Let $VABC$ be a triangular pyramid with $VA<VB<VC$. Prove that there is a point $P$ inside the triangle $ABC$ such that $VP= \frac{VA+VB+VC}{3}$.

The drawing

Let $$VABC$$ be a triangular pyramid with $$VA. Prove that there is a point $$P$$ inside the triangle $$ABC$$ such that $$VP= \frac{VA+VB+VC}{3}$$.

The idea

Let $$VO \perp (ABC)$$

The first thought I got is clearly that $$AO because $$VA and using the Phytaghora Theorem we get this inequality

I tried imaging P being in different posts such as the midpoint of $$AB,BC,AC$$ or as the center of the circumscribed circle or inscribed circle of the triangle, in some of them I got that P doesn't satisfy the equality or got to nothing useful to demonstrate.

Hope one of you can help me! Thank you!

• Are you sure the formula is correct? Maybe $\frac{VA+VB+VC}3$?. Commented Feb 13 at 17:08
• @Andrei I checked again in my math book and I wrote it correctly... Commented Feb 13 at 17:12
• Imagine a tiny triangle $ABC$, and $V$ about the same distance $d$ from $A$, $B$, $C$. Then every point $P$ inside is also about distance $d$ from $V$. Andrei is right; there may be a typo in the book Commented Feb 13 at 17:50
• Indeed, Andrei is right : here is another counter-example : take $A(0,0,0),B(1,0,0),C(0,2,0)$ and $D(0,0,3\varepsilon)$. It will be very very difficult to find $P$ inside the triangle verifying $VP=\varepsilon+\sqrt{2}+\sqrt{5}$... because the maximum value that $VP$ can take is $AC=\sqrt{5}$... Commented Feb 13 at 18:03
• I’m voting to close this question because the result to be established is erroneous (printer's typo ?) Commented Feb 14 at 13:09

Let $$D$$ be a point on the line segment $$AB$$ such that $$VD=\frac{VA+VB}{2}$$.

(Such a point $$D$$ exists. The reason is as follows : Let $$A'$$ be a point on the half line $$VA$$ such that $$VA'=\frac{VA+VB}{2}$$. Let $$B'$$ be a point on the half line $$VB$$ such that $$VB'=\frac{VA+VB}{2}$$. Then, we have $$VA'\gt VA$$ since $$2VA\lt VA+VB\implies VA\lt \frac{VA+VB}{2}=VA'$$ Similarly, $$VB'\lt VB$$ since $$2VB\gt VA+VB\implies VB\gt\frac{VA+VB}{2}=VB'$$ Now on the plane $$VAB$$, let us consider a circle whose center is $$V$$ with radius $$\frac{VA+VB}{2}$$. We see that $$A',B'$$ are on the circle. Since $$VA'\gt VA$$ and $$VB'\lt VB$$, there is a point $$D$$ on the line segment $$AB$$ such that $$D$$ is on the circle. For such a point $$D$$, we have $$VD=\frac{VA+VB}{2}$$.)

Let $$E$$ be a point on the line segment $$VD$$ such that $$VE=\frac 23VD$$.

Let $$F$$ be a point on the line segment $$VC$$ such that $$VF=\frac 13VC$$.

Let $$G$$ be a point on the half line $$VD$$ such that $$VG=VE+VF$$.

Let $$H$$ be a point on the half line $$VC$$ such that $$VH=VE+VF$$.

Now, since $$VD=\frac{VA+VB}{2}$$, we have $$\frac{VA+VB+VC}{3}=\frac{2VD+VC}{3}=VE+VF$$

We have $$VG\gt VD$$ since \begin{align}&VC-VA+VC-VB\gt 0 \\&\implies VC\gt \frac{VA+VB}{2} \\&\implies VC\gt VD \\&\implies \frac 13VC\gt \frac 13VD \\&\implies VF\gt ED \\&\implies VF+VE\gt ED+VE \\&\implies VG\gt VD\end{align}

Also, we have $$VC\gt VH$$ since \begin{align}\frac 23VC\gt\frac 23VD&\implies FC\gt VE \\&\implies FC+VF\gt VE+VF \\&\implies VC\gt VH\end{align}

Now, on the plane $$VCD$$, let us consider a circle whose center is $$V$$ with radius $$VE+VF$$.

$$G,H$$ are on the circle with $$VG\gt VD$$ and $$VH\lt VC$$.

So, there is a point $$P$$ on the line segment $$CD$$ such that $$P$$ is on the circle.

For such a point $$P$$, we have $$VP=VE+VF=\frac{VA+VB+VC}{3}. \blacksquare$$