|
|
| (91 intermediate revisions by 22 users not shown) |
| Line 1: |
Line 1: |
| | {{AoPSWiki:Sandbox/header}} <!-- Please do not delete this line --> | | {{AoPSWiki:Sandbox/header}} <!-- Please do not delete this line --> |
| − | | + | [[<><><><><><><><> |
| − | ==Solution==
| |
| − | | |
| − | Note: All solutions compiled + written by Aayush Gupta!
| |
| − | | |
| − | We want to find the area of this figure:
| |
| − | | |
| − | <asy> | |
| − | path rt,tt, tri;
| |
| − | real x, y;
| |
| − | y = 1+sqrt(2);
| |
| − | x = y+(6/1.7);
| |
| − | tt=(0,0)..(y,1)--(y,-1)..cycle;
| |
| − | rt=(y,-1)--(x,-1)--(x,1)--(y,1);
| |
| − | tri=(y,-1)--(y-1,0)--(y,1);
| |
| − | draw(rt);
| |
| − | draw(tt);
| |
| − | draw(tri);
| |
| − | label("1.7", (x, 0), E);
| |
| − | label("3", (y+(3/1.7), -1), S);
| |
| − | label("C", (y-1, -0.1), S);
| |
| − | </asy> | |
| − | | |
| − | We label the circle as circle C. We can break the figure into three parts, shown as the 3/4 circle, the triangle, and the rectangle.
| |
| − | | |
| − | Lets first take a look at the rectangle.
| |
| − | | |
| − | <asy> | |
| − | path rt;
| |
| − | real x, y;
| |
| − | y = 1+sqrt(2);
| |
| − | x = y+(6/1.7);
| |
| − | rt=(y,-1)--(x,-1)--(x,1)--(y,1)--cycle;
| |
| − | draw(rt);
| |
| − | label("1.7", (x, 0), E);
| |
| − | label("3", (y+(3/1.7), -1), S);
| |
| − | </asy> | |
| − | | |
| − | It has an area of <math> 3 * 1.7 = 5.1</math> .
| |
| − | | |
| − | | |
| − | Lets now take a look at the triangle, after drawing the height.
| |
| − | | |
| − | <asy> | |
| − | unitsize(0.8inch);
| |
| − | path tri, lin;
| |
| − | real x, y;
| |
| − | y = 1+sqrt(2);
| |
| − | x = y+(6/1.7);
| |
| − | tri=(y,-1)--(y-1,0)--(y,1)--cycle;
| |
| − | lin=(y-1, 0)--(y,0);
| |
| − | draw(tri);
| |
| − | draw(lin);
| |
| − | label("1.7", (y, 0), E);
| |
| − | label("C", (y-1, -0.1), S);
| |
| − | </asy> | |
| − | | |
| − | We see that both the radii are the two shorter sides of the triangle, making this a isosceles 45-45-90 triangle.
| |
| − | | |
| − | We also see that the height that we drew is half the hypotenuse(note the two smaller 45-45-90 isosceles triangles).
| |
| − | | |
| − | Hence, the area of the triangle is <math>\frac{1.7 * \frac{1.7}{2}}{2} = 0.7225</math> .
| |
| − | | |
| − | Now, let's take a look at the 3/4 circle. We know it is 3/4 because there is a 90 degree triangle cut out of it.
| |
| − | | |
| − | <asy>
| |
| − | unitsize(0.6inch);
| |
| − | path tt, tri;
| |
| − | real x, y;
| |
| − | y = 1+sqrt(2);
| |
| − | x = y+(6/1.7);
| |
| − | tt=(0,0)..(y,1)--(y-1, 0)--(y,-1)..cycle;
| |
| − | tri=(y, 1)--(y,-1);
| |
| − | draw(tt);
| |
| − | draw(tri);
| |
| − | fill(tt, gray(0.6));
| |
| − | label("1.7", (y, 0), E);
| |
| − | label("C", (y-1, -0.1), S);
| |
| − | </asy>
| |
| − | | |
| − | We find the radius using the 45-45-90 triangle. Since the ratios of the sides are 1:1:<math>\sqrt{2}</math>, we can find the radius to be <math>\frac{1.7}{\sqrt{2}} = \frac{1.7 \sqrt{2}}{2}</math> .
