Difference between revisions of "2002 AMC 10B Problems/Problem 20"
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− | == Problem == | + | ==Problem== |
− | |||
Let a, b, and c be real numbers such that <math>a-7b+8c=4</math> and <math>8a+4b-c=7</math>. Then <math>a^2-b^2+c^2</math> is | Let a, b, and c be real numbers such that <math>a-7b+8c=4</math> and <math>8a+4b-c=7</math>. Then <math>a^2-b^2+c^2</math> is | ||
<math> \mathrm{(A)\ }0\qquad\mathrm{(B)\ }1\qquad\mathrm{(C)\ }4\qquad\mathrm{(D)\ }7\qquad\mathrm{(E)\ }8 </math> | <math> \mathrm{(A)\ }0\qquad\mathrm{(B)\ }1\qquad\mathrm{(C)\ }4\qquad\mathrm{(D)\ }7\qquad\mathrm{(E)\ }8 </math> | ||
− | == Solution == | + | ==Solution== |
+ | ===Solution 1=== | ||
− | <math>a+8c=7b+4</math> and <math>8a-c=7-4b</math> | + | Rearranging, we get <math>a+8c=7b+4</math> and <math>8a-c=7-4b</math> |
Squaring both, <math>a^2+16ac+64c^2=49b^2+56b+16</math> and <math>64a^2-16ac+c^2=16b^2-56b+49</math> are obtained. | Squaring both, <math>a^2+16ac+64c^2=49b^2+56b+16</math> and <math>64a^2-16ac+c^2=16b^2-56b+49</math> are obtained. | ||
− | Adding the two equations and dividing by <math>65</math> gives <math>a^2+c^2=b^2+1</math>, so <math>a^2-b^2+c^2= | + | Adding the two equations and dividing by <math>65</math> gives <math>a^2+c^2=b^2+1</math>, so <math>a^2-b^2+c^2=\boxed{(\text{B})1}</math>. |
− | + | ||
+ | ===Solution 2=== | ||
+ | The easiest way is to assume a value for <math>a</math> and then solve the system of equations. For <math>a = 1</math>, we get the equations | ||
+ | <math>-7b + 8c = 3</math> and | ||
+ | <math>4b - c = -1</math> | ||
+ | Multiplying the second equation by <math>8</math>, we have | ||
+ | <math>32b - 8c = -8</math> | ||
+ | Adding up the two equations yields | ||
+ | <math>25b = -5</math>, so <math>b = -\frac{1}{5}</math> | ||
+ | We obtain <math>c = \frac{1}{5}</math> after plugging in the value for <math>b</math>. | ||
+ | Therefore, <math>a^2-b^2+c^2 = 1-\frac{1}{25}+\frac{1}{25}=\boxed{1}</math> which corresponds to <math>\text{(B)}</math>. | ||
+ | This time-saving trick works only because we know that for any value of <math>a</math>, <math>a^2-b^2+c^2</math> will always be constant (it's a contest), so any value of <math>a</math> will work. This is also called [[without loss of generality]] or WLOG. | ||
==See Also== | ==See Also== | ||
− | {{AMC10 box|year=2002|ab=B| | + | {{AMC10 box|year=2002|ab=B|num-b=19|num-a=21}} |
− | + | {{MAA Notice}} |
Revision as of 20:59, 10 August 2022
Problem
Let a, b, and c be real numbers such that and . Then is
Solution
Solution 1
Rearranging, we get and
Squaring both, and are obtained.
Adding the two equations and dividing by gives , so .
Solution 2
The easiest way is to assume a value for and then solve the system of equations. For , we get the equations and Multiplying the second equation by , we have Adding up the two equations yields , so We obtain after plugging in the value for . Therefore, which corresponds to . This time-saving trick works only because we know that for any value of , will always be constant (it's a contest), so any value of will work. This is also called without loss of generality or WLOG.
See Also
2002 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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.