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These problems were extremely difficult and I could not solve them. Help is appreciated.

 

1. In the diagram below, \(\overline{PQ}\) is tangent at \(P\) to the circle with center \(O\), point \(S\) is inside the circle, and \(\overline{QS}\) intersects the circle at \(R\). If \(QR = RS = 3\)\(OS = 2\), and \(PQ = 6\), then find the radius of the circle.

 

(Here's an image)

 

2. In right triangle \(ABC\), the length of side \(\overline{AC}\) is \(8\), the length of side \(\overline{BC}\) is \(6\), and \(\angle C = 90^\circ.\) The circumcircle of triangle \(ABC\) is drawn. The angle bisector of \(\angle ACB\) meets the circumcircle at point \(M.\) Find the length \(CM.\)

 

 

 

 

Thanks a lot!

 Apr 8, 2020
 #1
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Problem #2:     If you are allowed to use coordinate geometry:

 

Since triangle(ACB) is a right triangle, the center of the circumcenter is the midpoint of the hypotenuse.

Place the diagram on a coordinate plane with the midpoint of the hypotenuse the origin.

Then point C will be the point (-4, -3).

The equation of the bisector of angle(C) has a slope of 1.

This line has an equation of y + 3  =  1(x + 4)   --->   y  =  x + 1.

 

The circumcenter has its center at (0, 0) and a radius of 5   --->   x2 + y2  =  25.

 

Combining these two equations will give us the coordinates of point M.

x2 + y2  =  25     and     y  =  x + 1     --->     x2 + (x + 1)2  =  25     

                                                          --->     x2 + x2 + 2x + 1  =  25

                                                          --->     2x2 + 2x - 24  =  0

                                                          --->     x2 + x - 12  =  0

                                                          --->     (x - 4)(x + 3)  =  0

Either  x = 4  or  x = -3  (impossible)

Since  x = 4, by substitution,   x = 3   --->   M  =  (3, 4)

 

Using the distance formula to find the distance from C(-4, -3) to M(3, 4),

we get the distance of CM to be:  7·sqrt(2).

 Apr 8, 2020
 #2
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Thank you!

 Apr 8, 2020
 #3
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Any tips on #1

 Apr 9, 2020
 #4
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For #1, extend QRS across the circle to meet the other side at T and drop a perpendicular from the centre O onto this line meeting it at N.

TQ.RQ = PQ^2, so (TR + 3).3 = 36, so TR = 9.

NS = 9/2 - 3 = 3/2.

From the triangle ONS, use Pythagoras to calculate ON^2, and then from the triangle ONT use Pythagoras again to calculate OT, the radius of the circle.

 

There's a trig method for #2.

Join M to B and A.

Since angle C is a right angle, BA, (length 10) will be a diameter of the circle and in which case angle BMA is also a right angle.

Angles ABM and BAM are both equal to 45 deg, (angles on the same arc), so BM = AM.

Now use Pythagoras in the triangle ABM, BM^2 + AM^2 = 2BM^2 = 10^2 = 100, BM = sqrt(50) = 5sqrt(2).

Call angle CBA theta, then using the sine rule in triangle CBM, 

5sqrt(2)/ sin(45) = CM/ sin(theta + 45),

so CM = 5sqrt(2)(sin(theta).cos(45) + cos(theta).sin(45))/sin(45).

Simplify that to get 7sqrt(2), (same as geno).

 Apr 9, 2020

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