Difference between revisions of "1997 JBMO Problems/Problem 4"
Rockmanex3 (talk | contribs) (WIP Solution) |
Rockmanex3 (talk | contribs) (Solution to Problem 4) |
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\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> | ||
+ | Since all values in the inequality are positive, squaring both sides would not change the inequality sign, so | ||
+ | <cmath>0 < b^2 + 2bc + c^2 \le 4bc</cmath> | ||
+ | <cmath>-4bc < b^2 - 2bc + c^2 \le 0</cmath> | ||
+ | <cmath>-4bc < (b-c)^2 \le 0</cmath> | ||
+ | By the [[Trivial Inequality]], <math>(b-c)^2 \ge 0</math> for all <math>b</math> and <math>c,</math> so the only values of <math>b</math> and <math>c</math> that satisfies is when <math>(b-c)^2 = 0</math>. Thus, <math>b = c</math>. Since <math>-4bc < 0</math> for positive <math>b</math> and <math>c</math>, 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 == |
Revision as of 14:24, 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,
. 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 |