Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2017 AMC 10A Problems/Problem 17"

m (→‎Solution)
m (→‎Solution 2)
 
(33 intermediate revisions by 15 users not shown)
Line 2: Line 2:
 
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>\mathrm{\textbf{(A)}}\ 3\qquad\mathrm{\textbf{(B)}}\ 5\qquad\mathrm{\textbf{(C)}}\ 3\sqrt{5}\qquad\mathrm{\textbf{(D)}}\ 7\qquad\mathrm{\textbf{(E)}}\ 5\sqrt{2}</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==
+
==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 the square root of something, because they are both irrational. The greatest value of <math>PQ</math> happens when it <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. 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> 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>
+
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 00:42, 17 August 2023

Problem

Distinct points $P$, $Q$, $R$, $S$ lie on the circle $x^{2}+y^{2}=25$ and have integer coordinates. The distances $PQ$ and $RS$ are irrational numbers. What is the greatest possible value of the ratio $\frac{PQ}{RS}$?

$\textbf{(A) } 3 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 3\sqrt{5} \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 5\sqrt{2}$

Solution 1

Because $P$, $Q$, $R$, and $S$ are lattice points, there are only a few coordinates that actually satisfy the equation. The coordinates are $(\pm 3,\pm 4), (\pm 4, \pm 3), (0,\pm 5),$ and $(\pm 5,0).$ We want to maximize $PQ$ and minimize $RS.$ They also have to be non perfect squares, because they are both irrational. The greatest value of $PQ$ happens when $P$ and $Q$ are almost directly across from each other and are in different quadrants. For example, the endpoints of the segment could be $(-4,3)$ and $(3,-4)$ because the two points are almost across from each other. Another possible pair could be $(-4,3)$ and $(5,0)$. 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 $(-4,3)$ to $(4,-3)$. The distance between $(3,-4)$ and $(-4,3)$ is greater than the distance between $(5,0)$ and $(4,-3)$. Therefore, the segment from $(3,-4)$ to $(-4,3)$ is the longest attainable (the other possible coordinates for $P$ and $Q$ are $(4,3)$ and $(-3, -4)$, $(3, 4)$ and $(-4, -3)$, $(-3, 4)$ and $(4, -3)$. The least value of $RS$ is when the two endpoints are in the same quadrant and are very close to each other. This can occur when, for example, $R$ is $(3,4)$ and $S$ is $(4,3).$ They are in the same quadrant and no other point on the circle with integer coordinates is closer to the point $(3,4)$ than $(4,3)$ and vice versa. Using the distance formula, we get that $PQ$ is $\sqrt{98}$ and that $RS$ is $\sqrt{2}.$ $\frac{\sqrt{98}}{\sqrt{2}}=\sqrt{49}=\boxed{\mathrm{\textbf{(D)}}\ 7}$

Solution 2

We can look at the option choices. Since we are aiming for the highest possible ratio, let's try using $7$ (though $5 \sqrt{2}$ actually is the highest ratio.) Now, looking at the problem alone, we know that to have the largest ratio possible, we have to let $RS$ be the minimum possible value while at the same time using integer coordinates. Thus, the smallest possible value of $RS$ is $\sqrt{1^{2}+1^{2}} = \sqrt{2}$. Assuming that $\frac{PQ}{RS} = 7$, we plug in $RS = \sqrt{2}$ and solve for $PQ$: $PQ=7\sqrt{2}$. 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 $x$ and $y$ does $\sqrt{x^{2}+y^{2}}=7\sqrt2$? We see that this can easily be made into a $45-45-90$ triangle. But, instead of substituting $y=x$ into the equation and then using a whole lot of algebra, we can save time and use the little trick, that if in a $45-45-90$ triangle, the two $45$ degree sides have side length $s$, then the hypotenuse is $s\sqrt2$. Using this, we can see that $s=7$, and since our equation does in fact yield a sensible solution, we can be assured that our answer is $\boxed{\mathrm{\textbf{(D)}}\ 7}$.

Quality Control by fasterthanlight

(Note by Carrot_Karen: 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 $\textbf{(E)}$ 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 $\textbf{(D)}$ is the answer.

Solution 3

By inspection, when $R$ is at $(3, 4)$ and $S$ is at $(4, 3),$ it makes $RS$ as small as possible with a distance of $\sqrt{2}$. The greatest possible length of $PQ$ arises when $P$ is at $(-3, 4)$ and $Q$ is at $(4, -3).$ Using the distance formula, we find that $PQ$ has a length of $7\sqrt{2}.$ The requested fraction is then $\dfrac{PQ}{RS} = \dfrac{7\sqrt{2}}{\sqrt{2}} = \boxed{\mathrm{\textbf{(D)}}\ 7}$.

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_17&oldid=197091"