Difference between revisions of "2002 AMC 10B Problems/Problem 20"
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− | == Problem == | + | ==Problem== |
− | + | Let <math>a</math>, <math>b</math>, and <math>c</math> 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 == | + | ==Solutions== |
+ | ===Solution 1=== | ||
Rearranging, we get <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> | ||
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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>. | 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 == | + | ===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>-7b + 8c = 3</math> and | ||
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<math>4b - c = -1</math> | <math>4b - c = -1</math> | ||
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Multiplying the second equation by <math>8</math>, we have | Multiplying the second equation by <math>8</math>, we have | ||
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<math>32b - 8c = -8</math> | <math>32b - 8c = -8</math> | ||
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Adding up the two equations yields | Adding up the two equations yields | ||
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<math>25b = -5</math>, so <math>b = -\frac{1}{5}</math> | <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>. | 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. | ||
− | + | ===Solution 3 (fakesolve)=== | |
+ | Notice that the coefficients of <math>a</math> and <math>c</math> are pretty similar (15s for reading and noticing), so let <math>b=0</math> gives <math>a+8c=4</math>, and <math>8a-c=7</math> (10s writing). Since the desired quantity simplifies to <math>a^2+c^2</math>, the <math>ac</math> term of the quadratics after squaring gets canceled by adding up the squares of the two equations because they have the same coefficients but opposite sign (15s mind-binom). This simplifies to <math>65(a^2+c^2)=16+49</math>, or <math>a^2-b^2+c^2=\frac{65}{65}=\boxed{1}</math>(15s writing and addition and fraction simplification and (B) circling and submission) | ||
− | + | ===Video Solution=== | |
+ | https://www.youtube.com/watch?v=3Oq21r5OezA ~David | ||
==See Also== | ==See Also== |
Latest revision as of 19:02, 14 October 2023
Contents
Problem
Let ,
, and
be real numbers such that
and
. Then
is
Solutions
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.
Solution 3 (fakesolve)
Notice that the coefficients of and
are pretty similar (15s for reading and noticing), so let
gives
, and
(10s writing). Since the desired quantity simplifies to
, the
term of the quadratics after squaring gets canceled by adding up the squares of the two equations because they have the same coefficients but opposite sign (15s mind-binom). This simplifies to
, or
(15s writing and addition and fraction simplification and (B) circling and submission)
Video Solution
https://www.youtube.com/watch?v=3Oq21r5OezA ~David
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.