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# tough geometry questions :(

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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