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# need help on a few tough geo questions :(

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1. In triangle $$ABC$$, point $$X$$ is on side $$\overline{BC}$$ such that $$AX = 13, BX = 14, CX = 4,$$ and the circumcircles of triangles $$ABX$$ and $$ACX$$ have the same radius. Find the area of triangle $$ABC$$.

2. Let $$\mathcal{R}$$ be the circle centered at $$(0,0)$$ with radius $$10.$$ The lines $$x=6$$ and $$y=5$$ divide $$\mathcal{R}$$ into four regions $$\mathcal{R}_1$$$$\mathcal{R}_2$$$$\mathcal{R}_3$$, and $$\mathcal{R}_4$$. Let $$[\mathcal{R}_i]$$ denote the area of region $$\mathcal{R}_i$$. If $$[\mathcal{R}_1] > [\mathcal{R}_2] > [\mathcal{R}_3] > [\mathcal{R}_4],$$ then find $$[\mathcal{R}_1] - [\mathcal{R}_2] - [\mathcal{R}_3] + [\mathcal{R}_4]$$.

Thank you so much for your help!

May 30, 2020

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For problem 1,  I placed the diagram on a coordinate axis, with P(0,0), C(4,0), and B(-14,0).

Since PA = 13, it is on a circle with center P and radius 13:  x2 + y2  =  13

-- I called the x-value of this point 'a', so the y-value became  sqrt(169 - a2).

I used the theorem that the center of a circle is on the perpendicular bisectors of the chords.

Since one chord is BP, its midpoint has x-value -7 and its y-value is on the line x = -7.

Since another chord is PC, its midpoint has x-value 2 and its y-value is on the line x = 2.

Another chord is AP: using its endpoints, I found its midpoint and its slope.

From these values, I could find the equation of its perpendicular bisector (not a lot of fun, for it had a lot of a-terms and square roots).

Then, I found the intersection of this perpendicular biscector with the line x = -7 and the intersection of this perpendicular bisector with the line x = 2 (there were still a lot of a-terms and square roots).

With these two points (the centers of the circumcircles), I could find the distance from one center to point P and the distance from the other center to point P (By this time, I was on a firt-name basis with the a-terms and square roots).

Since these are equal, by solving this equation, I could find the value of a, which was 8.

This is the x-value of point A, allowing me to find the y-value of point A, 12, which is the height of triangle ABC.

With the height 8 + 8 = 16 and the base 12, the area is 96.

May 30, 2020
#2
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2. By calculus,

$$[\mathcal{R}_1] = 30 + \frac{1}{4} \cdot \pi \cdot 10^2 + \int_0^5 \sqrt{100 - x^2} \ dx + \int_0^6 \sqrt{100 - x^2} \ dx.$$

We can write out the areas similarly, to get [R_1] - [R_2] - [R_3] + [R_4] = 84.

May 31, 2020