Difference between revisions of "1976 AHSME Problems/Problem 27"
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==Solution 1== | ==Solution 1== | ||
− | Let <math>x=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}</math> and <math>y=\sqrt{3-2\sqrt{2}}.</math> | + | Let <math>x=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}</math> and <math>y=\sqrt{3-2\sqrt{2}}.</math> |
Note that | Note that | ||
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&=\frac{\left(\sqrt{5}+2\right)+2+\left(\sqrt{5}-2\right)}{\sqrt{5}+1} \\ | &=\frac{\left(\sqrt{5}+2\right)+2+\left(\sqrt{5}-2\right)}{\sqrt{5}+1} \\ | ||
&=\frac{2\sqrt{5}+2}{\sqrt{5}+1} \\ | &=\frac{2\sqrt{5}+2}{\sqrt{5}+1} \\ | ||
− | &=2 | + | &=2. |
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | + | Since <math>x>0,</math> we have <math>x=\sqrt{2}.</math> | |
On the other hand, note that | On the other hand, note that | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | y&= | + | y^2&=3-2\sqrt{2} \\ |
− | &= | + | &=2-2\sqrt{2}+1 \\ |
− | &= | + | &=\left(\sqrt{2}-1\right)^2. |
− | |||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | Finally, the answer is < | + | Since <math>y>0,</math> we have <math>y=\sqrt{2}-1.</math> |
+ | |||
+ | Finally, the answer is <cmath>N=x-y=\boxed{\textbf{(A) }1}.</cmath> | ||
~Someonenumber011 (Fundamental Logic) | ~Someonenumber011 (Fundamental Logic) | ||
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==Solution 2== | ==Solution 2== | ||
− | Let <math>x=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}</math> and <math>y=\sqrt{3-2\sqrt{2}}.</math> | + | Let <math>x=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}</math> and <math>y=\sqrt{3-2\sqrt{2}}.</math> |
Note that | Note that | ||
Line 41: | Line 42: | ||
We rewrite each term in the numerator separately: | We rewrite each term in the numerator separately: | ||
<ol style="margin-left: 1.5em;"> | <ol style="margin-left: 1.5em;"> | ||
− | <li>Let <math>\sqrt{a}+\sqrt{b}=\sqrt{3+\sqrt{5}}</math> for some nonnegative rational numbers <math>a</math> and <math>b.</math> We square both sides of this equation, then | + | <li>Let <math>\sqrt{a}+\sqrt{b}=\sqrt{3+\sqrt{5}}</math> for some nonnegative rational numbers <math>a</math> and <math>b.</math> We square both sides of this equation, then rearrange: |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
a+b+2\sqrt{ab}&=3+\sqrt{5} \\ | a+b+2\sqrt{ab}&=3+\sqrt{5} \\ | ||
Line 57: | Line 58: | ||
On the other hand, we have <cmath>y=\sqrt2-1</cmath> by the argument of either Solution 1 or Solution 2. | On the other hand, we have <cmath>y=\sqrt2-1</cmath> by the argument of either Solution 1 or Solution 2. | ||
− | Finally, the answer is < | + | Finally, the answer is <cmath>N=x-y=\boxed{\textbf{(A) }1}.</cmath> |
~MRENTHUSIASM | ~MRENTHUSIASM |
Latest revision as of 13:41, 8 September 2021
Contents
Problem
If then
equals
Solution 1
Let and
Note that
Since
we have
On the other hand, note that
Since
we have
Finally, the answer is
~Someonenumber011 (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Solution 2
Let and
Note that
We rewrite each term in the numerator separately:
- Let
for some nonnegative rational numbers
and
We square both sides of this equation, then rearrange:
It follows that
By inspection, we get
Alternatively, we conclude that
and
are the solutions to the quadratic equation
by Vieta's Formulas, in which
Therefore, we obtain
- Similarly, we obtain
Substituting these results into we have
On the other hand, we have
by the argument of either Solution 1 or Solution 2.
Finally, the answer is
~MRENTHUSIASM
See also
1976 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 26 |
Followed by Problem 28 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.