2007 iTest Problems/Problem 15
Problem
Form a pentagon by taking a square of side length 1 and an equilateral triangle of side length 1, and placing the triangle so that one of its sides coincides with a side of the square. Then "circumscribe" a circle around the pentagon, passing through three of its vertices, so that the circle passes through exactly one of the vertices of the equilateral triangle, and through exactly two vertices of the square. What is the radius of the circle?
Solutions
Solution 1
Let be points of the equilateral triangle, be points of the square, and be the center of the circle.
By the Base Angle Theorem, , so . Since , by SAS Congruency, , so .
By SSS Congruency, , so .
Finally, by SSS Congruency, , so . Thus, , making equilateral, so the radius of the circle is .
Solution 2
Draw lines connecting the top vertex of the equilateral triangle to the bottom two vertices of the square (as seen in the diagram). This means that the circle is the circumcircle of the triangle shown above.
From the Law of Cosines, the distance of each of the remaining sides is .
Now we need to find the length of the circumradius. It can be done in a number of ways, but we’re going to be using the area formula .
The area of the triangle is . The product of the sides of the triangle is . Plugging the values in the area formula yields
The length of the radius of the circle is .
See Also
2007 iTest (Problems, Answer Key) | ||
Preceded by: Problem 14 |
Followed by: Problem 16 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • TB1 • TB2 • TB3 • TB4 |