Difference between revisions of "1976 AHSME Problems/Problem 27"
MRENTHUSIASM (talk | contribs) m |
MRENTHUSIASM (talk | contribs) m |
||
(16 intermediate revisions by the same user not shown) | |||
Line 8: | Line 8: | ||
\textbf{(E) }\text{none of these} </math> | \textbf{(E) }\text{none of these} </math> | ||
− | ==Solution== | + | ==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> Note that | + | 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 | ||
+ | <cmath>\begin{align*} | ||
+ | x^2&=\frac{\left(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}\right)^2}{\left(\sqrt{\sqrt{5}+1}\right)^2} \\ | ||
+ | &=\frac{\left(\sqrt{5}+2\right)+2\cdot\left(\sqrt{\sqrt{5}+2}\cdot\sqrt{\sqrt{5}-2}\right)+\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} \\ | ||
+ | &=2. | ||
+ | \end{align*}</cmath> | ||
+ | Since <math>x>0,</math> we have <math>x=\sqrt{2}.</math> | ||
+ | |||
+ | On the other hand, note that | ||
+ | <cmath>\begin{align*} | ||
+ | y^2&=3-2\sqrt{2} \\ | ||
+ | &=2-2\sqrt{2}+1 \\ | ||
+ | &=\left(\sqrt{2}-1\right)^2. | ||
+ | \end{align*}</cmath> | ||
+ | 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) | ||
+ | |||
+ | ~MRENTHUSIASM (Reconstruction) | ||
+ | |||
+ | ==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> | ||
+ | |||
+ | Note that | ||
+ | <cmath>x=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\cdot\frac{\sqrt{\sqrt{5}-1}}{\sqrt{\sqrt{5}-1}}=\frac{\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}}{2}. \hspace{15mm} (\bigstar)</cmath> | ||
+ | We rewrite each term in the numerator separately: | ||
+ | <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 rearrange: | ||
+ | <cmath>\begin{align*} | ||
+ | a+b+2\sqrt{ab}&=3+\sqrt{5} \\ | ||
+ | a+b+\sqrt{4ab}&=3+\sqrt{5}. | ||
+ | \end{align*}</cmath> | ||
+ | It follows that | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | + | a+b&=3, \\ | |
+ | ab&=\frac54. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
+ | By inspection, we get <math>\{a,b\}=\left\{\frac12,\frac52\right\}.</math> Alternatively, we conclude that <math>a</math> and <math>b</math> are the solutions to the quadratic equation <math>t^2-3t+\frac54=0</math> by Vieta's Formulas, in which <math>t=\frac12,\frac52.</math> <p> Therefore, we obtain <cmath>\sqrt{3+\sqrt{5}}=\sqrt{\frac12}+\sqrt{\frac52}=\frac{\sqrt2}{2}+\frac{\sqrt{10}}{2}.</cmath></li><p> | ||
+ | <li>Similarly, we obtain <cmath>\sqrt{7-3\sqrt{5}}=\frac{3\sqrt2}{2}-\frac{\sqrt{10}}{2}.</cmath></li><p> | ||
+ | </ol> | ||
+ | Substituting these results into <math>(\bigstar),</math> we have <cmath>x=\frac{\left(\frac{\sqrt2}{2}+\frac{\sqrt{10}}{2}\right)+\left(\frac{3\sqrt2}{2}-\frac{\sqrt{10}}{2}\right)}{2}=\sqrt2.</cmath> | ||
+ | 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 <cmath>N=x-y=\boxed{\textbf{(A) }1}.</cmath> | |
− | ~MRENTHUSIASM | + | ~MRENTHUSIASM |
== See also == | == See also == |
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.