Difference between revisions of "1997 JBMO Problems/Problem 4"
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== Problem == | == Problem == | ||
| − | + | 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 == | == Solution == | ||
| − | == See | + | 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 |
| + | <cmath>\begin{align*} | ||
| + | A &= \frac{abc}{4 \cdot \tfrac{a\sqrt{bc}}{b+c} } \\ | ||
| + | &= \frac{abc}{4a\sqrt{bc}} \cdot (b+c) \\ | ||
| + | &= \frac{(b+c)\sqrt{bc}}{4}. | ||
| + | \end{align*}</cmath> | ||
| + | We also know that <math>A = \tfrac{1}{2}ab \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*} | ||
| + | \tfrac{1}{2}ab \sin(\theta) &= \frac{(b+c)\sqrt{bc}}{4} \\ | ||
| + | 2\sqrt{bc} \cdot \sin(\theta) &= b+c \\ | ||
| + | \sin(\theta) &= \frac{b+c}{2\sqrt{bc}} | ||
| + | \end{align*}</cmath> | ||
| + | Since <math>\theta</math> is inside a triangle, <math>0 < \sin{\theta} \le 1</math>. | ||
| + | |||
| + | |||
| + | == See Also == | ||
{{JBMO box|year=1997|num-b=3|num-a=5}} | {{JBMO box|year=1997|num-b=3|num-a=5}} | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
Revision as of 13:58, 4 August 2018
Problem
Determine the triangle with sides 
and circumradius 
for which 
.
Solution
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, 
.
See Also
| 1997 JBMO (Problems • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 | ||
| All JBMO Problems and Solutions | ||