| |
| − | | |
| − | Hence, the area of the whole circle is <math>\pi r^2</math>, and the area of the 3/4 circle is <math>\frac{270}{360} * \pi * (\frac{1.7 \sqrt{2}}{2})^2 = \frac{3}{4} * \pi * \frac{2.89}{2}</math>.
| |
| − | | |
| − | Adding it all up, we find the answer to be <math>\frac{3}{4} * \pi * \frac{2.89}{2} + 0.7225 + 5.1</math> .
| |
| − | | |
| − | Just plug it into your calculator. And please understand where all the numbers came from before writing the answer down.
| |
| − | | |
| − | | |
| − | ---------------------------------------------------
| |
| − | | |
| − | We want to find the area of the intersection of the circles in this figure:
| |
| − | | |
| − | <asy>
| |
| − | //import graph;
| |
| − | draw(circle((0,0), 2));
| |
| − | draw(circle((2,0), 2));
| |
| − | path inter;
| |
| − | inter=(1,-sqrt(3))..(2,0)..(1, sqrt(3))--cycle;
| |
| − | draw(inter);
| |
| − | fill(inter, gray(0.6));
| |
| − | path inter2;
| |
| − | inter2=(1,-sqrt(3))..(0,0)..(1, sqrt(3))--cycle;
| |
| − | draw(inter2);
| |
| − | fill(inter2, gray(0.6));
| |
| − | </asy>
| |
| − | | |
| − | Lets call the radius of each of the circles 1, because we are calculating probability.
| |
| − | To do this, we need to divide the intersection area into smaller parts that we can find the area of.
| |
| − | | |
| − | <asy>
| |
| − | //import graph;
| |
| − | draw(circle((0,0), 2));
| |
| − | draw(circle((2,0), 2));
| |
| − | path a, b, c;
| |
| − | a = (0,0)--(1, sqrt(3))--(2,0)--cycle;
| |
| − | b = (0,0)--(1, -sqrt(3))--(2,0);
| |
| − | c = (0,0)..(2-sqrt(3), 1)..(1, sqrt(3))--cycle;
| |
| − | draw(a);
| |
| − | fill(a, gray(0.6));
| |
| − | draw(b);
| |
| − | label("A", (0,0), W);
| |
| − | label("B", (2,0), E);
| |
| − | label("C", (1,sqrt(3)), N);
| |
| − | label("D", (1,-sqrt(3)), S);
| |
| − | label("1", (1.5,sqrt(3)/2), NE);
| |
| − | label("1", (1,0), S);
| |
| − | </asy>
| |
| − | | |
| − | Lets examine the shaded <math>\triangle ABC</math>. Since each of its sides are radii, it is equilateral.
| |
| − | | |
| − | Hence, each of its angles measure 60 degrees. Namely, <math>\angle ABC</math> measures 60 degrees.
| |
| − | | |
| − | Important note: The area of an equilateral triangle is calculated by <math>\frac{s^2\sqrt{3}}{4}</math>, where <math>s</math> is the side length of a triangle. This is an EXTREMELY important formula to know, and it it also extremely simple to prove and memorize.
| |
| − | | |
| − | After this, we can deduce that the area of triangle ABC with side length 1 is <math>\frac{1^2\sqrt{3}}{4} = \frac{\sqrt{3}}{4}</math>
| |
| − | | |
| − | Let us now look at the encompassing sector.
| |
| − | | |
| − | <asy>
| |
| − | //import graph;
| |
| − | draw(circle((0,0), 2));
| |
| − | draw(circle((2,0), 2));
| |
| − | path a, b, c;
| |
| − | a = (0,0)--(1, sqrt(3))--(2,0)--cycle;
| |
| − | b = (0,0)--(1, -sqrt(3))--(2,0);
| |
| − | c = (0,0)..(2-sqrt(3), 1)..(1, sqrt(3))--cycle;
| |
| − | draw(a);
| |
| − | fill(a, gray(0.6));
| |
| − | draw(c);
| |
| − | fill(c, gray(0.6));
| |
| − | draw(b);
| |
| − | label("A", (0,0), W);
| |
| − | label("B", (2,0), E);
| |
| − | label("C", (1,sqrt(3)), N);
| |
| − | label("D", (1,-sqrt(3)), S);
| |
| − | label("1", (1.5,sqrt(3)/2), NE);
| |
| − | label("1", (1,0), S);
| |
| − | </asy>
| |
| − | We then find that the area of the entire sector <math>ABC</math> is <math>\frac{60}{360} * \pi * 1^2 = \frac{\pi}{6}</math> .
