Nice catch. I thought there was something off, since 15/16 is less than 1, and b is clearly larger than a, so b/a = 15/16 couldn't be right. Shows how important it is for everybody to check their work.
At 16:06 , you figured that x+h =15/2 . At 16:46 , you made an error when you said x+h=15/4 . You should have continued x+h=b+b=2b=15/2 ===> b=15/4 ===> b/a=(15/4)/2=15/8 .
Extend UN until it intersects the bottom part of the circle and label the intersection as point W. UW and AC are intersecting chords, intersecting at N, By the intersecting chords theorem, (AN)(CN) = (UN)(NW). AN = CN = 6.5, UN = x + h = 15/4, therefore NW = 169/15 = 11 4/15 = 11.267 approx. However, UN = 15/4 = 3.75. So, point O must be on the longer NW segment and not the UN segment. Furthermore, UW is a diameter and the radius x is half the diameter, or (1/2)(UW) = (1/2)(11.267 + 3.75) = 7.508 approx., so h = 3.75 - 7.507 = -3.758. However, h is length and can not be negative. The solution needs to be redone with point O on the NW segment. I believe that the answer will be the same. The equations that Math Booster generated will produce the correct answer for b/a when h is allowed to take on a negative value. Note that AB is tangent to the circle centered at P. At 8:50, the point of tangency is labelled E. There is also a point of tangency between the circle centered at P and the large circle centered at O. This point is labelled at 11:10 as S. These are 2 different points and, because they are not the same point, they make the problem very challenging to solve!
YOU MADE A MISTAKE .... X+h =15/2 and not 15/8. So b/a = 15/8 and not 15/16. Because b > a in the figure that means b/a must > 1. So the right answer is b/a = 15/8
Il ne résulte pas de la figure que le point de tangence (N dans le corrigé) du grand cercle de rayon b avec l’hypothénuse (AC dans le corrigé) du triangle rectangle soit le milieu de cette hypothénuse.
Hang on... You never told us that circle 'a' was tangent to the big circle, only that it was tangent to the short leg of triangle ABC... What gives? All you said that we had 2 circles 'inside' a big circle.
I supposed b might be 13/4 because we could draw the 5, 12,13 triangle below the diameter of a circle of radius 13/2 and put the middle sized circle in the top half and the rightangled triangle would fit tidily just in the lower half of the largest circle and B would not need to stick out , but b does need to be bigger to get two distinct tangents to the circle centre P so close. It turned out that the 15 was the 13 + the 2. The concentration of thinking out how this all fits is mind-blowing. ⚛❇🌀🌠 (The spirographs have lit up with christmas lights! I wonder what I ate for lunch!) x+h =( x.x-h.h) / (x-h) = 49/4 / (x-h) = 15/2 so x-h =(49/4) / (15/2) = 49/30 2h= 225/30 -49/30 = 176/30 h=44/15 adding x+h +x-h =2x =225/30 + 49/30 = 274/30 x= 137/30, radius of largest circle between 4 and 5 leaves us wondering what land of chalk drawings we have arrived at???
I was watching the video but then saw that OP and PS are collinear and perpendicular in AB so that would make point O be situated inside the ABC triangle so according to original drawing h needs to be negative. I am wondering if you got that result. This problem isn't simple
That was horrible, not 'very nice' 🤣 The diagram is so out of scale it's unreal. I constructed it in GeoGebra, and it turns out the point O lies below line BC, with △ABC being fully contained within the large circle (i.e. B is a point inside the large circle). I would really need convincing that circle 'a' is actually tangent to the large circle, because I'm not sure that is possible.
@@MathBooster I appreciate that, and I'm all for diagrams not being to scale, so one is forced to concentrate on the mathematics. However, I don't think that the the diagram as shown is possible, to scale or not. In GeoGebra, I constructed a 5-12-13 triangle, then the incircle of radius a = 2 (all very straightforward). I then added a perpendicular bisector of the hypotenuse AC, and along this line (denoted as UN in your workings) constructed a circle of radius b = 15/8 (= 1.875), tangent to AC at point N. Then using the 'Circle through 3 points' function, constructed a large circle passing through A, U and C. In no way is circle 'a' anywhere near to being tangent to the large circle, i.e. point S does not exist. I would invite you to construct the diagram to scale yourself, and see if your findings agree with mine.
@@MathBooster Ah OK, that makes much more sense. With the correct value for b as you state above, then the diagram becomes valid, and point S indeed exists. Still not a 'very nice' problem IMO! 🤣
@@Grizzly01-vr4pn For me, the 'not very nice' part is that we aren't told in the setup that circle 'a' is tangent to the big circle, only that circle 'a' and circle 'b' are "inside" the big circle, and that circle 'a' is the incircle for triangle ABC. Can one prove, from what is given, that circle 'a' is tangent to the big circle at S? Given the fact that you (Grizzly) actually performed the construction, and, I take it, circle 'a' appears to be tangent to the big circle, there must be a way to prove it... yes?
