Difference between revisions of "1997 AIME Problems/Problem 12"
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== Solution == | == Solution == | ||
=== Solution 1 === | === Solution 1 === | ||
− | First, we use the fact that <math>f(f(x)) = x</math> for all <math>x</math> in the domain. Substituting the function definition, we have <math> | + | First, we use the fact that <math>f(f(x)) = x</math> for all <math>x</math> in the domain. Substituting the function definition, we have <math>\frac {a\frac {ax + b}{cx + d} + b}{c\frac {ax + b}{cx + d} + d} = x</math>, which reduces to |
− | <cmath> | + | <cmath>\frac {(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)} =\frac {px + q}{rx + s} = x. </cmath> |
− | \frac {px + q}{rx + s} = x | ||
In order for this fraction to reduce to <math>x</math>, we must have <math>q = r = 0</math> and <math>p = s\not = 0</math>. From <math>c(a + d) = b(a + d) = 0</math>, we get <math>a = - d</math> or <math>b = c = 0</math>. The second cannot be true, since we are given that <math>a,b,c,d</math> are nonzero. This means <math>a = - d</math>, so <math>f(x) = \frac {ax + b}{cx - a}</math>. | In order for this fraction to reduce to <math>x</math>, we must have <math>q = r = 0</math> and <math>p = s\not = 0</math>. From <math>c(a + d) = b(a + d) = 0</math>, we get <math>a = - d</math> or <math>b = c = 0</math>. The second cannot be true, since we are given that <math>a,b,c,d</math> are nonzero. This means <math>a = - d</math>, so <math>f(x) = \frac {ax + b}{cx - a}</math>. | ||
The only value that is not in the range of this function is <math>\frac {a}{c}</math>. To find <math>\frac {a}{c}</math>, we use the two values of the function given to us. We get <math>2(97)a + b = 97^2c</math> and <math>2(19)a + b = 19^2c</math>. Subtracting the second equation from the first will eliminate <math>b</math>, and this results in <math>2(97 - 19)a = (97^2 - 19^2)c</math>, so | The only value that is not in the range of this function is <math>\frac {a}{c}</math>. To find <math>\frac {a}{c}</math>, we use the two values of the function given to us. We get <math>2(97)a + b = 97^2c</math> and <math>2(19)a + b = 19^2c</math>. Subtracting the second equation from the first will eliminate <math>b</math>, and this results in <math>2(97 - 19)a = (97^2 - 19^2)c</math>, so | ||
− | <cmath> | + | <cmath>\frac {a}{c} = \frac {(97 - 19)(97 + 19)}{2(97 - 19)} = 58 . </cmath> |
Alternatively, we could have found out that <math>a = -d</math> by using the fact that <math>f(f(-b/a))=-b/a</math>. | Alternatively, we could have found out that <math>a = -d</math> by using the fact that <math>f(f(-b/a))=-b/a</math>. | ||
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<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | + | 19 &= \frac{b - \frac da}{19 - e} + e\\ | |
97 &= \frac{b - \frac da}{97 - e} + e\\ | 97 &= \frac{b - \frac da}{97 - e} + e\\ | ||
− | b - \frac da &= ( | + | b - \frac da &= (19 - e)^2 = (97 - e)^2\\ |
− | + | 19 - e &= \pm (97 - e) | |
\end{align*}</cmath> | \end{align*}</cmath> | ||
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We may represent the real number <math>x/y</math> as | We may represent the real number <math>x/y</math> as | ||
− | <math>\begin{pmatrix}x \\ y\end{pmatrix}</math>, with two such [[column vectors]] | + | <math>\begin{pmatrix}x \\ y\end{pmatrix}</math>, with two such [[vector|column vectors]] |
considered equivalent if they are scalar multiples of each other. Similarly, | considered equivalent if they are scalar multiples of each other. Similarly, | ||
we can represent a function <math>F(x) = \frac{Ax + B}{Cx + D}</math> as a matrix | we can represent a function <math>F(x) = \frac{Ax + B}{Cx + D}</math> as a matrix | ||
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<cmath> \frac{c x^2 + (d-a) x - b}{cx+d} = g(x) = \frac{c(x-19)(x-97)}{cx+d}. </cmath> | <cmath> \frac{c x^2 + (d-a) x - b}{cx+d} = g(x) = \frac{c(x-19)(x-97)}{cx+d}. </cmath> | ||
Thus <math>-19 - 97 = \frac{d-a}{c} = -\frac{2a}{c}</math>, so <math>a/c = (19+97)/2 = 58</math>, | Thus <math>-19 - 97 = \frac{d-a}{c} = -\frac{2a}{c}</math>, so <math>a/c = (19+97)/2 = 58</math>, | ||