| |
| − | | |
| − | Subtracting the area of <math>\triangle ABC</math> from sector <math>ABC</math> gives us the shaded part below:
| |
| − | | |
| − | <asy>
| |
| − | //import graph;
| |
| − | draw(circle((0,0), 2));
| |
| − | draw(circle((2,0), 2));
| |
| − | path a, b, c;
| |
| − | a = (0,0)--(1, sqrt(3))--(2,0)--cycle;
| |
| − | b = (0,0)--(1, -sqrt(3))--(2,0);
| |
| − | c = (0,0)..(2-sqrt(3), 1)..(1, sqrt(3))--cycle;
| |
| − | draw(a);
| |
| − | draw(c);
| |
| − | fill(c, gray(0.6));
| |
| − | draw(b);
| |
| − | label("A", (0,0), W);
| |
| − | label("B", (2,0), E);
| |
| − | label("C", (1,sqrt(3)), N);
| |
| − | label("D", (1,-sqrt(3)), S);
| |
| − | label("1", (1.5,sqrt(3)/2), NE);
| |
| − | label("1", (1,0), S);
| |
| − | </asy>
| |
| − | | |
| − | Plugging in the values we know, we find that this has area <math>\frac{\pi}{6} - \frac{\sqrt{3}}{4}</math> .
| |
| − | | |
| − | We see that there are 4 such small curved areas.
| |
| − | | |
| − | <asy>
| |
| − | //import graph;
| |
| − | draw(circle((0,0), 2));
| |
| − | draw(circle((2,0), 2));
| |
| − | path a, b, c, d, e, f;
| |
| − | a = (0,0)--(1, sqrt(3))--(2,0)--cycle;
| |
| − | b = (0,0)--(1, -sqrt(3))--(2,0);
| |
| − | c = (0,0)..(2-sqrt(3), 1)..(1, sqrt(3))--cycle;
| |
| − | d = (0,0)..(2-sqrt(3), -1)..(1, -sqrt(3))--cycle;
| |
| − | e = (2,0)..(sqrt(3), -1)..(1, -sqrt(3))--cycle;
| |
| − | f = (2,0)..(sqrt(3), 1)..(1, sqrt(3))--cycle;
| |
| − | draw(a);
| |
| − | //draw(c);
| |
| − | fill(c, gray(0.6));
| |
| − | //draw(d);
| |
| − | fill(d, gray(0.6));
| |
| − | //draw(e);
| |
| − | fill(e, gray(0.6));
| |
| − | //draw(f);
| |
| − | fill(f, gray(0.6));
| |
| − | draw(b);
| |
| − | label("A", (0,0), W);
| |
| − | label("B", (2,0), E);
| |
| − | label("C", (1,sqrt(3)), N);
| |
| − | label("D", (1,-sqrt(3)), S);
| |
| − | label("1", (1,0), S);
| |
| − | </asy>
| |
| − | | |
| − | Multiplying the value we got for an individual small curved part by 4, we find the area of the above shaded region to be <math>\frac{4\pi}{6} - \frac{4\sqrt{3}}{4}</math> .
| |
| − | | |
| − | We then need to add back the areas of the equilateral triangles.
