Difference between revisions of "1983 AIME Problems/Problem 15"
m (→Solution 4) |
Alexlikemath (talk | contribs) |
||
Line 129: | Line 129: | ||
<cmath>DC^2 = (\sqrt{5})^2 + 5^2 -10\sqrt{5} \cdot \frac{2\sqrt{5}}{5} = 10</cmath>Therefore, <math>DC = \sqrt{10} \Longleftrightarrow BA = \sqrt{2}</math>. Now, drop the altitude from <math>O</math> to <math>BA</math> and call its intersection with <math>BA</math> <math>Z</math>. Then, by the Pythagorean theorem, <math>OZ = \frac{7\sqrt{2}}{2}</math>. Thus, <math>\sin (BOZ) = \frac{\sqrt{2}}{10}</math> and <math>\cos (BOZ) = \frac{7\sqrt{2}}{10}</math>. As a result, <math>\sin (BOA) = \sin (2 BOZ) = 2\sin(BOZ)\cos(BOZ) = \frac{7}{25}</math>. <math>7 \cdot 25 = \boxed{175}</math>. | <cmath>DC^2 = (\sqrt{5})^2 + 5^2 -10\sqrt{5} \cdot \frac{2\sqrt{5}}{5} = 10</cmath>Therefore, <math>DC = \sqrt{10} \Longleftrightarrow BA = \sqrt{2}</math>. Now, drop the altitude from <math>O</math> to <math>BA</math> and call its intersection with <math>BA</math> <math>Z</math>. Then, by the Pythagorean theorem, <math>OZ = \frac{7\sqrt{2}}{2}</math>. Thus, <math>\sin (BOZ) = \frac{\sqrt{2}}{10}</math> and <math>\cos (BOZ) = \frac{7\sqrt{2}}{10}</math>. As a result, <math>\sin (BOA) = \sin (2 BOZ) = 2\sin(BOZ)\cos(BOZ) = \frac{7}{25}</math>. <math>7 \cdot 25 = \boxed{175}</math>. | ||
+ | == Solution 5== | ||
+ | |||
+ | [[Image:Dgram.png|thumb|none|800px]] | ||
+ | |||
+ | Let I be the intersection of AD and BC. | ||
+ | |||
+ | Lemma: <math>AI = ID</math> if and only if <math>\angle AIO = 90</math>. | ||
+ | |||
+ | Proof: If AI = ID, we get AO = OD, and thus IO is a perpendicular bisector of AD. If <math>\angle AIO = 90</math>, We can get <math>\triangle AIO \cong \triangle OID</math> | ||
+ | |||
+ | Let be this the circle with diameter AO. | ||
+ | |||
+ | Thus, we get <math>\angle AIO = 90</math>, implying I must lie on <math>\omega</math>. I also must lie on BC. If both of these conditions are satisfied, The extension of AI intersecting with circle O will create a point D such that AI = ID. Thus, since AD is the only chord starting at A, there must be only 1 possible I, implying the circle with diameter AO is tangent to BC. | ||
+ | |||
+ | Let Y be a point such that <math>\angle DOY = 90</math>, and minor arc DY contains A,B. | ||
+ | Let X be the center of <math>\omega</math>. Since <math>\omega</math>'s radius is <math>\frac{5}{2}</math>, the altitude from X to OY is 1.5, since the altitude from I to OY is 4. XO is 2.5, so <math>sin(XOY) = sin(AOY) = \frac{3}{5}</math>. <math>sin(BOD) = \frac{3}{5}</math>. If we let <math>sin(\theta) = \frac{3}{5}</math>, we can find that what we are looking for is <math>sin(90 - 2\theta)</math>, which we can evaluate and get <math>\frac{7}{25} \implies \boxed{175}</math> | ||
== See Also == | == See Also == | ||
{{AIME box|year=1983|num-b=14|after=Last question}} | {{AIME box|year=1983|num-b=14|after=Last question}} | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] |
Revision as of 12:39, 23 August 2020
Contents
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 central angle of minor arc is a rational number. If this number is expressed as a fraction in lowest terms, what is the product ?
Solution
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 at point .
Let be the midpoint of the chord . From right triangle , we have . This gives .
Notice that the distance equals , where is the radius of circle .
Hence (where represents the radius, , of the large circle given in the question). Therefore, since is clearly acute, we see that
Next, notice that . We can therefore apply the subtraction formula for to obtain It follows that , such that the answer is .
