Difference between revisions of "1976 AHSME Problems/Problem 27"
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Note that | 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> | ||
+ | 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 simplify: | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | + | a+b+2\sqrt{ab}&=3+\sqrt{5} \\ | |
+ | a+b+\sqrt{4ab}&=3+\sqrt{5}. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
+ | It follows that | ||
+ | <cmath>\begin{align*} | ||
+ | a+b&=3, \\ | ||
+ | ab&=\frac54. | ||
+ | \end{align*}</cmath> | ||
+ | By inspection, we conclude that <math>\{a,b\}=\left\{\frac12,\frac52\right\},</math> from which <cmath>\sqrt{3+\sqrt{5}}=\sqrt{\frac12}+\sqrt{\frac52}=\frac{\sqrt2}{2}+\frac{\sqrt{10}}{2}.</cmath> | ||
== See also == | == See also == |
Revision as of 02:06, 8 September 2021
Contents
Problem
If then equals
Solution 1
Let and Clearly, and are both positive.
Note that from which
On the other hand, note that Finally, the answer is
~Someonenumber011 (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Solution 2
Let and Clearly, and are both positive.
Note that Let for some nonnegative rational numbers and We square both sides of this equation, then simplify: It follows that By inspection, we conclude that from which
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 |
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