Difference between revisions of "1997 JBMO Problems/Problem 4"
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Determine the triangle with sides <math>a,b,c</math> and circumradius <math>R</math> for which <math>R(b+c) = a\sqrt{bc}</math>. | Determine the triangle with sides <math>a,b,c</math> and circumradius <math>R</math> for which <math>R(b+c) = a\sqrt{bc}</math>. | ||
− | == Solution == | + | == Solutions == |
+ | |||
+ | ===Solution 1=== | ||
Solving for <math>R</math> yields <math>R = \tfrac{a\sqrt{bc}}{b+c}</math>. We can substitute <math>R</math> into the area formula <math>A = \tfrac{abc}{4R}</math> to get | Solving for <math>R</math> yields <math>R = \tfrac{a\sqrt{bc}}{b+c}</math>. We can substitute <math>R</math> into the area formula <math>A = \tfrac{abc}{4R}</math> to get | ||
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&= \frac{(b+c)\sqrt{bc}}{4}. | &= \frac{(b+c)\sqrt{bc}}{4}. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | We also know that <math>A = \tfrac{1}{2} | + | We also know that <math>A = \tfrac{1}{2}bc \sin(\theta)</math>, where <math>\theta</math> is the angle between sides <math>b</math> and <math>c.</math> Substituting this yields |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | \tfrac{1}{2} | + | \tfrac{1}{2}bc \sin(\theta) &= \frac{(b+c)\sqrt{bc}}{4} \\ |
2\sqrt{bc} \cdot \sin(\theta) &= b+c \\ | 2\sqrt{bc} \cdot \sin(\theta) &= b+c \\ | ||
\sin(\theta) &= \frac{b+c}{2\sqrt{bc}} | \sin(\theta) &= \frac{b+c}{2\sqrt{bc}} | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | Since <math>\theta</math> is inside a triangle, <math>0 < \sin{\theta} \le 1</math>. | + | Since <math>\theta</math> is inside a triangle, <math>0 < \sin{\theta} \le 1</math>. Substitution yields |
+ | <cmath>0 < \frac{b+c}{2\sqrt{bc}} \le 1.</cmath> | ||
+ | Note that <math>2\sqrt{bc}</math>, so multiplying both sides by that value would not change the inequality sign. This means | ||
+ | <cmath>0 < b+c \le 2\sqrt{bc}.</cmath> | ||
+ | However, by the [[AM-GM Inequality]], <math>b+c \ge 2\sqrt{bc}</math>. Thus, the equality case must hold, so <math>b = c</math> where <math>b, c > 0</math>. When plugging <math>b = c</math>, the inequality holds, so the value <math>b=c</math> truly satisfies all conditions. | ||
+ | <br> | ||
+ | That means <math>\sin(\theta) = \frac{2b}{2\sqrt{b^2}} = 1,</math> so <math>\theta = 90^\circ.</math> That means the only truangle that satisfies all the conditions is a 45-45-90 triangle where <math>a</math> is the longest side. In other words, <math>(a,b,c) \rightarrow \boxed{(n\sqrt{2},n,n)}</math> for all positive <math>n.</math> | ||
== See Also == | == See Also == |
Latest revision as of 15:48, 23 February 2021
Contents
Problem
Determine the triangle with sides and circumradius for which .
Solutions
Solution 1
Solving for yields . We can substitute into the area formula to get We also know that , where is the angle between sides and Substituting this yields Since is inside a triangle, . Substitution yields Note that , so multiplying both sides by that value would not change the inequality sign. This means However, by the AM-GM Inequality, . Thus, the equality case must hold, so where . When plugging , the inequality holds, so the value truly satisfies all conditions.
That means so That means the only truangle that satisfies all the conditions is a 45-45-90 triangle where is the longest side. In other words, for all positive
See Also
1997 JBMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 | ||
All JBMO Problems and Solutions |