Solution 2
This solution, while similar to Solution 1, is far more motivated and less contrived.
Firstly, we note the statement in the problem that " is the only chord starting at and bisected by " – what is its significance? 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 scale factor and center . Thus, the locus is the result of the dilation with scale factor and centre of circle . Let the center of this circle be .
Now, 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 straightforward.
Our goal is to find , where is the midpoint of . We have and . Let be the projection of onto , and similarly let be the projection of onto . Then it remains to find so that we can use the addition formula for .
As is a radius of circle , , and similarly, . Since , we have . Thus .
Further, we see that is a dilation of about center with scale factor , so .
Lastly, we apply the formula: Thus the answer is .
Solution 3 (coordinate geometry)
Let the circle have equation , with centre . Since , we can calculate (by the Pythagorean Theorem) that the distance from to the line is . Therefore, we can let and . Now, assume that is any point on the major arc BC, and any point on the minor arc BC. We can write , where is the angle measured from the positive axis to the ray . It will also be convenient to define .
Firstly, since must lie in the minor arc , we see that . However, since the midpoint of must lie on , and the highest possible -coordinate of is , we see that the -coordinate cannot 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 they must be perpendicular. Therefore, suppose that is the intersection point of and , so that by the theorem, is perpendicular to . So, if is the only chord starting at which is bisected by , this means that is the only point on the chord such that is perpendicular to . Now suppose that , where . The fact that must be perpendicular to is equivalent to the following equation:
which becomes
This rearranges to
Given that this equation must have only one real root , we study the following function:
First, by the fact that the equation has real solutions, its discriminant must be non-negative, so we calculate
It is obvious that this is in fact non-negative. If it is actually zero, then , and . In this case, , so we have found a possible solution. We thus calculate by the subtraction formula for . This means that the answer is .
Addendum to Solution 3
Since this is an AIME problem, we can assume that we are done since we have found one possible case. However, in a full-solution contest, we could not assume that the answer is unique, but would need to prove that this is the unique solution. This can be proven as follows.
Suppose that , which would mean that there could be two real roots of , one lying in the interval , and another outside of it. We also see, by Vieta's Formulas, that the average of the two roots is , which is non-negative, so the root outside of must be no less than . By considering the graph of , which is a "U-shaped" parabola, it is now evident that and . We can just use the second inequality:
so
The only way for this inequality to be satisfied is when (by applying the Cauchy-Schwarz inequality, or just plotting the line to see that point can't be above this line), which does not make sense in the original problem statement. (For it would mean that the point lies in the half-plane above the line , inclusive, and the half-plane below the line , exclusive. This can be seen to be impossible by drawing the lines and observing that the intersection of the two half-planes does not share any point with the circle.)
Solution 4
Let the center of the circle be . Fix and . Then, as moves around the circle, the locus of the midpoints of is clearly a circle. Since the problems gives that is the only chord starting at bisected by , it follows that the circle with diameter and is tangent to .
Now, let the intersection of and be and let the midpoint of (the center of the circle tangent to that we described beforehand) be . Drop the altitude from to and call its intersection with . Drop the perpendicular from to and call its intersection with . Clearly, and since is radius, it equals . The same applies for , which also equals . By the Pythagorean theorem, we deduce that , so . This is very important information! Now we know that , so by Power of a Point, .
We’re almost there! Since by the Pythagorean theorem, , we deduce that . , so . Furthermore, since , we know that . By the law of cosines, Therefore, . Now, drop the altitude from to and call its intersection with . Then, by the Pythagorean theorem, . Thus, and . As a result, . .
Solution 5
Let I be the intersection of AD and BC.
Lemma: if and only if .
Proof: If AI = ID, we get AO = OD, and thus IO is a perpendicular bisector of AD. If , We can get
Let be this the circle with diameter AO.
Thus, we get , implying I must lie on . I also must lie on BC. If both of these conditions are satisfied, The extension of AI intersecting with circle O will create a point D such that AI = ID. Thus, since AD is the only chord starting at A, there must be only 1 possible I, implying the circle with diameter AO is tangent to BC.
Let Y be a point such that , and minor arc DY contains A,B. Let X be the center of . Since 's radius is , the altitude from X to OY is 1.5, since the altitude from I to OY is 4. XO is 2.5, so . . If we let , we can find that what we are looking for is , which we can evaluate and get
See Also
1983 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |