Can you calculate the Green circle Radius? | (Rectangle) | #math #maths | #geometry

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Can you calculate the Green circle Radius? | (Rectangle) | #math #maths | #geometry

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[LawtonDigital]

Thank you! These puzzles are so much more fun than Sudoku.

[AzouzNacir]

Let a be the distance between the points of contact of the circle with radius 6 and the circle with radius r with the length of the rectangle, then we get the equations (6+r)²=a²+(6-r)², (3+r)²=(6√6-a)²+(9-r)². Solving these two equations, we geta=4√6, r=4.

[alanthayer]

NICE multiple break downs ! Like these xtra challenging types sir ! Thanku

[jamestalbott]

Nice! Using our constant friend, the Pythagorean theorem, redefining two variables into a common variable and solving. Great problem for a Saturday.

[zupitoxyt]

What an incredible job 👏. Nice question sir 😊

[alexundre]

Bom dia Mestre
Obrigado pela aula
Desejo-lhe um sábado abençoado

[cyruschang]

6√6 = √[(6 + r)^2 - (6 - r)^2] + √[(3 + r)^2 - (9 - r)^2] = √(24r)+ √(24r - 72)
3 = √(r)+ √(r - 3)
9 - 6√(r) + r = r - 3
r = 4

[kimchee]

It's hard to see how the manipulation ends up correct. You could start in a different paths and ended up nowhere multiple times like many problems. It's not the math manipulation/mechanics. It's the thought process and visualization that are more important driving the math manipulation. If you don't have a clear thought process you could try this and try that until you ended up with the correct answer. Excellent video once again.

[uwelinzbauer]

Hallo!
I was afraid that we would have to deal with biquadratic or quartic equations and expressions longer than a sheet of paper, but all the difficult things canceled out.
Thanks for sharing this nice geometry question!
Best wishes ❤️

[wackojacko]

@ 10:53 the intense ping pong of math begins. First we square...then take square root to find x ... then square root to find y ... then square to find r ... I'm having a coffee and donut this morning and can't the difference between 'em. I better go back to bed. 😊

[angeluomo]

I was completely with you until 11:13, but then you went down a rabbit hole of complicated algebra and substitution. How about this? y^2=24r and x^2=24r-72. How elegant! So x^2=y^2-72, and y=6*sqrt6-x. Squaring y gives you y^2=216-12*sqrt6*x+x^2. Plug that into the equation and simplify: x=12/sqrt6. Stick this value for x into your equation 2, and you get r=4. Much simpler. The value of y is conveniently 24/sqrt6.

[rey-dqnx]

x=√((3+r)²-(9-r)²)
x=√(9+6r+r²-81+18r-r²)
x=√(24r-72)
y=√(6+r)²-(6-r)²
y=√36+12r+r²-36+12r-r²
y=√24r
9+6√6=3+6+√(24r-72)+√24r
6√6=√(24r-72)+√24r
6√6=2√6√(r-3)+2√6√r
3=√(r-3)+√r
(3-√r)² = (√(r-3)²
9-6√r+r=r-3
12=6√r
r=4

[marcgriselhubert]

We use an orthonormal center K and first axis (KB), we have Q(0; 3) O(6.sqrt(6); 6) P(a; 12-r) with r the radius of the green circle.
*We have VectorQP(a, 9- r) and QP^2 = (a^2) + (81-18.r+(r^2)), but also QP^2 = (r+3)^2 = (r^2)+ 6.r +9, so (a^2) +81 -18.r +(r^2) = (r^2) + 6.r +9;
We simplify and get: (a^2) -24.r + 72 = 0 (Eq1)
*We also have VectorOP(a -6.sqrt(6); 6 -r) and OP^2 = (a^2) -12.sqrt(6).a + 216 +36 -12.r +(r^2). = (a^2) -12.sqrt(6).a - 12.r +(r^2) +252, but also OP^2 = (r +6)^2
= (r^2) + 12.r +36, so (a^2) -12.sqrt(6).a -12.r +(r^2) + 252 = (r^2) +12.r + 36. We simplify and get: (a^2) -12.sqrt(6).a -24.r + 216 = 0 (Eq2)
*In (Eq2) we replace (a^2) -24.r +72 by 0 (which is given y (Eq1)) and get: -12.sqrt(6).a + 144 = 0 and then a = 144/(12.sqrt(6)) = 12/sqrt(6)
We replace a by this value in (Eq1) and get: 24.r = ((12/sqrt(6))^2 - 72 = (144/6) - 72 = 96, so finally r = 96/24 = 4.

