Mickey's Super-ears
Two lines intersect each other forming a right angle. A unit circle (radius 1) is tangent to both of these lines. Two equal size circles with radius
R
>
1
are:
(1) tangent to this circle,
(2) tangent to each other, and
(3) each tangent to one of the lines,
as shown in the figure above.
Calculate R to two decimal places.
The answer is 1.39.
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2 solutions
Cant you use Pythagorean theorem by saying (r+1)^2 + (r+1)^2 = (2r)^2
I agree with Brandon Lopez, 2R = sqrt((R+1)^2+(R+1)^2) = sqrt(2(R+1)^2) = sqrt(2) sqrt((R+1)^2) = sqrt(2) (R+1) = sqrt(2)R + sqrt(2). 2R - sqrt(2)R = sqrt(2). R(2-sqrt(2)) = sqrt(2). R = (sqrt(2)) / (2 - sqrt(2)). R = 2.41
Nice question and solution. I suppose "the other" solution for R would be for the case where the two identical circles are situated within the region between the radius 1 circle and the two axes.
The distance between origin and touching point of the both larger circles is:
2 + ( R + 1 ) 2 − R 2
Otherwise this distance is given by:
1 + ( R + 1 ) 2 − ( R − 1 ) 2
Equalization delivers the result!
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Would you please explain more clearly?
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First case deals with the distance as a sum of distance between (0, 0) and midpoint (1, 1) of the smaller circle and distance between (1, 1) and the touching point of the two larger circles.
Second case treats this distance via 2 rectangular triangles clamped by the origin, the touching point of the larger right positioned circle with the x axis, the midpoint of the larger right positioned circle and the touching point of the two larger circles.
One quickly realizes that the x coordinate of the larger right positioned circle is equal the distance between origin and touching point of the two larger circles.
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@Andreas Wendler – In case 1 how did you get the second part , that is the distance between (1,1) and the distance between the touching points of two points ?? And please also explain case 2 . Sorry for troubling you . Thank you.
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@Chirayu Bhardwaj – Think on Pythagoras' theorem and that tangent and radius are perpendicular for touching points in each case!
We could also represent the distance as ( 1 + 2 ) R .
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How is that determined?? Wrong!
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@Andreas Wendler – It is correct, Andreas. The equation ( 1 + 2 ) R = 2 + ( R + 1 ) 2 − R 2 yields the desired solution.
My value for the distance is obtained by forming a triangle with vertices (i) the origin, (ii) the point of tangency of the two identical circles, (call this point ( a , a ) ), and (iii) the point where the perpendicular bisector to y = x at ( a , a ) intersects the x -axis. This is an isosceles right triangle, and so vertex (iii) has coordinates ( 2 a , 0 ) . The leg from ( a , a ) to ( 2 a , 0 ) can be broken into two segments, one of length R from ( a , a ) to the center of the lower of the two identical circles, and another of length 2 R from this center to the point ( 2 a , 0 ) . As our right triangle is isosceles we can then say that the distance you indicate is ( 1 + 2 ) R as well.
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@Brian Charlesworth – Thank you! I see.
And so we realize: There's more than one way to skin a cat.
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@Andreas Wendler – Haha Yes, indeed. :) This problem in particular seems to lend itself to a multitude of different approaches.
Great way of thinking. Up voted.
What I find is 2 ( ( R + 1 ) 2 − ( R − 1 ) 2 − ( R − 1 ) ) = 2 R , which can be simplified into 2 R − R + 1 = 2 R . The quadratic formula gives R ≈ 1 . 1 7 9 6 (the other value is R ≈ − 0 . 3 5 1 2 , which is smaller than zero thus not possible). Squaring that yields R ≈ 1 . 3 9 1 , supporting Marta Reece's answer.
And the center also lays on the parabola y=(x-1)^2/4
Because of symmetry the left big circle radius R, will have its center on a line that makes 45/2=22.5 degrees with + x-axis. So its center will be at
t
a
n
2
2
.
5
R
distance along x-axis from origin.
The distance of this center and center of small circle along x-axis is
(
R
+
1
)
2
−
(
R
−
1
)
2
)
=
2
∗
R
.
Pythagoras Theorem.
But the center of the small circle is 1 away from origin on x-axis.
⟹
t
a
n
2
2
.
5
R
=
1
+
2
∗
R
.
Solving so that R> 1, we get R=1.3914.
The two straight lines can be thought of as coordinate axes, in which case the equation of the original circle is
( x − 1 ) 2 + ( y − 1 ) 2 = 1
The centers of the two circles radius R which are tangent to this circle must lay on the circle described by
( x − 1 ) 2 + ( y − 1 ) 2 = ( 1 + R ) 2
The lower one of these two circles is tangent to the x-axis, therefore the y coordinate of its center is equal R.
( x − 1 ) 2 + ( R − 1 ) 2 = ( 1 + R ) 2
And since both circles are equal in size, they will also be tangent to the line
y = x
The center of the lower one will therefore have to be on the line
y = R = x × t a n ( 2 2 . 5 )
Substituting this into the third one of the equations above will give us a quadratic equation for R. One of its solutions is
R = 1 . 3 9 1 4
The other, R = 0 . 1 2 3 3 1 , is smaller than 1 and therefore is not an answer to the given problem. What it describes is the radius of one of two circles squeezed in between the unit circle and the two lines.