1983 AIME Problems/Problem 14
Contents
- 1 Problem
- 2 Solution
- 2.1 Solution 1
- 2.2 Solution 2 (Easiest)
- 2.3 Solution 3
- 2.4 Solution 4 (quickest)
- 2.5 Solution 5 (Pythagorean Theorem and little algebraic manipulation)
- 2.6 Solution 6 (Only simple geometry and algebra needed)
- 2.7 Solution 7
- 2.8 Full Proof that R, A, B are collinear
- 2.9 Solution 8 (Coordinate Bash)
- 2.10 Solution 9 (basic solution)
- 3 See Also
Problem
In the adjoining figure, two circles with radii and are drawn with their centers units apart. At , one of the points of intersection, a line is drawn in such a way that the chords and have equal length. Find the square of the length of .
Solution
Note that some of these solutions assume that lies on the line connecting the centers, which is not true in general. It is true here only because the perpendicular from passes through through the point where the line between the centers intersects the small circle. This fact can be derived from the application of the Midpoint Theorem to the trapezoid made by dropping perpendiculars from the centers onto .
Solution 1
Firstly, notice that if we reflect over , we get . Since we know that is on circle and is on circle , we can reflect circle over to get another circle (centered at a new point , and with radius ) that intersects circle at . The rest is just finding lengths, as follows.
Since is the midpoint of segment , is a median of . Because we know , , and , we can find the third side of the triangle using Stewart's Theorem or similar approaches. We get . Now we have a kite with , , and , and all we need is the length of the other diagonal . The easiest way it can be found is with the Pythagorean Theorem. Let be the length of . Then
Solving this equation, we find that , so
Solution 2 (Easiest)
Draw additional lines as indicated. Note that since triangles and are isosceles, the altitudes are also bisectors, so let .
Since triangles and are similar. If we let , we have .
Applying the Pythagorean Theorem on triangle , we have . Similarly, for triangle , we have .
Subtracting, .
Solution 3
Let . Angles , , and must add up to . By the Law of Cosines, . Also, angles and equal and . So we have
Taking the cosine of both sides, and simplifying using the addition formula for as well as the identity , gives .
Solution 4 (quickest)
Let . Extend the line containing the centers of the two circles to meet , and to meet the other side of the large circle at a point .
The part of this line from to the point nearest to where it intersects the larger circle has length . The length of the diameter of the larger circle is .
Thus by Power of a Point in the circle passing through , , and , we have , so .
Solution 5 (Pythagorean Theorem and little algebraic manipulation)
Note that the midpoint of is Also, since bisect and respectively, is the midpoint of Thus, let This means that From the median formula, Thus, Also, since , from the Pythagorean Theorem, Thus, We conclude that ~pinkpig
Solution 6 (Only simple geometry and algebra needed)
Looking at Drawing 2 (by the way, we don't need point ), we set and , and the desired length. We know that a radius perpendicular to a chord bisects the chord, so and . Draw line and , and we see that they are radii of Circles and , respectively. We can write the Pythagorean relationships for triangle and for triangle . We also translate segment down so that coincides with , and form another right traingle. From that triangle, you can see that the shorter leg is on the left side, having length , the longer leg is the same as , and the hypotenuse is . We can write the Pythagorean relationship . Solving the system of 3 unknowns and 3 equations (One of the best ways to do it is to solve for in the first equation and in the second equation, and substitute into the third equation, get an equation only in terms of , and solve), you find that , so .
Solution by Kinglogic
Solution 7
The centers are collinear, you can prove it (but that is already given in the later section [Proof that R,A, and B are collinear]). Drop a perpendicular from to the lines that the centers are on. You then have 2 separate segments, separated by the foot of the altitude of . Call them and respectively. Call the measure of the foot of the altitude of . You then have 3 equations:
(this is given by the fact that the distance between the centers is 12.
. This is given by the fact that P is on the circle with radius 8.
. This is given by the fact that P is on the circle with radius 6.
Subtract (3) from (2) to get that . As per (1), then you have (4). Add (1) and (4) to get that . Then substitute into (1) to get . Substitute either a or b into (2) or (3) to get . Then to get it is just .
-dragoon
Full Proof that R, A, B are collinear
Let and be the feet of the perpendicular from to and to respectively. It is well known that a perpendicular from the center of a circle to a chord of that circle bisects the chord, so , since the problem told us .
We will show that lies on .
Let be the intersection of circle centered at with . Then .
Let ' be the foot of the perpendicular from to . Then is a midline (or midsegment) in trapezoid , so coincides with (they are both supposed to be the midpoint of ). In other words, since , then .
Thus, subtends a degree arc. So arc in circle is , so is a diameter, as desired. Thus , , are collinear.
NOTE: Note this collinearity only follows from the fact that is half of in the problem statement. The collinearity is untrue in general.
Solution 8 (Coordinate Bash)
We use coordinate geometry to approach this problem. Let the center of larger circle be the origin , the smaller circle be , and the x-axis be . Hence, we can get the two circle equations: and .
Let point be . Noting that it lies on both circles, we can plug the coordinates into both equations:
Substituting into equation 2 and solving for , we get . The problem asks us to find , which is congruent to . Using the distance formula for and (by Solution 7's collinear proof), we get . Using , we find that . Plugging the variables and in, we get ~SoilMilk
Solution 9 (basic solution)
Let the center of the circle with radius be and let the center of the one with radius be Also, let Using law of cosines on triangle we have that Angle chasing gives that so its cosines must be the same. Applying law of cosines again on triangle we have which gives that
~happypi31415
See Also
1983 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |