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

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. 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

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