− | our answer. | + | our answer. |
=== Solution 4 === | === Solution 4 === | ||
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This solution follows in the same manner as the last paragraph of the first solution. | This solution follows in the same manner as the last paragraph of the first solution. | ||
+ | |||
+ | === Solution 5 === | ||
+ | Since <math>f(f(x))</math> is <math>x</math>, it must be symmetric across the line <math>y=x</math>. Also, since <math>f(19)=19</math>, it must touch the line <math>y=x</math> at <math>(19,19)</math> and <math>(97,97)</math>. <math>f</math> a hyperbola that is a scaled and transformed version of <math>y=\frac{1}{x}</math>. Write <math>f(x)= \frac{ax+b}{cx+d}</math> as <math>\frac{y}{cx+d}+z</math>, and z is our desired answer <math>\frac{a}{c}</math>. Take the basic hyperbola, <math>y=\frac{1}{x}</math>. The distance between points <math>(1,1)</math> and <math>(-1,-1)</math> is <math>2\sqrt{2}</math>, while the distance between <math>(19,19)</math> and <math>(97,97)</math> is <math>78\sqrt{2}</math>, so it is <math>y=\frac{1}{x}</math> scaled by a factor of <math>39</math>. Then, we will need to shift it from <math>(-39,-39)</math> to <math>(19,19)</math>, shifting up by <math>58</math>, or <math>z</math>, so our answer is <math>\boxed{58}</math>. Note that shifting the <math>x</math> does not require any change from <math>z</math>; it changes the denominator of the part <math>\frac{1}{x-k}</math>. | ||
+ | |||
+ | === Solution 6 (Short) === | ||
+ | |||
+ | From <math>f(f(x))=x</math>, it is obvious that <math>\frac{-d}{c}</math> is the value not in the range. First notice that since <math>f(0)=\frac{b}{d}</math>, <math>f(\frac{b}{d})=0</math> which means <math>a(\frac{b}{d})+b=0</math> so <math>a=-d</math>. Using <math>f(19)=19</math>, we have that <math>b=361c+38d</math>; on <math>f(97)=97</math> we obtain <math>b=9409c+194d</math>. Solving for <math>d</math> in terms of <math>c</math> leads us to <math>d=-58c</math>, so the answer is <math>\boxed{058}</math>. | ||
+ | - solution by mathleticguyyy | ||
+ | |||
+ | === Solution 7=== | ||
+ | Begin by finding the inverse function of <math>f(x)</math>, which turns out to be <math>f^{-1}(x)=\frac{19d-b}{a-19c}</math>. Since <math>f(f(x))=x</math>, <math>f(x)=f^{-1}(x)</math>, so substituting 19 and 97 yields the system, <math>\begin{array}{lcl} \frac{19a+b}{19c+d} & = & \frac{19d-b}{a-19c} \\ \frac{97a+b}{97c+d} & = & \frac{97d-b}{a-97c} \end{array}</math>, and after multiplying each equation out and subtracting equation 1 from 2, and after simplifying, you will get <math>116c=a-d</math>. Coincidentally, then <math>116c+d=a</math>, which is familiar because <math>f(116)=\frac{116a+b}{116c+d}</math>, and since <math>116c+d=a</math>, <math>f(116)=\frac{116a+b}{a}</math>. Also, <math>f(f(116))=\frac{a(\frac{116a+b}{a})+b}{c(\frac{116a+b}{a})+d}=116</math>, due to <math>f(f(x))=x</math>. This simplifies to <math>\frac{116a+2b}{c(\frac{116a+b}{a})+d}=116</math>, <math>116a+2b=116(c(\frac{116a+b}{a})+d)</math>, <math>116a+2b=116(c(116+\frac{b}{a})+d)</math>, <math>116a+2b=116c(116+\frac{b}{a})+116d</math>, and substituting <math>116c+d=a</math> and simplifying, you get <math>2b=116c(\frac{b}{a})</math>, then <math>\frac{a}{c}=58</math>. Looking at <math>116c=a-d</math> one more time, we get <math>116=\frac{a}{c}+\frac{-d}{c}</math>, and substituting, we get <math>\frac{-d}{c}=\boxed{58}</math>, and we are done. | ||
+ | |||
+ | === Solution 8 (shorter than solution 6) === | ||
+ | Because there are no other special numbers other than <math>19</math> and <math>97</math>, take the average to get <math>\boxed{58}</math>. (Note I solved this problem the solution one way but noticed this and this probably generalizes to all <math>f(x)=x, f(y)=y</math> questions like these) | ||
== See also == | == See also == | ||
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[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 12:30, 9 August 2020
Problem
The function defined by , where ,, and are nonzero real numbers, has the properties , and for all values except . Find the unique number that is not in the range of .
