Difference between revisions of "2017 AMC 10A Problems/Problem 17"
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==Problem== | ==Problem== | ||
− | Distinct points <math>P</math>, <math>Q</math>, <math>R</math>, <math>S</math> lie on the circle <math>x^2+y^2=25</math> and have integer coordinates. The distances <math>PQ</math> and <math>RS</math> are irrational numbers. What is the greatest possible value of the ratio <math>\frac{PQ}{RS}</math>? | + | Distinct points <math>P</math>, <math>Q</math>, <math>R</math>, <math>S</math> lie on the circle <math>x^{2}+y^{2}=25</math> and have integer coordinates. The distances <math>PQ</math> and <math>RS</math> are irrational numbers. What is the greatest possible value of the ratio <math>\frac{PQ}{RS}</math>? |
+ | |||
+ | <math>\textbf{(A) } 3 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 3\sqrt{5} \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 5\sqrt{2}</math> | ||
+ | |||
+ | ==Solution 1== | ||
+ | |||
+ | Because <math>P</math>, <math>Q</math>, <math>R</math>, and <math>S</math> are lattice points, there are only a few coordinates that actually satisfy the equation. The coordinates are <math>(\pm 3,\pm 4), (\pm 4, \pm 3), (0,\pm 5),</math> and <math>(\pm 5,0).</math> We want to maximize <math>PQ</math> and minimize <math>RS.</math> They also have to be non perfect squares, because they are both irrational. The greatest value of <math>PQ</math> happens when <math>P</math> and <math>Q</math> are almost directly across from each other and are in different quadrants. For example, the endpoints of the segment could be <math>(-4,3)</math> and <math>(3,-4)</math> because the two points are almost across from each other. Another possible pair could be <math>(-4,3)</math> and <math>(5,0)</math>. 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 <math>(-4,3)</math> to <math>(4,-3)</math>. The distance between <math>(3,-4)</math> and <math>(-4,3)</math> is greater than the distance between <math>(5,0)</math> and <math>(4,-3)</math>. Therefore, the segment from <math>(3,-4)</math> to <math>(-4,3)</math> is the longest attainable (the other possible coordinates for <math>P</math> and <math>Q</math> are <math>(4,3)</math> and <math>(-3, -4)</math>, <math>(3, 4)</math> and <math>(-4, -3)</math>, <math>(-3, 4)</math> and <math>(4, -3)</math>. The least value of <math>RS</math> is when the two endpoints are in the same quadrant and are very close to each other. This can occur when, for example, <math>R</math> is <math>(3,4)</math> and <math>S</math> is <math>(4,3).</math> They are in the same quadrant and no other point on the circle with integer coordinates is closer to the point <math>(3,4)</math> than <math>(4,3)</math> and vice versa. Using the distance formula, we get that <math>PQ</math> is <math>\sqrt{98}</math> and that <math>RS</math> is <math>\sqrt{2}.</math> <math>\frac{\sqrt{98}}{\sqrt{2}}=\sqrt{49}=\boxed{\mathrm{\textbf{(D)}}\ 7}</math> | ||
+ | |||
+ | ==Solution 2== | ||
+ | We can look at the option choices. Since we are aiming for the highest possible ratio, let's try using <math>7</math> (though <math>5 \sqrt{2}</math> actually is the highest ratio.) Now, looking at the problem alone, we know that to have the largest ratio possible, we have to let <math>RS</math> be the minimum possible value while at the same time using integer coordinates. Thus, the smallest possible value of <math>RS</math> is <math>\sqrt{1^{2}+1^{2}} = \sqrt{2}</math>. Assuming that <math>\frac{PQ}{RS} = 7</math>, we plug in <math>RS = \sqrt{2}</math> and solve for <math>PQ</math>: <math>PQ=7\sqrt{2}</math>. 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 <math>x</math> and <math>y</math> does <math>\sqrt{x^{2}+y^{2}}=7\sqrt2</math>? We see that this can easily be made into a <math>45-45-90</math> triangle. But, instead of substituting <math>y=x</math> into the equation and then using a whole lot of algebra, we can save time and use the little trick, that if in a <math>45-45-90</math> triangle, the two <math>45</math> degree sides have side length <math>s</math>, then the hypotenuse is <math>s\sqrt2</math>. Using this, we can see that <math>s=7</math>, and since our equation does in fact yield a sensible solution, we can be assured that our answer is <math>\boxed{\mathrm{\textbf{(D)}}\ 7}</math>. | ||
+ | |||
+ | Quality Control by fasterthanlight | ||
+ | |||
+ | (Note by Carrot_Karen: We tried <math>7</math>, 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 the problem can only have one correct answer, 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. | ||
+ | |||
+ | ==Solution 3== | ||
+ | By inspection, when <math> R</math> is at <math>(3, 4)</math> and <math> S</math> is at <math>(4, 3),</math> it makes <math>RS</math> as small as possible with a distance of <math>\sqrt{2}</math>. The greatest possible length of <math>PQ</math> arises when <math> P</math> is at <math>(-3, 4)</math> and <math> Q</math> is at <math>(4, -3).</math> Using the distance formula, we find that <math>PQ</math> has a length of <math>7\sqrt{2}.</math> The requested fraction is then <math>\dfrac{PQ}{RS} = \dfrac{7\sqrt{2}}{\sqrt{2}} = \boxed{\mathrm{\textbf{(D)}}\ 7}</math>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/umr2Aj9ViOA?t=162 | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2017|ab=A|num-b=16|num-a=18}} | {{AMC10 box|year=2017|ab=A|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] |
Latest revision as of 01:42, 17 August 2023
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 non perfect squares, 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 greater than the distance between
and
. Therefore, the segment from
to
is the longest attainable (the other possible coordinates for
and
are
and
,
and
,
and
. 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
and vice versa. Using the distance formula, we get that
is
and that
is
Solution 2
We can look at the option choices. Since we are aiming for the highest possible ratio, let's try using (though
actually is the highest ratio.) 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
:
. 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 Carrot_Karen: We tried , 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 the problem can only have one correct answer, so if we have an option that already works, we can conclude that none of the others work and
is the answer.
Solution 3
By inspection, when is at
and
is at
it makes
as small as possible with a distance of
. The greatest possible length of
arises when
is at
and
is at
Using the distance formula, we find that
has a length of
The requested fraction is then
.
Video Solution
https://youtu.be/umr2Aj9ViOA?t=162
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.