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SOURCE:COMPETITION Number of Problems: 14. FOR PRINT ::: (Book)
A right circular cylinder with its diameter equal to its height is inscribed in a right circular cone. The cone has diameter and altitude , and the axes of the cylinder and cone coincide. Find the radius of the cylinder.
Let the diameter of the cylinder be . Examining the cross section of the cone and cylinder, we find two similar triangles. Hence, which we solve to find . Our answer is .
In rectangle and . Points and are on so that and . Lines and intersect at . Find the area of .
because The ratio of to is since and from subtraction. If we let be the height of
The height is so the area of is .
Points and lie, in that order, on , dividing it into five segments, each of length 1. Point is not on line . Point lies on , and point lies on . The line segments and are parallel. Find .
As is parallel to , angles FHD and FGA are congruent. Also, angle F is clearly congruent to itself. From SSS similarity, ; hence . Similarly, . Thus, .
Points and are located on square so that is equilateral. What is the ratio of the area of to that of ?
Since triangle is equilateral, , and and are congruent. Thus, triangle is an isosceles right triangle. So we let . Thus . If we go angle chasing, we find out that , Thus . . Thus , or . Thus , and , and . Thus the ratio of the areas is .
Right has and Square is inscribed in with and on on and on What is the side length of the square?
There are many similar triangles in the diagram, but we will only be using If is the altitude from to and is the sidelength of the square, then is the altitude from to By similar triangles,
Find the length of the altitude of Since it is a right triangle, the area of is
The area can also be expressed as so
Substitute back into
A circle of radius 1 is tangent to a circle of radius 2. The sides of are tangent to the circles as shown, and the sides and are congruent. What is the area of ?
Note that . Using the first pair of similar triangles, we write the proportion:
By the Pythagorean Theorem we have that .
Now using ,
The area of the triangle is .
Consider the -sided polygon , as shown. Each of its sides has length , and each two consecutive sides form a right angle. Suppose that and meet at . What is the area of quadrilateral ?
We can obtain the solution by calculating the area of rectangle minus the combined area of triangles and .
We know that triangles and are similar because . Also, since , the ratio of the distance from to to the distance from to is also . Solving with the fact that the distance from to is 4, we see that the distance from to is .
The area of is simply , the area of is , and the area of rectangle is .
Taking the area of rectangle and subtracting the combined area of and yields .
Extend and and call their intersection .
The triangles and are clearly similar with ratio , hence and thus . The area of the triangle is .
The triangles and are similar as well, and we now know that the ratio of their dimensions is .
Draw altitudes from onto and , let their feet be and . We get that . Hence .
Then the area of is , and the area of can be obtained by subtracting the area of , which is . Hence the answer is .
Rectangle has and . Point is the midpoint of diagonal , and is on with . What is the area of ?
By the Pythagorean theorem we have , hence .
The triangles and have the same angle at and a right angle, thus all their angles are equal, and therefore these two triangles are similar.
The ratio of their sides is , hence the ratio of their areas is .
And as the area of triangle is , the area of triangle is .
Points and lie on a circle centered at , and . A second circle is internally tangent to the first and tangent to both and . What is the ratio of the area of the smaller circle to that of the larger circle?
Let be the center of the small circle with radius , and let be the point where the small circle is tangent to . Also, let be the point where the small circle is tangent to the big circle with radius .
Then is a right triangle, and a triangle at that. Therefore, .
Since , we have , or , or .
Then the ratio of areas will be squared, or .