An interesting geometry competition problem.

Рет қаралды 36,357

3 жыл бұрын

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Пікірлер: 187

@akhtariitk 3 жыл бұрын

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)

@DitDede 3 жыл бұрын

yup, exactly my approach. (Although, this does not directly give us the family of possible rectangles, I suppose.)

@TheLetterW736 3 жыл бұрын

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).

@michaelempeigne3519 3 жыл бұрын

i agree, the british flag theorem would result exactly to this equation you gave.

@michaelempeigne3519 3 жыл бұрын

Furthermore, I also did this approach

@koenth2359 2 жыл бұрын

Yup I solved in 20 sec, without pen and paper: t=sqrt(196-100+25)=11

@austinm271 3 жыл бұрын

Small typo at the end, should have sqrt(t^2 - 75), which means t is on [sqrt(75), 10].

@cernejr 3 жыл бұрын

Correct.

@ThePfilip 3 жыл бұрын

And then the t must be from interval sqrt(75) up to 10.

@cernejr 3 жыл бұрын

@@ThePfilip Yes, Mr/Ms Optimality already said that.

@michaelempeigne3519 2 жыл бұрын

How is someone supposed to make a typo with no computer keyboard?

@stmisbehavin662 2 жыл бұрын

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.

@warmpianist 3 жыл бұрын

t = 0 makes w1 become complex number, which is invalid. The range of t is actually [sqrt(75),10].

@red0guy 3 жыл бұрын

The errors are the salt of the channel

@warmpianist 3 жыл бұрын

@John Tse which side? I think you misread sqrt(196-t^2)

@GiornoYoshikage 3 жыл бұрын

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

@kevinmartin7760 3 жыл бұрын

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).

@ManuelRuiz-xi7bt 2 жыл бұрын

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.

@nicolascamargo8339 Жыл бұрын

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

@emiltonklinga3035 3 жыл бұрын

Why don't we just notice at 8:20 that x² - 196 = -75 and then solve for x?

@yoav613 3 жыл бұрын

You can find x very quickly without all of this, but he wanted to show the parameterization with t at the end

@adityaekbote8498 3 жыл бұрын

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)

@user-cr4fc3nj3i 3 жыл бұрын

@@adityaekbote8498 Not really, they don't have to be integer at all.

@theartisticactuary 3 жыл бұрын

Just looking through the comments to see whether this had already been spotted and yes it has 👍🏻

@VaradMahashabde 3 жыл бұрын

It is all for family (of solution rectangles)

@goodplacetostop2973 3 жыл бұрын

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.

@a_llama 3 жыл бұрын

*give

@goodplacetostop2973 3 жыл бұрын

@@a_llama Indeed, fixed. Thanks

@Horinius 2 жыл бұрын

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" 😄

@Bennici 2 жыл бұрын

Hey now. You're OUR entertaining clown posting timestamps.

@jimschneider799 3 жыл бұрын

@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 ].

@kuzuma4523 3 жыл бұрын

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).

@littlefermat 3 жыл бұрын

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 😅)

@alanclarke4646 2 жыл бұрын

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.

@MrMeztar 2 жыл бұрын

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.

@alanclarke4646 2 жыл бұрын

@@MrMeztar looked a whole lot more complicated to me.

@MrMeztar 2 жыл бұрын

@@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.

@hb1338 2 жыл бұрын

@@MrMeztar Ask yourself whether a typical student aged 16 would find simultaneous equations or row reduction easier to understand.

@MrMeztar 2 жыл бұрын

@@hb1338 row reduction once they understand the notation

@tomatrix7525 3 жыл бұрын

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!

@manucitomx 3 жыл бұрын

What a fantastic problem! I love Geometry and Linear Algebra, so this was great. Thank you, professor

@beautyofmath6821 3 жыл бұрын

Well this can be simplified just by relating the equations formed by using pythagoras theorem. But still, a great video!

@noahtawil8793 2 жыл бұрын

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

@qwadratix 2 жыл бұрын

Exactly.

@dalan1999 2 жыл бұрын

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

@beautyofmath6821 2 жыл бұрын

@@dalan1999 Not sure, but I think matrixes can be used to solve system of equations. (Not sure, also a high school student)

@Paul-sj5db 2 жыл бұрын

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.

@williamrutherford553 3 жыл бұрын

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 :)

@macnolds4145 3 жыл бұрын

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).

