Use the equivalent of the Pythagorean Theorem for 3D to find the distance from the origin to the center of the sphere in the octant with all positive coordinates. This would be . Add on an additional radius to reach to the outside edge of a sphere, giving . 51 Recall that a right triangle is inscribed in a circle. Thus, an acute triangle would have vertex outside the circle that has the midpoint of segment as its C AB center. To be acute, you would also need the angles at and to be less than A B 90°. Therefore, find the area of the region inside the large triangle below. Then subtract the areas of the small triangle and the circle. You would have 6 The area of the rectangle is 144, so the square must be 12 x 12. Draw the radii to the points of tangency. The triangles are then similar. Extend the lines in the drawing to the left until they intersect. That point of intersection is . The slope of the line through the centers is , which will be the tangent of the angle, , formed at . You then need . Use the Law of Cosines and solve the quadratic equation for . Use the property of secants to find that = 1. DC Due to congruent angles, we know that with a ratio of similitude of 8:1. Using the areas of the three regions to be and as marked in the figure, M, N, K, we then know that = 64N. We also know by M AA~ Theorem that with a ratio of similitude of 2:1. This means that + = 4(K + Substituting, M K N). we get , so and . Draw . Use the ratios of special right triangles. Then the area of the square is . Note: For fun, TNYWR = 28!
Use the equivalent of the Pythagorean Theorem for 3D to find the distance from the origin to the center of the sphere in the octant with all positive coordinates. This would be . Add on an additional radius to reach to the outside edge of a sphere, giving . 51 Recall that a right triangle is inscribed in a circle. Thus, an acute triangle would have vertex C outside the circle that has the midpoint of segment AB as its center. To be acute, you would also need the angles at A and B to be less than 90°. Therefore, find the area of the region inside the large triangle below. Then subtract the areas of the small triangle and the circle. You would have 6 The area of the rectangle is 144, so the square must be 12 x 12. Draw the radii to the points of tangency. The triangles are then similar. Extend the lines in the drawing to the left until they intersect. That point of intersection is . The slope of the line through the centers is , which will be the tangent of the angle, , formed at . You then need . Use the Law of Cosines and solve the quadratic equation for . Use the property of secants to find that DC = 1. Due to congruent angles, we know that with a ratio of similitude of 8:1. Using the areas of the three regions to be M, N, and K, as marked in the figure, we then know that M = 64N. We also know by AA~ Theorem that with a ratio of similitude of 2:1. This means that M + K = 4(K + N). Substituting, we get , so and . Draw . Use the ratios of special right triangles. Then the area of the square is . Note: For fun, TNYWR = 28!
FAVORITE_CONTEST_GEOMETRY_PROBS_10-13__2_.docx