
5. Given three positive real numbers $o, \rho, v$, construct a triangle $A B C$ with perimeter equal to $o$, the radius of its $A$-excircle equal to $\rho$, and the length of its $A$ altitude equal to $v$. Determine the number of solutions (non-congruent triangles) in terms of the given lengths.

(Patrik Bak)


Solution. Let $P$ be a point on the ray opposite to $B C$ such that $B P=B A$ and, similarly, let $Q$ be a point on the ray opposite to $C B$ such that $C Q=C A$. Then $P Q=o$.

Let $I_{a}$ be the $A$-excenter of $\triangle A B C$ (Fig. 4). Since $I_{a} B$ is the bisector of angle $A B P$ and the triangle $A B P$ is $B$-isosceles, line $I_{a} B$ is the perpendicular bisector of $A P$. Thus, $I_{a}$ lies on the perpendicular bisector of $A P$ and similarly for $A Q$. Point $I_{a}$ is therefore the circumcenter of $\triangle A P Q$ and it lies on the perpendicular bisector of $P Q$ too.

![](https://cdn.mathpix.com/cropped/2024_04_17_447c393b5df10bbb25bcg-08.jpg?height=533&width=959&top_left_y=335&top_left_x=271)

Fig. 4

With this insight, we can construct $\triangle A B C$ as follows: Let $P Q$ be a segment with length $o$ and let $M$ be its midpoint. Point $I_{a}$ satisfies $M I_{a} \perp P Q$ and $M I_{a}=\rho$. Point $A$ has to lie on a circle $k$ with center $I_{a}$ and radius $I_{a} P=I_{a} Q$ and also on a line $l$ parallel with $P Q$ with distance $v$ from $P Q$ in the opposite half-plane determined by $P Q$ than point $I_{a}$. Points $B, C$ can then be constructed in many ways, for example as the intersections of segment $P Q$ with the perpendicular bisectors of $A P$, $A Q$, respectively. Any triangle $A B C$ constructed in this way then satisfies all three requirements.

It remains to determine the number of solutions in terms of $o, \rho, v$. This is the same as asking for the number of intersections of $k$ and $l$.

Let $r=I_{a} P=\sqrt{\rho^{2}+(o / 2)^{2}}$ be the radius of $k$. When $r>\rho+v$, circle $k$ intersects line $l$ at two points $A_{1}, A_{2}$ and we get two triangles $A_{1} B_{1} C_{1}, A_{2} B_{2} C_{2}$ (Fig. 5). We count them as two separate solutions since they differ e.g. in the angle by $B$.

When $r=\rho+v$, circle $k$ is tangent to line $l$, we get a single intersection $A$ and a single solution $\triangle A B C$ (by symmetry, it will be isosceles).

Finally, when $r\rho+v: & \ldots 2 \text { solutions, } \\
\sqrt{\rho^{2}+\frac{1}{4} o^{2}}=\rho+v: & \ldots 1 \text { solution, } \\
\sqrt{\rho^{2}+\frac{1}{4} o^{2}}<\rho+v: & \ldots 0 \text { solutions. }
\end{array}
$$

![](https://cdn.mathpix.com/cropped/2024_04_17_447c393b5df10bbb25bcg-09.jpg?height=529&width=961&top_left_y=129&top_left_x=222)

Fig. 5