@wowZhenek 3 жыл бұрын

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

@nikitakipriyanov7260 2 жыл бұрын

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.

@yasinmohammadi8 3 жыл бұрын

That was great.can you please solve some questions of IOI

@Joshua123456791Funny6791 3 жыл бұрын

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

@henk7747 3 жыл бұрын

Nice video. The point is not just to solve the problem but to highlight interesting consequences. This is a highschool competition problem just solving it is boring, but I guess KZfaq commenters are really proud of solving it easily.

@TheSmilodon2 3 жыл бұрын

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.

@georgemaxwell3997 2 жыл бұрын

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.

@alexismiller2349 3 жыл бұрын

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 :)

@Aqua17292 2 жыл бұрын

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.

@ZedaZ80 3 жыл бұрын

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

@asutosh2429 2 жыл бұрын

x² + z² = y² + w² where x y z w are in circular order. xz and yw are opposite to each other

@Bodyknock 3 жыл бұрын

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. 🤷‍♂️

@wernergamper6200 3 жыл бұрын

I tried here: kzfaq.info/get/bejne/kM2VmM-XksvOeo0.html

@michaelact 3 жыл бұрын

@@wernergamper6200 Thanks! Is it possible to keep the lower left corner at (0,0) so it is easier to see P move?

@wernergamper6200 3 жыл бұрын

@@michaelact Since the lower left "radius" always is 5, P moves on a quarter circle. kzfaq.info/get/bejne/nc6JhMl_rdCRqH0.html

@Bodyknock 2 жыл бұрын

@@wernergamper6200 Neat, thanks. 🙂

@sawyerw5715 3 жыл бұрын

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

@jonpress6773 2 жыл бұрын

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.

@OrenLikes 2 жыл бұрын

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

@OrbitTheSun 2 жыл бұрын

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.

@travishayes6037 2 жыл бұрын

Very cool problem.

@zygoloid 2 жыл бұрын

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.

@Johan-yy9pk 3 жыл бұрын

where can i buy that chalkboard

@wyboo2019 Жыл бұрын

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

@jimmyh2137 2 жыл бұрын

How did you do the starting Matrix? 4:18

@andrewbuchanan5342 2 жыл бұрын

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?

@ciberiada01 2 жыл бұрын

My thoughts exactly (for the square).

@rskissack 2 жыл бұрын

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...

@rskissack 2 жыл бұрын

...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!

@lnx0007 2 жыл бұрын

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.

@Grizzly01 3 жыл бұрын

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.

@MichalMarsalek 3 жыл бұрын

This...

@patrickpablo217 3 жыл бұрын

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.)

@wernergamper6200 3 жыл бұрын

kzfaq.info/get/bejne/nc6JhMl_rdCRqH0.html a quarter circle

@patrickpablo217 3 жыл бұрын

@@wernergamper6200 Awesome! Thanks for the great animation!

@danielleza908 2 жыл бұрын

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.

@user-yp2hg6hz8n Жыл бұрын

It follows quickly from the British flag theorem. 10^2+x^2=14^2+5^2, so x^2=121 and x=11.

@andreivila7607 3 жыл бұрын

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.

@cengizerol4095 2 жыл бұрын

Theorem belongs to Mustafa Yağcı '' height theorem'' x^2 - 5^2 = 14^2 - 10^2 used for similar questions

@DavidCorneth 2 жыл бұрын

"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) kzfaq.info/get/bejne/rdaPo8V2yN2UeGw.html

@carrotfacts 2 жыл бұрын

Why orthogonal over perpendicular?

@nuranichandra2177 3 жыл бұрын

Good problem

@general9064 3 жыл бұрын

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....

@ApresSavant 3 жыл бұрын

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.

@engineersspace2024 3 жыл бұрын

How is the first matrix calculated? I can't figure it out

@YOM2_UB 2 жыл бұрын

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 ]

@engineersspace2024 2 жыл бұрын

@@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

@KarlFredrik 3 жыл бұрын

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 😞

@robertowusu8408 2 жыл бұрын

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.

@sldimaf 3 жыл бұрын

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 ))

@olau5478 3 жыл бұрын

is that true?

@TedHopp 3 жыл бұрын

@@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 3 жыл бұрын

@@TedHopp wow, i didnt know that. certainly seems useful in this problem

@TedHopp 3 жыл бұрын

@@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.

