Difference between revisions of "2017 AMC 10A Problems/Problem 17"
Carrot karen (talk | contribs) (→Solution 2) |
Carrot karen (talk | contribs) m (→Solution 2) |
||
Line 13: | Line 13: | ||
Quality Control by fasterthanlight | Quality Control by fasterthanlight | ||
− | (Note by CarrotKaren: We tried 7, but some might be confused why we concluded that it was the answer after verifying without trying the others, like why wasn't option <math>E</math> tried? This is because there can only be one answer in the problem that works, so if we have an option that already works, we can conclude that none of the others work and <math>D</math> is the answer. | + | (Note by CarrotKaren: We tried 7, but some might be confused why we concluded that it was the answer after verifying without trying the others, like why wasn't option <math>\textbf{(E)}</math> tried? This is because there can only be one answer in the problem that works, so if we have an option that already works, we can conclude that none of the others work and <math>\textbf{(D)}</math> is the answer. |
==Video Solution== | ==Video Solution== |
Revision as of 19:07, 23 December 2020
Problem
Distinct points ,
,
,
lie on the circle
and have integer coordinates. The distances
and
are irrational numbers. What is the greatest possible value of the ratio
?
Solution 1
Because ,
,
, and
are lattice points, there are only a few coordinates that actually satisfy the equation. The coordinates are
and
We want to maximize
and minimize
They also have to be the square root of something, because they are both irrational. The greatest value of
happens when
and
are almost directly across from each other and are in different quadrants. For example, the endpoints of the segment could be
and
because the two points are almost across from each other. Another possible pair could be
and
. To find out which segment is longer, we have to compare the distances from their endpoints to a diameter (which must be the longest possible segment). The closest diameter would be from
to
. The distance between
and
is shorter than the distance between
and
. Therefore, the segment from
to
is the longest attainable. The least value of
is when the two endpoints are in the same quadrant and are very close to each other. This can occur when, for example,
is
and
is
They are in the same quadrant and no other point on the circle with integer coordinates is closer to the point
than
Using the distance formula, we get that
is
and that
is
Solution 2
So what we can do is look at the option choices. Since we are aiming for the highest possible ratio, let's try using . Now, looking at the problem alone, we know that to have the largest ratio possible, we have to let
be the minimum possible value while at the same time using integer coordinates. Thus, the smallest possible value of
is
. Assuming that
, we plug in
and solve for PQ:
. Remember, we don't know if this is possible yet, we are only trying to figure out if it is. But for what values of
and
does
? We see that this can easily be made into a
triangle. But, instead of substituting
into the equation and then using a whole lot of algebra, we can save time and use the little trick, that if in a
triangle, the two
degree sides have side length
, then the hypotenuse is
. Using this, we can see that
, and since our equation does in fact yield a sensible solution, we can be assured that our answer is
.
Quality Control by fasterthanlight
(Note by CarrotKaren: We tried 7, but some might be confused why we concluded that it was the answer after verifying without trying the others, like why wasn't option tried? This is because there can only be one answer in the problem that works, so if we have an option that already works, we can conclude that none of the others work and
is the answer.
Video Solution
See Also
2017 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.