[johnbrennan]

Another way to do it is to let x=(6root6)-y etc.

[texitaliano]

Considerando i due trapezi JQPG e GPOE aventi basi rispettivamente JG=a GE=b.
La distanza JE=a+b=9+6√(6)-3-6=6√6
La distanza QP=3+r
La distanza PO=6+r

Utilizzando pitagora abbiamo che:
√((6+r)^2-(6-r)^2)=b
√((3+r)^2-(9-r)^2)=a


√(24r)+√(24r-72)=6√6
√(6r)+√(6r-18)=3√6
√(r)+√(r-3)=3
√(r)=-√(r-3)+3
√(r)=3-√(r-3)
r=9+6√(r-3)+r-3
r=6-6√(r-3)+r
0=6-6√(r-3)
6√(r-3)+r=6
√(r-3)=1
r-3=1
r=4

[unknownidentity]

Let's find the radius:
.
..
...
....


First of all we calculate the second side length of the rectangle:

AD = BC = 2*R(big white circle) = 2*6 = 12

Now let's add points R and S such that OPR and PQS are right triangles. In this case P, R and S are located on the same line. By applying the Pythagorean theorem we obtain:

OP² = PR² + OR²
PQ² = PS² + QS²

The green circle has exactly one point of intersection with the big white circle and it has exactly one point of intersection with the small white circle. So we obtain:

OP = R(green circle) + R(big white circle) = R + 6
PQ = R(green circle) + R(small white circle) = R + 3

From the given diagram we can conclude:

PR = AD − R − 6 = 12 − R − 6 = 6 − R
PS = AD − R − 3 = 12 − R − 3 = 9 − R

OR + QS = AB − 3 − 6 = 9 + 6√6 − 3 − 6 = 6√6 ⇒ QS = (6√6 − OR)

So we finally obtain:

OP² = PR² + OR²
PQ² = PS² + QS²

(R + 6)² = (6 − R)² + OR²
(R + 3)² = (9 − R)² + (6√6 − OR)²

R² + 12R + 36 = 36 − 12R + R² + OR²
R² + 6R + 9 = 81 − 18R + R² + 216 − (12√6)*OR + OR²

24R = OR²
24R = 288 − (12√6)*OR + OR²

OR² = 288 − (12√6)*OR + OR²
(12√6)*OR = 288
⇒ OR = 288/(12√6) = 24/√6 = 4*6/√6 = 4√6
⇒ OR² = (4√6)² = 96

24R = OR² = 96
⇒ R = 96/24 = 4

Best regards from Germany

[LuisdeBritoCamacho]

MY RESOLUTION PROPOSAL :


01) AD = BC = (2 * 6) = 12 lin un

02) Draw 2 Horizontal Lines and 2 Vertical Lines, passing by Point Q and Point O.

03) The Horizontal Distance between these two Vertical Lines is equal to (6sqrt(6)) lin un; (3 + 6) - (9 + 6sqrt(6)) = 9 - 9 + 6sqrt(6) = 6sqrt(6)

04) Now I have 2 Different Right Triangles.

05) Let the Radius of the Green Circle equal to X lin un.

06) Divide 6sqrt(6) in two Different Parts : 1) Y and 2) (6sqrt(6) - Y)

07) Using the Pythagorean Theorem I have :

08) (X + 3)^2 = (9 - X)^2 + Y^2. Note that (9 - X) = (12 - 3 - X)

09) (6 + X)^2 = (6 - X) + (6sqrt(6) - Y)^2

10) Now I have a System of 2 Nonlinear Equations with 2 Unknows

11) Solutions :

12) X = 4 lin un and Y = 2sqrt(6) lin un


Therefore,


MY BEST ANSWER IS :


The Radius of the Green Circle is equal to 4 Linear Units.