@gregy1570 2 жыл бұрын

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..

@ciberiada01 2 жыл бұрын

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

@typha 3 жыл бұрын

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 3 жыл бұрын

But does that work for ALL rectangles.

@typha 3 жыл бұрын

@@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.

@DitDede 3 жыл бұрын

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.

@alpham9877 3 жыл бұрын

Bizdede 1966 üss de var bu soru. Demekki çalışırken dünyayı takip etmek gerekliymiş

@drozfarnyline4940 3 жыл бұрын

well, this is a nice classical problem in Geometry called British Flag Theorem

@Happy_Abe 3 жыл бұрын

t can’t be 0 because then t^2-75 would be negative in the square root

@VadymDrozd 3 жыл бұрын

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

@tamarpeer261 2 жыл бұрын

I know this can be done with Pythagoras, but this looks like Ceva’s theorem

@mithutamang3888 2 жыл бұрын

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!!! 😁😁👍👍

@abdullahyousef3596 2 жыл бұрын

Can anyone please explain to me how did he get the first matrix in 4:19

@abdullahyousef3596 2 жыл бұрын

and also the second matrix if you can, and thanks.

@YOM2_UB 2 жыл бұрын

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.

@abdullahyousef3596 2 жыл бұрын

@@YOM2_UB Oh ok, thanks.

@satyabrataghosh703 3 жыл бұрын

British Flag Theorem

@angelogandolfo4174 3 жыл бұрын

Just after 8:10 (I think!) we have, x^2-196 = -75 and then we could just solve for x easily?? No?

@shohamsen8986 3 жыл бұрын

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.

@VadymDrozd 3 жыл бұрын

Why so complicated?

@lisandro73 3 жыл бұрын

Actually,there is a typo t can be bigger than root of 75

@sidimohamedbenelmalih7133 Жыл бұрын

Using linear combination would be easier

@chrismiller6579 2 жыл бұрын

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.

@NachiketVartak 2 жыл бұрын

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.

@MichalMarsalek 3 жыл бұрын

This is might be the most ridiculous way to solve that problem...

@kumarsaurav8885 3 жыл бұрын

It's just straight up British Flag Theorem.

@mcbeaulieu 3 жыл бұрын

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

@c8h182 3 жыл бұрын

Algebratic geometry xd

@abhishekpatil1063 2 жыл бұрын

It will be x^2= 14^2+5^2 -10^2 = 121 i.e x= 11

@jarikosonen4079 3 жыл бұрын

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

@bobh6728 3 жыл бұрын

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.

@jarikosonen4079 3 жыл бұрын

@@bobh6728 It looks they are independent enough. (Not enough of coupling among them?)

@shubhamsinghal9777 3 жыл бұрын

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!!

@bobh6728 3 жыл бұрын

That is what he did, but just used matrices to solve.

@nimennacnamme6328 3 жыл бұрын

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 ^^'

@hugh081 Жыл бұрын

Solution is 11 by the British Flag Theorem

@user-wi8iq3hn3k 3 жыл бұрын

Это устно решается за минуту, сложили два равенства и вычли третье...

@FootLettuce Жыл бұрын

This is the sickest proof of the British flag Theorem.

@user-vg1qo5gi3l 2 жыл бұрын

We can use british flag theorem

@damyankorena Жыл бұрын

British flag theorem has joined the chat 💀

@HaiNguyen-qx3db 3 жыл бұрын

Oh yeah, a classical geometry problem. But why Michael's answer look so complicated?

@failsmichael2542 3 жыл бұрын

It’s not about solving the problem. It’s about making a point.

@skylardeslypere9909 3 жыл бұрын

It's very elementary linear algebra

@swatiattam8498 3 жыл бұрын

See my solution

@kuzuma4523 3 жыл бұрын

When you want to give an answer to an indeterminate system of equations, you usually use parametric solutions.

@paulstillman4949 2 жыл бұрын

*sees SMT in the thumbnail* *clicks* My disappointment is immeasurable and my day is ruined.

@vinayakpandey1032 3 жыл бұрын

@billyd78 2 жыл бұрын

T=11

@michaelempeigne3519 3 жыл бұрын

this is perfect for the british flag theorem

@ApresSavant 3 жыл бұрын

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.

@grahamwarford5452 2 жыл бұрын

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.