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 15:24, 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 |