Difference between revisions of "1997 JBMO Problems/Problem 4"
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== Solution == | == Solution == | ||
− | + | NOTE(not by author): we can conclude b = c by noticing that b + c <=2sqrt(bc) but by AM-GM b = c must hold | |
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 | ||
<cmath>\begin{align*} | <cmath>\begin{align*} |
Revision as of 01:24, 21 December 2020
Problem
Determine the triangle with sides and circumradius
for which
.
Solution
NOTE(not by author): we can conclude b = c by noticing that b + c <=2sqrt(bc) but by AM-GM b = c must hold
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
Since all values in the inequality are positive, squaring both sides would not change the inequality sign, so
By the Trivial Inequality,
for all
and
so the only values of
and
that satisfies is when
. Thus,
. Since
for positive
and
, 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 |