1. A triangle has side lengths of \(10, 24,\) and \(26.\) Let \(a\) be the area of the circumcircle. Let \(b\) be the area of the incircle. Compute \(a-b.\)

2. In triangle \(ABC, AB =BC=17\) and \(AC =16.\) Find the inradius of triangle \(ABC.\)

3. In triangle \(ABC,M\) is the midpoint of \(\overline{AB}\). Let \(D\) be the point on \(\overline{BC}\) such that \(\overline{AD}\) bisects \(\angle{BAC},\) and let the perpendicular bisector of \(\overline{AB}\) intersect \(\overline{AD}\) at \(E.\) If \(AB=44\) and \(ME=12,\) then find the distance from \(E\) to line \(AC.\)

Divineology Feb 22, 2020

#1**+1 **

I'll try the first one.

Since the triangle has side lengths of 10, 24, and 26, this triangle is a right triangle (10^{2} + 24^{2} = 26^{2}).

The center of the circumcenter of a right triangle is the midpoint of the hypotenuse.

Since the hypotenuse has a length of 26, the distance from this point to any of the vertices is 13.

Therefore, the area of the circumcircle is: pi·13^{2} = 169·pi.

The center of the incircle is the point of intersection of the angle bisectors.

Call the vertex between the 10 and 24 sides = C (it is a right angle).

Call the vertex between the 24 and 26 sides = A.

Call the center of the incircle X.

Draw the perpendicular from X to side AC; call this point Y; this is a radius of the incircle;

call its value 'h'.

Angle C = 90^{o}; therefore, angle(XCY) = 45^{o}.

I will need to find the size of angle(XAY).

I will use the sin(A) in triangle ABC) and find one-half of that value.

sin(A) = 10/26 ---> A = sin^{-1}(10/26) = 22.61986^{o} ---> angle(XAY) = 11.31^{o}.

Now, to find the value of h:

Call the distance from C to Y 'x'; therefore, the distance from A to Y is '24 - x'.

In triangle(XCY), tan(45^{o}) = h/x ---> h = x·tan(45^{o}).

In triangle (XAY), tan(11.31^{o}) = h/(24 - x) ---> h = (24 - x)·tan(11.31^{o}).

Setting these two equation equal to each other: x·tan(45^{o}) = (24 - x)·tan(11.31^{o})

---> x·tan(45^{o}) = 24·tan(11.31^{o}) - x·tan(11.31^{o})

---> x·tan(45^{o}) + x·tan(11.31^{o}) = 24·tan(11.31^{o})

---> x·( tan(45^{o}) + tan(11.31^{o}) ) = 24·tan(11.31^{o})

---> x = 24·tan(11.31^{o}) / [ tan(45^{o}) + tan(11.31^{o}) ]

---> x = 4

Therefore, the area of the incircle is: pi·4^{2} = 16·pi.

169·pi - 16·pi = 153·pi

geno3141 Feb 22, 2020

#2**+1 **

#2: Let M be the midpoint of side AC. Draw BM.

Since triangle ABC is an isosceles triangle BM is the angle bisector of angle(B) and also the altitude from B to side AC.

This makes angle(AMB) a right angle.

Since AC = 16, AM = 8.

Using the Pythagorean Theorem on triangle(AMB), we can find that BM = 15.

In right triangle(AMB), cos(A) = 8/17 ---> angle(A) = cos^{-1}(8/17) = 61.9275^{o}.

Draw the bisector of angle(A). Call the point where this angle bisector intersects BM point 'X'.

Angle(MAX) = ½·angle(A) = ½·61.9275^{o} = 30.96^{o}.

To find XM of triangle(MAX) ---> tan(30.96^{o}) = XM/8 ---XM = 8·tan(30.96^{o}) = 4.8.

The radius of the incircle is 4.8.

geno3141 Feb 22, 2020