Given a point $P$ inside the $\triangle ABC$, the pedal triangle of $P$ is the triangle whose polygon vertices are the feet of the perpendiculars from $P$ to the side lines.
As we know that if $P$ is the orthocenter of $\triangle ABC$, then the perimeter of the pedal triangle of $P$ takes the maximum, and if $P$ is the circumcenter of $\triangle ABC$, then the area of the pedal triangle of $P$ takes the maximum.
Naturally, I think that there are the inverse problems of the above problem:
Given the distances $p, q$, and $r$ between point $P$ and the three sides of $\triangle ABC$, then find the maximum and minimum values of the area and perimeter of $\triangle ABC$.
At first, I thought the problem was simple, but I underestimated its difficulty So far, I haven't made any substantial progress. Is there a good solution to this problem?
Best Answer
Let $D,E,F$ such that $PD\perp BC, PE\perp CA,PF\perp AB$ with $PD=r,PE=p,PF=q$.
Also, $\angle{EPF}=\alpha$ and $\angle{FPD}=\beta$.
Here, $\alpha,\beta$ have to satisfy $$0\lt \alpha\lt\pi,\qquad 0\lt\beta\lt\pi,\qquad 0\lt 2\pi-\alpha-\beta\lt\pi$$ i.e. $$0\lt \pi-\beta\lt\alpha\lt\pi$$
Let us first consider the quadrilateral $AFPE$.
Let $\theta=\angle{FEP}$.
By the law of cosines, $$EF^2=p^2+q^2-2pq\cos\alpha$$ and $$q^2=p^2+EF^2-2pEF\cos\theta$$ so we have $$EF\cos\theta=\frac{p^2+EF^2-q^2}{2p}=p-q\cos\alpha$$
By the law of sines in $\triangle{AEF}$, $$AF=\frac{EF\sin(\frac{\pi}{2}-\theta)}{\sin(\pi-\alpha)}=\frac{EF\cos\theta}{\sin\alpha}=\frac{p-q\cos\alpha}{\sin\alpha}$$ Similarly, we get $$AE=\frac{q-p\cos\alpha}{\sin\alpha}$$ $$BF=\frac{r-q\cos\beta}{\sin\beta},\qquad BD=\frac{q-r\cos\beta}{\sin\beta}$$ $$CE=\frac{r-p\cos(2\pi-\alpha-\beta)}{\sin(2\pi-\alpha-\beta)}=\frac{-r+p\cos(\alpha+\beta)}{\sin(\alpha+\beta)},\qquad CD=\frac{-p+r\cos(\alpha+\beta)}{\sin(\alpha+\beta)}$$
(1) Perimeter
Let $L(\alpha,\beta)$ be the perimeter of $\triangle{ABC}$ where $0\lt\pi-\beta\lt\alpha\lt \pi$.
$$L(\alpha,\beta)=\frac{(p+q)(1-\cos\alpha)}{\sin\alpha}+\frac{(q+r)(1-\cos\beta)}{\sin\beta}-\frac{(p+r)(1-\cos(\alpha+\beta))}{\sin(\alpha+\beta)}$$
Let $F(\alpha)=\frac{\partial L}{\partial\alpha}$. Then, we have $$F(\alpha)=\frac{p+q}{1+\cos\alpha}-\frac{p+r}{1+\cos(\alpha+\beta)}$$ $$F'(\alpha)=\frac{(p+q)\sin\alpha}{(1+\cos\alpha)^2}+\frac{(p+r)\overbrace{(-\sin(\alpha+\beta))}^{\text{positive}}}{(1+\cos(\alpha+\beta))^2}\gt 0$$ So, $F(\alpha)$ is increasing with $$\lim_{\alpha\to\pi-\beta}F(\alpha)=-\infty,\qquad \lim_{\alpha\to\pi}F(\alpha)=+\infty$$
So, there is only one $\alpha$ such that $F(\alpha)=0$.
Letting $G(\beta)=\frac{\partial L}{\partial\beta}$, we have $$G(\beta)=\frac{q+r}{1+\cos\beta}-\frac{p+r}{1+\cos(\alpha+\beta)}$$ $$G'(\beta)=\frac{(q+r)\sin\beta}{(1+\cos\beta)^2}+\frac{(p+r)\overbrace{(-\sin(\alpha+\beta))}^{\text{positive}}}{(1+\cos(\alpha+\beta))^2}>0$$ So, $G(\beta)$ is increasing with $$\lim_{\beta\to\pi-\alpha}G(\beta)=-\infty,\qquad \lim_{\beta\to \pi}G(\beta)=+\infty$$
which implies that there is only one $\beta$ such that $G(\beta)=0$.
We see that there is no maximum (Paul Sinclair has already provided a good answer on this).
The minimum of $L(\alpha,\beta)$ is $L(\alpha_0,\beta_0)$ where $(\alpha_0,\beta_0)$ is the only one solution of $$\frac{p+q}{1+\cos\alpha}=\frac{q+r}{1+\cos\beta}=\frac{p+r}{1+\cos(\alpha+\beta)}$$
Example : If $p=2,q=3$ and $r=4$, then $$\alpha_0\approx 2.19202,\qquad \beta_0\approx 1.99857$$(see here). So, the minimum of $L(\alpha,\beta)$ is approximately $30.983$.
(2) Area
Let $S(\alpha,\beta)$ be the area of $\triangle{ABC}$ where $0\lt\pi-\beta\lt\alpha\lt \pi$.
We have $$S(\alpha,\beta)=\frac{qAF}{2}+\frac{pAE}{2}+\frac{qBF}{2}+\frac{rBD}{2}+\frac{pCE}{2}+\frac{rCD}{2}$$ $$=\frac{2pq-(p^2+q^2)\cos\alpha}{2\sin\alpha}+\frac{2qr-(q^2+r^2)\cos\beta}{2\sin\beta}-\frac{2pr-(p^2+r^2)\cos(\alpha+\beta)}{2\sin(\alpha+\beta)}$$
Let $H(\alpha)=\frac{\partial S}{\partial\alpha}$. Then, $$H(\alpha)=\frac{p^2+q^2-2pq\cos\alpha}{2\sin^2\alpha}-\frac{p^2+r^2-2pr\cos(\alpha+\beta)}{2\sin^2(\alpha+\beta)}$$ with $$\lim_{\alpha\to \pi-\beta}H(\alpha)=-\infty,\qquad \lim_{\alpha\to \pi}H(\alpha)=+\infty$$ So, there is at least one $\alpha$ such that $H(\alpha)=0$.
Let $I(\beta)=\frac{\partial S}{\partial\beta}$. Then, $$I(\beta)=\frac{q^2+r^2-2qr\cos\beta}{2\sin^2\beta}-\frac{p^2+r^2-2pr\cos(\alpha+\beta)}{2\sin^2(\alpha+\beta)}$$ with $$\lim_{\beta\to\pi-\alpha}I(\beta)=-\infty,\qquad \lim_{\beta\to\pi}I(\beta)=+\infty$$
which implies that there is at least one $\beta$ such that $I(\beta)=0$.
We see that there is no maximum.
The minimum of $S(\alpha,\beta)$ is $S(\alpha_0,\beta_0)$ where $(\alpha_0,\beta_0)$ is one of the solutions of $$\frac{p^2+q^2-2pq\cos\alpha}{2\sin^2\alpha}=\frac{q^2+r^2-2qr\cos\beta}{2\sin^2\beta}=\frac{p^2+r^2-2pr\cos(\alpha+\beta)}{2\sin^2(\alpha+\beta)}$$