Triangle $ABC$ is given. Prove that $\frac{R}{r} > \frac{a}{h}$, where $R$ is the radius of the circumscribed circle, $r$ is the radius of the inscribed circle, $a$ is the length of the longest side, $h$ is the length of the shortest altitude.
Problem
Source: Tournament of Towns Spring 2003 - Junior A-Level - Problem 2
Tags: trigonometry, inequalities, geometry, circumcircle, triangle inequality, geometry unsolved
applepi2000
15.06.2011 05:07
Nice problem (In the solution, $s$ is semiperimeter and $A$ is area.)
Since $2R=\frac{a}{\sin A}$ and $\sin A\le1$, we have:
\[2R\ge a\]
By the triangle inequality, $a<b+c$, or $2a<a+b+c\rightarrow \frac{a}{2}<\frac{s}{2}$. Thus:
\[Rs>\frac{a^2}{2}\]
Since $A=rs=\frac{ah}{2}$, we divide both sides by $A$ to get:
\[\frac{R}{r}>\frac{a}{h}\]
as desired.
tuan119
15.06.2011 05:18
note that: $a$ is the length of the longest side and $h$ is the length of the shortest altitude.????? Ex: $ A=\frac{ah}{2} $?? if $a=BC, h=h_c$?? then $A \neq \frac{ah}{2} $; with $h_c =CH \perp AB, H \in AB$ Your solution is true, if $ h=h_a$, but $h_a $ isn't the length of the shortest altitude, $a$ too.
Amir Hossein
15.06.2011 12:48
The longest side of the triangle is a chord of a circumscribed circle and thus it does not exceeds its diameter: $a \leq 2R$. Projection of incircle onto the shortest altitude is contained strictly inside of the projection of the triangle onto this altitude. So $2r < h$. Since all the numbers are positive we can multiply these inequalities: $2r \cdot a < h \cdot 2R$ which implies $\frac ah < \frac Rr$.
See here for a similar problem.
Akiyama
15.06.2011 15:53
amparvardi wrote:
The longest side of the triangle is a chord of a circumscribed circle and thus it does not exceeds its diameter: $a \leq 2R$. Projection of incircle onto the shortest altitude is contained strictly inside of the projection of the triangle onto this altitude. So $2r < h$. Since all the numbers are positive we can multiply these inequalities: $2r \cdot a < h \cdot 2R$ which implies $\frac ah < \frac Rr$.
See here for a similar problem. wow! very nice solution! Here is my ugly solution Put $r=\frac{abc}{4Rs}$, and $h=\frac{abc}{2aR}$. Then original inequality becomes $4R^{2}s>2a^{2}R$. $\Longleftrightarrow 2Rs>a^{2}$ $\Longleftrightarrow \sin A+\sin B+\sin C>\sin ^{2}A$ It is obvious held inequality.($\because \sin A \le 1 \implies \sin ^{2}A \le \sin A$)
applepi2000
15.06.2011 16:33
tuan119 wrote: note that: $a$ is the length of the longest side and $h$ is the length of the shortest altitude.????? Ex: $ A=\frac{ah}{2} $?? if $a=BC, h=h_c$?? then $A \neq \frac{ah}{2} $; with $h_c =CH \perp AB, H \in AB$ Your solution is true, if $ h=h_a$, but $h_a $ isn't the length of the shortest altitude, $a$ too. Well, since $2A=ah_a=bh_b=ch_c$, having $a$ the longest side implies $h_a$ is the shortest altitude.
gold46
15.06.2011 21:01
$R=\frac{abc}{4A} , r=\frac{2A}{a+b+c} , h=\frac{2A}{a} , A=\sin \alpha bc $ . So we need to prove $\frac{a+b+c}{2\sin \alpha}\ge $. But $\frac{a+(b+c)}{2 \sin \alpha} > \frac{a+a}{2}=a$