*sin(theta), h1=sqrt(25-t^2)=5*cos(theta), w2=sqrt(75+t^2), h2=sqrt(121-t^2) and sweep 0
I think you are asking why isn’t there one solution for the four equations with four unknowns. The four equations are independent, leading to infinite solutions. You can see that by looking at the drawing and changing one of the segments. If you try to change h1, you either have to change h2 or you wouldn’t have a rectangle.
@@bobh6728 It looks they are independent enough. (Not enough of coupling among them?)
Oh yeah, a classical geometry problem.
But why Michael's answer look so complicated?
It’s not about solving the problem. It’s about making a point.
It's very elementary linear algebra
See my solution
When you want to give an answer to an indeterminate system of equations, you usually use parametric solutions.
Actually,there is a typo t can be bigger than root of 75
*sees SMT in the thumbnail*
*clicks*
My disappointment is immeasurable and my day is ruined.
T=11
The minimum t is sqrt(75) = 8.66, not zero, otherwise w1 is imaginary
Just saw others have already spotted this!
Algebratic geometry xd
I have very short method
Name rectangle PQRS and point inside rectangle O.
Put P as origin in Cartesian plane, Q as (0,b), R as (a,b), S as (0,a), O as (c,d).
Write distance formula for line segments OP, OQ, OR, OS as follows,
c^2 + d^2 = 25 ....(1)
c^2 + (b-d)^2 = t^2 ......(2)
(a-c)^2 + ( b-d)^2 = 196 ......(3)
(a-c)^2 + d^2 = 100 ......(4)
Add 2 and 4 eq. and put value from 1 and 3 eq.
We get t=11.
comment if I have done something wrong!!
That is what he did, but just used matrices to solve.
Stafford has gone downhill if this is an example.. did a long winded way to solve ( didn't need half of that equation) and got it wrong anyway..🤣🤣🤣 ime not impressed. Solved it in my head while waiting for the Internet to buffer..Oxford would laugh at this 10th grade problem.
this is perfect for the british flag theorem
Это устно решается за минуту, сложили два равенства и вычли третье...
You might be wrong this time, t^2 is smaller than 75 according to your solution
Was this mistake planned?
11
Can anyone please explain to me how did he get the first matrix in 4:19
and also the second matrix if you can, and thanks.
The frist matrix is made by taking the coefficients from the system of equations. The vector multiplying the matrix tells which variables have what coefficients, with the top element corresponding to the left-most column and the bottom to the right-most.
The second matrix is just a shorthand for the first equation. It adds the vector on the right-hand side of the equation to the matrix and hides the multiplying vector to make row operations quicker, as they effect both sides of the equation but not the multiplying vector.
@@YOM2_UB Oh ok, thanks.
イギリスがnatoから離脱すると聞いてホッとしました。
British flag theorem
May I just use: eq1 + eq3 = eq2 + eq4?? It seems simpler..
Would be interesting to know why we overcomplicated a 2 line solution.
Hehehe quadratures
I tried solving it using al kashi's cosin law ....... only tried ..... thank god im still Alive. 🥴
You could have used _goku_ theorem instead