Prof, you have found that h+x = 15/2 !! So 2b = h+x = 15/2 and b = 15/4 with final result b/a = 15/8
Nice catch. I thought there was something off, since 15/16 is less than 1, and b is clearly larger than a, so b/a = 15/16 couldn't be right.
Shows how important it is for everybody to check their work.
Precisely.
In trangle ONC - OC=OS=R , CN=13/2 , ON=h=2b-R , by Pythagoras theorem - OC*2=ON*2+CN*2 , R*2=(2b-R)*2+(13/2)*2 , 4b*2-4Rb+169/4=0 (1) . In trangle POT - OP=R-2 , PT=7/2 , OT=2+h , h=ON=2b-R , by Pythagoras theorem OP*2=OT*2+PT*2 , (R-2)*2=(2+2b-R)*2+(7/2)*2 - 4b*2+8b-4Rb+49/4=0 (2) , from eguation (2) subtract (1) - 8b-120/4=0 , b=30/8=15/4 , b/а=15/(4х2)=15/8 .
At 16:06 , you figured that x+h =15/2 . At 16:46 , you made an error when you said x+h=15/4 . You should have continued x+h=b+b=2b=15/2 ===> b=15/4 ===> b/a=(15/4)/2=15/8 .
Nice, but there are some concerns. Simple math shows that there must be a mistake:
AC = 13 → AF = AO + FO = 2r >13; UN = 2b = r + n → n > 0 →
r < 2b → b > r/2 > 13/4 > 15/8
See details below.
φ = 30° → sin(3φ) = 1; ∆ ABC → AB = AH + SH + BS = 3/2 + 3/2 + 2
SP = JP = MP = 2; BC = BJ + EJ + EC = 2 + (10 - t) + t
AC = AN + CN; sin(CNE) = 1; AF = AO + FO = r + r
UG = UQ + QO + NO + NE + GE = b + (b - n) + n + k + GE; sin(ANU) = 1 = sin(ABC)
BCS = θ → sin(θ) = √37/37 → cos(θ) = 6√37/37
BCA = δ → sin(δ) = 5/13 → cos(δ) = 12/13 → SCA = δ - θ →
sin(δ - θ) = sin(δ)cos(θ) - sin(θ)cos(δ) = (18/13)(√37/37) → cos(δ - θ) = (77/13)(√37/37)
AS = AH + SH = 3/2 + 3/2 → ∆ AHO → HOA = δ - θ → OAS = 3φ - (δ - θ)
∆ ANO → sin(ANO) = 1; NAS = 3φ - δ → OAN = 3φ - (δ - θ) - (3φ - δ) = θ →
cos(δ) = 12/13 = 13/2t → t = 169/24 → sin(δ) = 5/13 = k/t → k = 65/24
tan(θ) = 1/6 = 2n/13 → n = 13/12 → sin(θ) = √37/37 = n/r → r = (13√37)/12 →
2b = r + n = (13/12)(√37 + 1) → b = (13/24)(√37 + 1) > 13/4 > 15/8
Please check again. It's a joke if b/a = 15/16 < 1
Extend UN until it intersects the bottom part of the circle and label the intersection as point W. UW and AC are intersecting chords, intersecting at N, By the intersecting chords theorem, (AN)(CN) = (UN)(NW). AN = CN = 6.5, UN = x + h = 15/4, therefore NW = 169/15 = 11 4/15 = 11.267 approx. However, UN = 15/4 = 3.75. So, point O must be on the longer NW segment and not the UN segment. Furthermore, UW is a diameter and the radius x is half the diameter, or (1/2)(UW) = (1/2)(11.267 + 3.75) = 7.508 approx., so h = 3.75 - 7.507 = -3.758. However, h is length and can not be negative. The solution needs to be redone with point O on the NW segment. I believe that the answer will be the same. The equations that Math Booster generated will produce the correct answer for b/a when h is allowed to take on a negative value.
Note that AB is tangent to the circle centered at P. At 8:50, the point of tangency is labelled E. There is also a point of tangency between the circle centered at P and the large circle centered at O. This point is labelled at 11:10 as S. These are 2 different points and, because they are not the same point, they make the problem very challenging to solve!
The problem is that he figures h+x=15/2 ; then , he mistakenly says x+h=15/4. The first figure is the correct one.
Amazing problem. But, in triangle ONC, NC is 6.5, and the instructor calculates h (side ON) + x (hypotenuse) to equal 3.75. Clarification, please.
I think you are wrong in assuming the small circle touches the large circle at S. It touches AB at 2 from B, but that is not S
YOU MADE A MISTAKE ....
