MHB Is the Area and Radius of a Triangle Always Proportional to its Side Lengths?

[Ackbach]

Here is this week's POTW:

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Let $T_1$ and $T_2$ be two acute-angled triangles with respective side lengths $a_1, b_1, c_1$ and $a_2, b_2, c_2$, areas $\Delta_1$ and $\Delta_2$, circumradii $R_1$ and $R_2$ and inradii $r_1$ and $r_2$. Show that, if $a_1\ge a_2, \; b_1\ge b_2, \; c_1\ge c_2,$ then $\Delta_1\ge\Delta_2$ and $R_1\ge R_2$, but it is not necessarily true that $r_1\ge r_2$.

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[Ackbach]

No one answered this week's POTW, which was Problem 454 in the MAA Challenges. The solution follows:

[sp]Let the angles of $T_1$ and $T_2$ be respectively $A_1, \; B_1,\;C_1$ and $A_2,\;B_2,\;C_2$. Since $A_1+B_1+C_1=180^{\circ}=A_2+B_2+C_2,$ it cannot happen that each angle of one exceeds the corresponding angle of the other. Consequently, by relabelling if necessary, we may suppose that
$$A_1\ge A_2\quad\text{and}\quad B_1\le B_2.$$
Then $\Delta_1=\frac12\,b_1 c_1 \sin(A_1)\ge\frac12\,b_2c_2 \sin(A_2)=\Delta_2$ (since $A_1$ and $A_2$ are acute). Also,
$$R_1=\frac{b_1}{2\sin(B_1)}\ge\frac{b_2}{2\sin(B_2)}=R_2.$$
The formula for the inradius of the triangle is $r=\dfrac{\Delta}{s},$ where $s$ is the semiperimeter. The easiest cases for calculating area are right-angled and isosceles triangles. Consequently, we try to make $T_1$ a right-angled triangle and $T_2$ an isosceles triangle with one side shorter for which $r_1<r_2$. If $(a_1, b_1, c_1)=(3,4,5)$ and $(a_2,b_2,c_2=(3,4,4)$ then $\Delta_1=6, \; s_1=6,\;r_1=1$ while
$$\Delta_2=\frac{3\sqrt{55}}{4},\quad s_2=\frac{11}{2},\quad r_2=\sqrt{\frac{45}{44}} \, >1.$$
Note. If just $T_1$ is acute, it still follows that $\Delta_1\ge\Delta_2$.
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