Difference between revisions of "2010 AMC 12A Problems/Problem 13"
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<math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 8</math> | <math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 8</math> | ||
− | == Solution == | + | == Solution 1== |
The image below shows the two curves for <math>k=4</math>. The blue curve is <math>x^2+y^2=k^2</math>, which is clearly a circle with radius <math>k</math>, and the red curve is a part of the curve <math>xy=k</math>. | The image below shows the two curves for <math>k=4</math>. The blue curve is <math>x^2+y^2=k^2</math>, which is clearly a circle with radius <math>k</math>, and the red curve is a part of the curve <math>xy=k</math>. | ||
Line 49: | Line 49: | ||
After multiplying through by <math> x^2 </math> and rearranging, we find <math> x^4-x^2k^2+k^2 \leq 0 </math>. | After multiplying through by <math> x^2 </math> and rearranging, we find <math> x^4-x^2k^2+k^2 \leq 0 </math>. | ||
We see this is a quadratic in <math> x^2 </math> and consider taking the determinant, which tells us that solutions are real when, after factoring: | We see this is a quadratic in <math> x^2 </math> and consider taking the determinant, which tells us that solutions are real when, after factoring: | ||
− | <math> k^2(k^2-4) \ | + | <math> k^2(k^2-4) \geq 0 </math> |
We plot this inequality on the number line to find it is satisfied for all values except: <math> (-1, 0, 1) </math>. | We plot this inequality on the number line to find it is satisfied for all values except: <math> (-1, 0, 1) </math>. | ||
We then eliminate 0 because it is extraneous as both <math> xy=0 </math> and <math> x^2+y^2=0 </math> are points which coincide. | We then eliminate 0 because it is extraneous as both <math> xy=0 </math> and <math> x^2+y^2=0 </math> are points which coincide. | ||
Therefore, there are a total of <math>\boxed{2\ \textbf{(C)}}</math> values. | Therefore, there are a total of <math>\boxed{2\ \textbf{(C)}}</math> values. | ||
− | == | + | == Solution 3 (Algebra) == |
+ | |||
+ | Since <math>xy=k</math>, multiply the equation by 2 on both sides to get <math>2xy=2k</math>. Now we can add the two equations to get <math>(x+y)^2=k^2+2k</math>, for which the only value of <math>k</math> that does not satisfy the equation is <math>-1</math>, as that makes the RHS negative. Similarly, if we subtract the two equations, we obtain <math>(x-y)^2=k^2-2k</math>, for which the only value of <math>k</math> that does not satisfy the equation is <math>1</math>, for the same reason above. | ||
+ | |||
+ | Thus, the only values are <math>k = 1, -1</math>, giving us a total of <math>\boxed{2\ \textbf{(C)}}</math> values. | ||
+ | |||
+ | ~ ccx09 (Roy Short) | ||
+ | |||
+ | == Solution 4 (Quick) == | ||
+ | Multiply <math>k=xy</math> by and substitute it into <math>k^2=x^2+y^2</math>. Then, <math>k=\frac{x^2+y^2}{xy}</math>. Recognize it? It's also <math>k=\frac{x}{y}+\frac{y}{x}</math>. The minimum of this function (more accurately the minimum absolute value of the function) is k=2, -2 (when x=y or x=-y). As long as k>2 or k<-2, the function is valid. As such, <math>k\neq1,-1 \implies \boxed{2\ \textbf{(C)}}</math>. Elegant, huh? | ||
+ | |||
+ | ~~BJHHar | ||
+ | |||
+ | == Solution 5== | ||
+ | |||
+ | Assume that <math>k\ge 0</math> since if <math>k</math> works then <math>-k</math> also works. Let <math>x = ka</math> and <math>y = kb</math>. Then the given equations become <math>ab = \frac{1}{k}</math> and <math>a^2 + b^2 = 1</math>, which we don't want to intersect. The points that are the closest on this graph are <math>(1/\sqrt{2}, 1/\sqrt{2})</math> and <math>(1/\sqrt{k})(1/\sqrt{k})</math>. We don't want the hyperbola to intersect or go inside the circle, so we require <math>1/\sqrt{k} > 1/\sqrt{2}</math>, so <math>k< 2\implies k = 0, 1</math>. Also, by symmetry <math>k = -1</math> also works.Obviously <math>k = 0</math> doesn't work, so we can discard that, leaving <math>\boxed{\mathbf{(C)}\quad 2}</math> integral solutions for <math>k</math>. | ||
{{AMC12 box|year=2010|num-b=12|num-a=14|ab=A}} | {{AMC12 box|year=2010|num-b=12|num-a=14|ab=A}} | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 15:55, 18 February 2022
Contents
Problem
For how many integer values of do the graphs of and not intersect?
Solution 1
The image below shows the two curves for . The blue curve is , which is clearly a circle with radius , and the red curve is a part of the curve .
In the special case the blue curve is just the point , and as , this point is on the red curve as well, hence they intersect.
The case is symmetric to : the blue curve remains the same and the red curve is flipped according to the axis. Hence we just need to focus on .
Clearly, on the red curve there will always be points arbitrarily far from the origin: for example, as approaches 0, approaches . Hence the red curve intersects the blue one if and only if it contains a point whose distance from the origin is at most .
At this point we can guess that on the red curve the point where is always closest to the origin, and skip the rest of this solution.
For an exact solution, fix and consider any point on the red curve. Its distance from the origin is . To minimize this distance, it is enough to minimize . By the Arithmetic Mean-Geometric Mean Inequality we get that this value is at least , and that equality holds whenever , i.e., .
Now recall that the red curve intersects the blue one if and only if its closest point is at most from the origin. We just computed that the distance between the origin and the closest point on the red curve is . Therefore, we want to find all positive integers such that .
Clearly the only such integer is , hence the two curves are only disjoint for and . This is a total of values.
Solution 2
From the graph shown above, we see that there is a specific point closest to the center of the circle. Using some logic, we realize that as long as said furthest point is not inside or on the graph of the circle. This should be enough to conclude that the hyperbola does not intersect the circle.
Therefore, for each value of k, we only need to check said value to determine intersection. Let said point, closest to the circle have coordinates derived from the equation. Then, all coordinates that satisfy intersect the circle. Squaring, we find After multiplying through by and rearranging, we find . We see this is a quadratic in and consider taking the determinant, which tells us that solutions are real when, after factoring: We plot this inequality on the number line to find it is satisfied for all values except: . We then eliminate 0 because it is extraneous as both and are points which coincide. Therefore, there are a total of values.
Solution 3 (Algebra)
Since , multiply the equation by 2 on both sides to get . Now we can add the two equations to get , for which the only value of that does not satisfy the equation is , as that makes the RHS negative. Similarly, if we subtract the two equations, we obtain , for which the only value of that does not satisfy the equation is , for the same reason above.
Thus, the only values are , giving us a total of values.
~ ccx09 (Roy Short)
Solution 4 (Quick)
Multiply by and substitute it into . Then, . Recognize it? It's also . The minimum of this function (more accurately the minimum absolute value of the function) is k=2, -2 (when x=y or x=-y). As long as k>2 or k<-2, the function is valid. As such, . Elegant, huh?
~~BJHHar
Solution 5
Assume that since if works then also works. Let and . Then the given equations become and , which we don't want to intersect. The points that are the closest on this graph are and . We don't want the hyperbola to intersect or go inside the circle, so we require , so . Also, by symmetry also works.Obviously doesn't work, so we can discard that, leaving integral solutions for .
2010 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
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