Difference between revisions of "2002 AMC 10B Problems/Problem 20"
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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. | 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)== | + | ===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) | 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) | ||
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== Video Solution== | == Video Solution== | ||
Revision as of 18:01, 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. 
