1983 AIME Problems/Problem 15
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Problem
The adjoining figure shows two intersecting chords in a circle, with on minor arc . Suppose that the radius of the circle is , that , and that is bisected by . Suppose further that is the only chord starting at which is bisected by . It follows that the sine of the minor arc is a rational number. If this fraction is expressed as a fraction in lowest terms, what is the product ?
Solution 1
-Credit to Adamz for diagram- Let be any fixed point on circle and let be a chord of circle . The locus of midpoints of the chord is a circle , with diameter . Generally, the circle can intersect the chord at two points, one point, or they may not have a point of intersection. By the problem condition, however, the circle is tangent to BC at point N.
Let M be the midpoint of the chord . From right triangle , . Thus, .
Notice that the distance equals (Where is the radius of circle P). Evaluating this, . From , we see that
Next, notice that . We can therefore apply the tangent subtraction formula to obtain , . It follows that , resulting in an answer of .
Motivation Behind Solution 1
The above solution works, but is quite messy and somewhat difficult to follow. This solution provides the motivation behind the solution.
First of all, where did the statement " is the only chord starting at and bisected by " come from? What is its significance in this problem? What is the criterion for this statement to be true?
We consider the locus of midpoints of the chords from . It is well known that this is the circle with diameter , where is the center of the circle. The proof is simple: every midpoint of a chord is a dilation of the endpoint with ratio with center . Thus, the locus is the result of the dilation with ratio of circle with center . Let the center of this circle be .
Aha! Now we see. is bisected by if they cross at some point on the circle. Moreover, since is the only chord, must be tangent to the circle .
The rest of this problem is straight forward.
Our goal is to find where is the midpoint of . Then we have and . Let be the projection of onto , and similarly be the projection of onto . Then it remains to find so we can use the sine addition formula.
As is a radius of circle , , and similarly, . Since , . Thus, .
From here, we see that is a dilation of about center with ratio , so .
Lastly, we apply the formula:
Thus, our answer is .
Solution 2 (with the help of coordinates)
Let the circle be , and its center be labeled . Since BC=6, we can calculate (by the Pythagorean Theorem) that the distance from to the line is . So we can let and . Now assume that is any point on the major arc BC, and any point on the minor arc BC. We may let , where is the angle measured from the positive axis to the ray . It will also be convenient to define as the measure of angle XOB.
Firstly, since B must lie in the minor arc AD, we see that . However, since the midpoint of AD must lie on BC, and the highest coordinate of is , we see that the coordinate can't be lower than , that is, .
Secondly, there is a theorem that says that, in a circle, if a chord is bisected by a radius, then the they must be perpendicular. Therefore, suppose that is the intersection between and , then is perpendicular to . So, if is the only chord starting at which is bisected by , it means that is the only point on the chord such that is perpendicular to . Now we are ready to make it an algebraic problem. Suppose , . The statement is perpendicular to is equivalent to the following equation:
It rearranges to the following:
Given that this equation has only one real root , we study the following function:
First, by the fact that has real solution, it is good to look at its discriminant: must be non-negative:
It is obvious that this is non-negative. If it is actually zero, then , and . In this case, . We found a possible case. So we calculate .
Note to Solution 2
Note: As an AIME problem, it is already done since we have found one possible case. However, it takes one more step to complete it if we need to say that this is the unique possibility, without appealing to the AIME Uniqueness Principle.
Suppose that , which means that there can be two real roots of , one lying in the interval , but another falling out of it. We also see that the average of the two root is , which is a quantity greater than 0, so the root outside of must be no less than 3. Spectating the parabolic curve of , which is a "U shaped" curve hitting the interval once and another time, it is evident that and . We can just work on the second one:
The only way to satisfy this equation is when (by working on Cauchy-Schwarz inequality, or just plotting the line to see that point can't go above this line), which does not make sense from the description of the problem. It means that the point lies in the half plane above the line , inclusive, and the half plane below the line , exclusive. It is obviously impossible, by drawing the lines and see that the intersection of the two half planes does not share any point with the circle.
See Also
1983 AIME (Problems • Answer Key • Resources) | ||
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