Many Students Failed To Solve This Geometry Problem

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Күн бұрын

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@WahranRai 9 ай бұрын

We have the 2 following equations : (x+1)^2 + (y+1)^2 = 9 (E1) and x*y= 1 (E2) E1 is the equation of circle with center = (-1,-1) and radius 3 E2 is the equation of hyperbola If you plot these 2 curves (with Desmos for example) you will see the 4 points of intersections which are symmetrical with respect to the 1st bisector

@soli9mana-soli4953 9 ай бұрын

I am one of those who fails... I tried with similarity and get xy=1 then went on with pythagorean identity, but getting a 4th power equation...😒 then tried with coordinate geometry calculating the intersections with axis of the generic line passing on point (1,1) setting thei distance equal to 3, but again a 4th power equation...😒 A little better with trigonometry doing: x+1=3*sin(alpha) => X = 3sin(alpha) -1 y+1=3*cos(alpha) => Y = 3cos(alpha) - 1 knowing that xy=1 we have: (3sin(alpha)-1)*(3cos(alpha)-1)= 1 getting 3sin(alpha)cos(alpha) = cos(alpha) + sin(alpha) multiplying all terms by 2 we get: 3*sin(2alpha) =2*(cos(alpha) + sin(alpha)) then doing the square of all terms 9*sin²(2alpha) =4*(cos²(alpha) + sin²(alpha) +2*sin(alpha)cos(alpha)) being cos²(alpha) + sin²(alpha)=1 and 2*sin(alpha)cos(alpha)= sin(2alpha) we get: 9sin²(2alpha) - 4sin(2alpha) - 4 = 0 solving the equation we get: sin2(alpha)= (2+2√ 10)/9 the positive root then doing a not so elegant arcsin we get alpha = 33,83 and with this value we can find x and y...

@RAG981 9 ай бұрын

I should think that "many students failed to solve this" is a vast understatement. Using x+1/x as the variable is very clever.

@triharders2456 9 ай бұрын

yh that was low key beautiful. i had to pause and write down the equation to clock what just happened

@shmuelzehavi4940 8 ай бұрын

It's indeed a very nice and clever way to solve this problem, but it's not the single one.

@bullerheden 2 ай бұрын

This is the famous "ladder and box" problem. Very good explanation and very clever substitution t= x + 1/x. There will be two symmetrical solutions because the ladder with the lenght 3 can be placed low or high. Approximate solutions x= 3/2 , y= 2/3 and vice versa.

@rick57hart 9 ай бұрын

35 years ago i found 2 much shorter ways: Method 1. 1/sin alpha + 1/cos alpha = 3 alpha = 33,83045° x = 0,670212 y = 1,492066 I prefer this method, because it is fast and there is almost no risk to make mistakes. The mathematicians in their ivory towers don't like it, because you have to try out numbers. 😇 Method 2. (x + 1)^2 + (1/x +1)^2 = 3^2 You can transform x^2 + (1/x) ^2 + 2x + 2/x + 2 into (x + 1/x + 1)^2 - 1 so you get (x + 1/x +1)^2 = 3^2 + 1 Now the sqrt on both sides We get x + 1/x +1 = sqrt10 x + 1/x = (sqrt10) - 1 Let us call the right side "b" x + 1/x = b multiply both sides with x x(x + 1/x) = bx x^2 - bx + 1 = 0 If i do (x - b/2)^2, i don't get exactly the same, because i get (b/2)^2 to much but 1 is missing. But when i write: (x - b/2)^2 - (b/2)^2 + 1 = 0, than it is right again. (x-b/2)^2 = {(b/2)^2} - 1 Now sqrt on both sides x - b/2 = sqrt {(b/2)^2} - 1 x = b/2 + or - sqrt {(b/ 2)^2} - 1 Remember: b = (sqrt 10) - 1 = 2,162278 b/2 = 1,081139 sqrt{(b/2)^2 - 1} = 0,410927 x1 = 1,081139 + 0,410927 = 1,492066 x2 = 1,081139 - 0,410927 = 0,670212 alpha = 33,83045° From here it should be easy 😜

@jimleahy3858 8 ай бұрын

Your answer is only an approximation.Not good enough.

@rick57hart 8 ай бұрын

@@jimleahy3858 I have more decimal places. I just didn't wrote all of them. 😜

@rick57hart 8 ай бұрын

I have that much decimal places of the angle, that the display shows just "3". 😇

@daddykhalil909 9 ай бұрын

Very good and detailed explanation As much as it seemed easy in the beginning as much as it came out to be very difficult I enjoyed your efforts Thank you

@shmuelzehavi4940 9 ай бұрын

Nice and elegant solution. I have a different approach. The original system of equations is: (x + 1)^2 + (y + 1)^2 = 3^2 (1) xy = 1 (2) Eq. (1) becomes: x^2 + y^2 + 2 (x + y) + 2 = 9 which, using eq. (2) becomes: (x + y)^2 + 2 (x + y) - 9 = 0 (3) We obtained a quadratic equation in terms of: (x + y). This equation ha 2 roots: -1 ± √10 . The single root that has a geometric meaning is the positive valued one and therefore we get: x + y = √10 - 1 (4) However, (x - y)^2 = (x + y)^2 - 4xy = 11 - 2√10 - 4 = 7 - 2√10 = (√5 - √2)^2 Therefore we obtain: x - y = ± (√5 - √2) The original system of equations defines a symmetric relation between (x,y) . Therefore, without loss of generality, we assume that: x ≥ y and get: x - y = √5 - √2 (5) We solve now the linear system of equations (4), (5) for x , y by adding and subtracting the equations, which gives us: x = 1/2 ((√10 - 1) + (√5 - √2)) = 1/2 (√5 - 1) (√2 + 1) y = 1/2 ((√10 - 1) - (√5 - √2)) = 1/2 (√5 + 1) (√2 - 1) A second pair of solutions is obtained by interchanging values between x ,y . The complete set of solutions is therefore: { (x , y) } = { ( (√5-1)(√2+1)/2 , (√5+1)(√2-1)/2 ) , ( (√5+1)(√2-1)/2 , (√5-1)(√2+1)/2 ) }

@Irtsak 6 ай бұрын

*@shmuelzehavi4940* *Nice Solution !* Another aproach to your solution using Vieta. (x + y)² + 2 (x + y) - 9 = 0 Let t=x+y t²+2t-9=0 => t=-1±√10 So ( x+y=-1+√10 Λ xy=1 ) system 1 or ( x+y=-1-√10 Λ xy=1 ) system 2 System 1 : by Vieta S=-1+√10 and P=1 So x,y are the routes of the equation w²-Sw+P=0 => w²-(-1+√10)w+1=0 ……… Δ=⋯=(√5-√2)²>0 And w=(√5-1)(√2+1)/2 or w=(√5+1)(√2-1)/2 so (x=(√5-1)(√2+1)/2 Λ y=(√5+1)(√2-1)/2 ) or ( x=(√5+1)(√2-1)/2 Λ y=(√5-1)(√2+1)/2 ) I work accordingly with system 2…….. Have a nice day

@pk2712 9 ай бұрын

Nice . I solved it before I watched the video , and came up with the fact that xy=1 . I loved the symmetry of your final solution set .

@shadrana1 10 күн бұрын

Triangles ADE and EFC are similar, Therefore,AD/DE=EF/FC Y/1=1/X Y=1/X Use Pythagoras on triangle ABC,let X=x and Y=y AB^2+BC^2=AC^2 (1+1/x)^2)+(1+x)^2=3^2=9 (x+1)^2/x^2+x^2+2x=8 (x+1)^2+x^4+2x^3=8x^2 x^2+2x+1+x^4+2x^3-8x^2=0 x^4+2x^3-7x^2+2x+1=0 and got the four roots from Wolfram Alpha. x1= 1/2(sqrt2-1)(sqrt5+1) x2= 1/2(sqrt2+1)(sqrt5-1) x1 and x2 are the reciprocal of each other X3 and x4 are both negative therefore they are rejected. Since y=1/x, y1=x2 and y2=x1. (x,y)=((sqrt2+1)(sqrt5-1)/2/,(sqrt2-1)(sqrt5+1)/2) and ((sqrt2-1)(sqr5+1)/2,(sqrt2+1)(sqrt5-1)/2) is the solution set for this puzzle.

@kimba381 9 ай бұрын

Nice. two points: You didn't need to appeal to Trig., similar triangles would have gotten you y = 1/x more simply. The problem is symmetrical about x, y. Once you have found the pair of x values, the y values must be the same, in the opposite order.

@arbenkellici3808 9 ай бұрын

Masterpiece professor! Keep going like this!

@RondoCarletti 4 ай бұрын

x =1 and y =1 (h²=pxq) - EFC also for ADE.

@imim3095 6 ай бұрын

I believe there is very simple solution The triangles EFC and ADE are congruent so X=Y

@murdock5537 9 ай бұрын

Awesome, many thanks, Sir!

@cybercat119 9 ай бұрын

Piece of art. Thanks!

@dantallman5345 9 ай бұрын

Nice. This was a tough, appreciate the thoroughness in your solution. I got xy=1 easily but kept ending up with a quartic equation. I tried various variable substitutions but missed t=x^2+1/x^2, hence nothing resolved cleanly.

@gregorygarber3073 9 ай бұрын

I think, x=sqrt(5) - 1, y=1. Use the fact that all 3 vеrtices of the triangle are equidistant from the midpoint of the hуpotenuse. Hence the shown triangle is similar to the small triangle whose vertices are 1) the midpoint of the hypotenuse, 2) the vertex at the right angle of the shown triangle, and 3) the midpoint of the vertical catet. That gives you y+1=2

@skwest 9 ай бұрын

I'm lost... what is the point of the exercise between 17:13 and 18:11? The result of this portion appears to show that: x² = 1.

@davidellis1079 9 ай бұрын

We have 2 solutions for x; let's call them x1 and x2. What this part of the video shows is that x2 is the reciprocal x1, by showing that their product is equal to 1. We already know that y is the reciprocal of x, so now we can see that the solutions are (x,y) = (x1,x2) and (x,y) = (x2,x1) . I hope this helps :)

@skwest 9 ай бұрын

@@davidellis1079 It did. Very helpful. Thank you. - s.west

@davidellis1079 9 ай бұрын

@@skwest 👍