Contents
Solution
Solution 1
First, we use the fact that for all in the domain. Substituting the function definition, we have , which reduces to In order for this fraction to reduce to , we must have and . From , we get or . The second cannot be true, since we are given that are nonzero. This means , so .
The only value that is not in the range of this function is . To find , we use the two values of the function given to us. We get and . Subtracting the second equation from the first will eliminate , and this results in , so
Alternatively, we could have found out that by using the fact that .
Solution 2
First, we note that is the horizontal asymptote of the function, and since this is a linear function over a linear function, the unique number not in the range of will be . . Without loss of generality, let , so the function becomes .
(Considering as a limit) By the given, . , so . as reaches the vertical asymptote, which is at . Hence . Substituting the givens, we get
Clearly we can discard the positive root, so .
Solution 3
We first note (as before) that the number not in the range of is , as is evidently never 0 (otherwise, would be a constant function, violating the condition ).
We may represent the real number as , with two such column vectors considered equivalent if they are scalar multiples of each other. Similarly, we can represent a function as a matrix . Function composition and evaluation then become matrix multiplication.
Now in general, In our problem . It follows that for some nonzero real . Since it follows that . (In fact, this condition condition is equivalent to the condition that for all in the domain of .)
We next note that the function evaluates to 0 when equals 19 and 97. Therefore Thus , so , our answer.
Solution 4
Any number that is not in the domain of the inverse of cannot be in the range of . Starting with , we rearrange some things to get . Clearly, is the number that is outside the range of .
Since we are given , we have that
All the quadratic terms, linear terms, and constant terms must be equal on both sides for this to be a true statement so we have that .
This solution follows in the same manner as the last paragraph of the first solution.
Solution 5
Since is , it must be symmetric across the line . Also, since , it must touch the line at and . a hyperbola that is a scaled and transformed version of . Write as , and z is our desired answer . Take the basic hyperbola, . The distance between points and is , while the distance between and is , so it is scaled by a factor of . Then, we will need to shift it from to , shifting up by , or , so our answer is . Note that shifting the does not require any change from ; it changes the denominator of the part .
Solution 6 (Short)
From , it is obvious that is the value not in the range. First notice that since , which means so . Using , we have that ; on we obtain . Solving for in terms of leads us to , so the answer is .
- solution by mathleticguyyy
Solution 7
Begin by finding the inverse function of , which turns out to be . Since , , so substituting 19 and 97 yields the system, , and after multiplying each equation out and subtracting equation 1 from 2, and after simplifying, you will get . Coincidentally, then , which is familiar because , and since , . Also, , due to . This simplifies to , , , , and substituting and simplifying, you get , then . Looking at one more time, we get , and substituting, we get , and we are done.
Solution 8 (shorter than solution 6)
Because there are no other special numbers other than and , take the average to get . (Note I solved this problem the solution one way but noticed this and this probably generalizes to all questions like these)
See also
1997 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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