## Task 6 - 260836

Let $A B C D$ be a triangular pyramid that satisfies the condition that the four triangles $A B C, A B D$, $A C D$ and $B C D$ are all mutually perimeter-equal.

Investigate whether this condition and the lengths

$$
\overline{A D}=p, \quad \overline{B D}=q, \quad \overline{C D}=r
$$

uniquely determine the lengths $\overline{B C}=a, \overline{C A}=b$ and $\overline{A B}=c$!

If this is the case, give these lengths $a, b, c$ in terms of $p, q, r$!
Due to the equality of the perimeters of $ABC$ and $ABD$, we have $a+b+c=p+q+c$ (1).

Due to the equality of the perimeters of $ABC$ and $ACD$, we have $a+b+c=p+b+r$ (2).

Due to the equality of the perimeters of $ABC$ and $BCD$, we have $a+b+c=a+q+r$ (3).

Claim: From this, it follows that $a, b, c$ are uniquely determined and specifically $a=p, b=q, c=r$.

1. One possible proof:

From (1) it follows that $a+b=p+q$ (4), from (3) and (2) it follows that $a+q+r=p+b+r$ and thus $a+q=p+b$. (5)

Adding (4) and (5) results in $2a+b+q=2p+b+q$ and thus $a=p$.

Substituting this into (1) or (2) results in $b=q$ and $c=r$.

2. Another possible proof:

Adding (1), (2), and (3) results in

$$
3(a+b+c)=a+b+c+2(p+q+r)
$$

thus $a+b+c=p+q+r$ (6).

From (6) together with (1), (2), and (3), it follows that $c=r$, $b=q$, and $a=p$.

Solutions from the third round of 1987 adopted from [5]

### 5.29 XXVII. Olympiad 1987

### 5.29.1 First Round 1987, Class 8