| |
| − | | |
| − | <asy>
| |
| − | //import graph;
| |
| − | draw(circle((0,0), 2));
| |
| − | draw(circle((2,0), 2));
| |
| − | path a, b, c, d, e, f;
| |
| − | a = (0,0)--(1, sqrt(3))--(2,0)--cycle;
| |
| − | b = (0,0)--(1, -sqrt(3))--(2,0)--cycle;
| |
| − | c = (0,0)..(2-sqrt(3), 1)..(1, sqrt(3))--cycle;
| |
| − | d = (0,0)..(2-sqrt(3), -1)..(1, -sqrt(3))--cycle;
| |
| − | e = (2,0)..(sqrt(3), -1)..(1, -sqrt(3))--cycle;
| |
| − | f = (2,0)..(sqrt(3), 1)..(1, sqrt(3))--cycle;
| |
| − | draw(a);
| |
| − | fill(a, gray(0.6));
| |
| − | //draw(c);
| |
| − | //fill(c, gray(0.6));
| |
| − | //draw(d);
| |
| − | //fill(d, gray(0.6));
| |
| − | //draw(e);
| |
| − | //fill(e, gray(0.6));
| |
| − | //draw(f);
| |
| − | //fill(f, gray(0.6));
| |
| − | draw(b);
| |
| − | fill(b, gray(0.6));
| |
| − | label("A", (0,0), W);
| |
| − | label("B", (2,0), E);
| |
| − | label("C", (1,sqrt(3)), N);
| |
| − | label("D", (1,-sqrt(3)), S);
| |
| − | label("1", (1,0), S);
| |
| − | </asy>
| |
| − | | |
| − | Recall that the area of each equilateral triangle was <math>\frac{\sqrt{3}}{4}</math>.
| |
| − | | |
| − | Hence the area of the above shaded region is <math>\frac{2\sqrt{3}}{4}</math> .
| |
| − | | |
| − | We then look at the total.
| |
| − | | |
| − | <asy>
| |
| − | //import graph;
| |
| − | draw(circle((0,0), 2));
| |
| − | draw(circle((2,0), 2));
| |
| − | path a, b, c, d, e, f;
| |
| − | a = (0,0)--(1, sqrt(3))--(2,0)--cycle;
| |
| − | b = (0,0)--(1, -sqrt(3))--(2,0)--cycle;
| |
| − | c = (0,0)..(2-sqrt(3), 1)..(1, sqrt(3))--cycle;
| |
| − | d = (0,0)..(2-sqrt(3), -1)..(1, -sqrt(3))--cycle;
| |
| − | e = (2,0)..(sqrt(3), -1)..(1, -sqrt(3))--cycle;
| |
| − | f = (2,0)..(sqrt(3), 1)..(1, sqrt(3))--cycle;
| |
| − | draw(a);
| |
| − | fill(a, gray(0.6));
| |
| − | draw(c);
| |
| − | fill(c, gray(0.6));
| |
| − | draw(d);
| |
| − | fill(d, gray(0.6));
| |
| − | draw(e);
| |
| − | fill(e, gray(0.6));
| |
| − | draw(f);
| |
| − | fill(f, gray(0.6));
| |
| − | draw(b);
| |
| − | fill(b, gray(0.6));
| |
| − | label("A", (0,0), W);
| |
| − | label("B", (2,0), E);
| |
| − | label("C", (1,sqrt(3)), N);
| |
| − | label("D", (1,-sqrt(3)), S);
| |
| − | label("1", (1,0), S);
| |
| − | </asy>
| |
| − | | |
| − | Hence the area of the total shaded region is <math>\frac{2\sqrt{3}}{4} + \frac{4\pi}{6} - \frac{4\sqrt{3}}{4} = \frac{2\pi}{3} - \frac{\sqrt{3}}{2}</math> .
| |
| − | | |
| − | We plug this into the calculator to find the final answer.
| |
| − | | |
| − | ----------------------------
| |
| − | | |
| − | In order to find the union(total) of the two circles, we examine the figure.
| |
| − | | |
| − | We want to find the area of the intersection of the circles in this figure:
| |
| − | | |
| − | <asy>
| |
| − | //import graph;
| |
| − | fill((circle((2,0), 1.98)), gray(0.6));
| |
| − | fill((circle((0,0), 1.98)), gray(0.6));
| |
| − | draw(circle((0,0), 2));
| |
| − | draw(circle((2,0), 2));
| |
| − | draw((0,0)--(2,0));
| |
| − | label("1", (1,0), S);
| |
| − | </asy>
| |
| − | | |
| − | Using <math>\pi r^2</math>, the area of the circle on the left is <math>\pi</math>. We also know that the area of the circle on the right is <math>\pi</math>. When we add these together, we are counting the entire area, but we are double-counting the overlap area. We thus need to subtract the overlap from the <math>2\pi</math> in order to find the union(total) of the figure.
| |
| − | | |
| − | Hence, the total is <math>2\pi - (\frac{2\pi}{3} - \frac{\sqrt{3}}{2}) = \frac{4\pi}{3} + \frac{\sqrt{3}}{2}</math>.
| |