X+h =15/2 and not 15/8. So b/a = 15/8 and not 15/16. Because b > a in the figure that means b/a must > 1. So the right answer is b/a = 15/8
При планировании атаки на "Звезду Смерти" главное - всё правильно рассчитать заранее!☝😁
Il ne résulte pas de la figure que le point de tangence (N dans le corrigé) du grand cercle de rayon b avec l’hypothénuse (AC dans le corrigé) du triangle rectangle soit le milieu de cette hypothénuse.
Hang on... You never told us that circle 'a' was tangent to the big circle, only that it was tangent to the short leg of triangle ABC... What gives? All you said that we had 2 circles 'inside' a big circle.
I supposed b might be 13/4 because we could draw the 5, 12,13 triangle below the diameter of a circle of radius 13/2 and put the middle sized circle in the top half
and the rightangled triangle would fit tidily just in the lower half of the largest circle and B would not need to stick out , but b does need to be bigger to get two distinct tangents to the circle centre P so close. It turned out that the 15 was the 13 + the 2. The concentration of thinking out how this all fits is mind-blowing.
⚛❇🌀🌠
(The spirographs have lit up with christmas lights! I wonder what I ate for lunch!)
x+h =( x.x-h.h) / (x-h) = 49/4 / (x-h) = 15/2 so x-h =(49/4) / (15/2) = 49/30 2h= 225/30 -49/30 = 176/30 h=44/15
adding x+h +x-h =2x =225/30 + 49/30 = 274/30 x= 137/30, radius of largest circle between 4 and 5 leaves us wondering what land of chalk drawings we have arrived at???
B=(0; 0)
A=(0; 5)
C=(12; 0)
P=(2; 2)
O=(x; y)
y^2+(x-12)^2=R^2
x^2+(y-5)^2=R^2
(x-2)^2+(y-2)^2=(R-2)^2
x^2+y^2>R^2
Решу, когда выйду на пенсию😂.
Respected Sir, Good evening... Some part of the triangle is in the outer part of the circle. So, Can we say the triangle is an incircle triangle?
Incircle of a triangle means circle inside a triangle tangent to it's sides. So the circle of radius a is incircle of the triangle.
I was watching the video but then saw that OP and PS are collinear and perpendicular in AB so that would make point O be situated inside the ABC triangle so according to original drawing h needs to be negative. I am wondering if you got that result.
This problem isn't simple
In no way can circle p be tangent to line AB and the big circle at the same point s.
That was horrible, not 'very nice' 🤣
The diagram is so out of scale it's unreal.
I constructed it in GeoGebra, and it turns out the point O lies below line BC, with △ABC being fully contained within the large circle (i.e. B is a point inside the large circle).
I would really need convincing that circle 'a' is actually tangent to the large circle, because I'm not sure that is possible.
The same diagram came in Russian Math Olympiad 2006. Note that in geometry problems, diagrams are only for reference (they may not be exact)
@@MathBooster I appreciate that, and I'm all for diagrams not being to scale, so one is forced to concentrate on the mathematics.
However, I don't think that the the diagram as shown is possible, to scale or not.
In GeoGebra, I constructed a 5-12-13 triangle, then the incircle of radius a = 2 (all very straightforward).
I then added a perpendicular bisector of the hypotenuse AC, and along this line (denoted as UN in your workings) constructed a circle of radius b = 15/8 (= 1.875), tangent to AC at point N.
Then using the 'Circle through 3 points' function, constructed a large circle passing through A, U and C.
In no way is circle 'a' anywhere near to being tangent to the large circle, i.e. point S does not exist.
I would invite you to construct the diagram to scale yourself, and see if your findings agree with mine.
b is actually 15/4 not 15/8
At 16:12 i have written x+h = 15/2 but at 16:47, by mistake i took it 15/4. Sorry for it.
@@MathBooster Ah OK, that makes much more sense.
With the correct value for b as you state above, then the diagram becomes valid, and point S indeed exists.
Still not a 'very nice' problem IMO! 🤣
@@Grizzly01-vr4pn
For me, the 'not very nice' part is that we aren't told in the setup that circle 'a' is tangent to the big circle, only that circle 'a' and circle 'b' are "inside" the big circle, and that circle 'a' is the incircle for triangle ABC.
Can one prove, from what is given, that circle 'a' is tangent to the big circle at S? Given the fact that you (Grizzly) actually performed the construction, and, I take it, circle 'a' appears to be tangent to the big circle, there must be a way to prove it... yes?
Ho calcolato a=2, facilmente...poi pero non riesco a disegnare bene la figura
I got b= 4
asnwer=15cm isit
asnwer=15/8 cm isit
Está errado
A resposta é: 1,857
@@GilmarJosédoNascimentoJosé 1.875??