A right triangle has side lengths $3$, $4$, and $5.$ A circle is then drawn with each side as a diameter. Find the area of the region contained in all three circles.

Answer 1

Let $O_1$, $O_2$, $O_3$ be the centers of the circles with diameters 3, 4, and 5, respectively.

[asy]
unitsize(0.4 cm);

pair A, B, C, O1, O2, O3;

A = (0,0);
C = (5,0);
O3 = extension(A, A + (C - A)/(C - A).rotate(90), C, C + (A - C)/(A - C).rotate(90));
B = intersectionpoint(arc(O3,5),arc(C,3));
O1 = (A + B)/2;
O2 = (C + B)/2;

draw(A--B--C--cycle);
draw(arc(O1,3/2,0,180));
draw(arc(O2,4/2,0,180));
draw(arc(O3,5/2,0,180));

label("$A$", A, SW);
label("$B$", B, N);
label("$C$", C, SE);
label("$3$", (A + B)/2, NW);
label("$4$", (B + C)/2, NE);
label("$5$", (A + C)/2, S);

dot("$O_1$", O1, N);
dot("$O_2$", O2, SW);
dot("$O_3$", O3, dir(270));
[/asy]

We rotate $\triangle ABC$ about $O_3$ by 180 degrees, to form a new triangle with side lengths 3, 4, and 5.

[asy]
unitsize(0.4 cm);

pair A, B, C, O1, O2, O3;

A = (0,0);
C = (5,0);
O3 = (0,0);
B = intersectionpoint(arc(O3,5),arc(C,3));
O1 = (A + B)/2;
O2 = (C + B)/2;

draw(A--B--C--cycle);
draw(O1--O3);
draw(O2--O3);

label("$A$", A, W);
label("$B$", B, N);
label("$C$", C, E);
label("$3$", (A + B)/2, NW);
label("$4$", (B + C)/2, NE);
label("$5$", (A + C)/2, S);

dot("$O_1$", O1, SB);
dot("$O_2$", O2, dir(120));
dot("$O_3$", O3, S);
dot("$(0,3)$", (0,3), N);
[/asy]

The centers of the circles with diameters 4 and 5 are $O_2 = (0,0)$ and $O_3 = (0,3)$, respectively. Rotating the triangle about $O_3$ also rotates the lines $O_2 O_3$ and $O_1 O_3,$ so the centers of the circles with diameters 5 and 3 are mapped to points $(0,-4)$ and $(0,6)$, respectively. The equation of the circle with diameter 5 is then
\[(x - 0)^2 + (y - 3)^2 = \left( \frac{5}{2} \right)^2 = \frac{25}{4},\]the equation of the circle with diameter 4 is
\[(x - 0)^2 + (y - 4)^2 = 2^2 = 4,\]and the equation of the circle with diameter 3 is
\[(x - 0)^2 + (y - 6)^2 = \left( \frac{3}{2} \right)^2 = \frac{9}{4}.\]We take the horizontal line $y = k,$ for $-4 \le k \le 3,$ to slice the region that we want.

[asy]
unitsize(0.4 cm);

int i;
pair A, B, C, O1, O2, O3;
real k;

A = (0,0);
C = (5,0);
O3 = (0,0);
B = intersectionpoint(arc(O3,5),arc(C,3));
O1 = (A + B)/2;
O2 = (C + B)/2;

k = 1;

draw(A--B--C--cycle);
draw(O1--O3);
draw(O2--O3);

draw(arc(O1,3/2,0,180));
draw(arc(O2,4/2,0,180));
draw(arc(O3,5/2,0,180));

label("$A$", A, SW);
label("$B$", B, dir(190));
label("$C$", C, SE);
label("$3$", (A + B)/2, NW);
label("$4$", (B + C)/2, NE);
label("$5$", (A + C)/2, S);

dot("$O_1$", O1, N);
dot("$O_2$", O2, SW);
dot("$O_3$", O3, S);
dot("$(0,3)$", (0,3), N);

draw((0,-4)--(0,6),dashed);

draw((0,3)--(intersectionpoint(arc(O1,3/2,0,90),arc(O3,5/2,0,90)))--(intersectionpoint(arc(O2,4/2,0,90),arc(O3,5/2,0,90)))--cycle);
draw((intersectionpoint(arc(O2,4/2,90,180),arc(O3,5/2,90,180)))--(intersectionpoint(arc(O1,3/2,90,180),arc(O3,5/2,90,180)))--(0,3)--cycle);
draw((0,3)--(intersectionpoint(arc(O1,3/2,180,270),arc(O3,5/2,180,270)))--(intersectionpoint(arc(O2,4/2,180,270),arc(O3,5/2,180,270)))--cycle);
draw((intersectionpoint(arc(O2,4/2,270,360),arc(O3,5/2,270,360)))--(intersectionpoint(arc(O1,3/2,270,360),arc(O3,5/2,270,360)))--(0,3)--cycle);

label("$k = -4$", ((0,-4) + (0,-4) + intersectionpoint(arc(O2,4/2,90,180),arc(O3,5/2,90,180)))/3, N);
label("$k = 0$", ((0,0) + (0,0) + intersectionpoint(arc(O2,4/2,0,90),arc(O3,5/2,0,90)))/3, N);
label("$k = 3$", (3/2,(0,3) + (0,3))/2, N);
[/asy]

The solution is the sum of the areas of the three shaded regions. The area of each shaded region is a quarter-circle minus a triangle.

The area of the lowest shaded region is
\[\frac{1}{4} \cdot \pi \cdot \left( \frac{5}{2} \right)^2 - \frac{1}{2} \cdot 4 \cdot 4 = \frac{25 \pi}{8} - 8.\]The area of the middle shaded region is
\[\frac{1}{4} \cdot \pi \cdot 2^2 - \frac{1}{2} \cdot 3 \cdot 4 = \frac{4 \pi}{4} - 6.\]The area of the top shaded region is
\[\frac{1}{4} \cdot \pi \cdot \left( \frac{3}{2} \right)^2 - \frac{1}{2} \cdot 3 \cdot 3 = \frac{9 \pi}{16} - \frac{9}{2}.\]Hence, the area that is contained in all three circles is
\[\boxed{\frac{25 \pi}{8} + \frac{4 \pi}{4} + \frac{9 \pi}{16} - 8 - 6 - \frac{9}{2}}.\]