## Task 6 - 170736

In a square $A B C D$, the diagonal $A C$ has a length of $10.0 \, \text{cm}$.

a) Construct such a square! Describe and justify your construction!

b) A rectangle $E F G H$ is said to be inscribed in the square $A B C D$ if, with appropriate labeling, $E$ lies on $A B$, $F$ lies on $B C$, $G$ lies on $C D$, and $H$ lies on $D A$. Additionally, $E F \| A C$.

Determine the perimeter of each such rectangle $E F G H$!
a) Since in every square the diagonals are perpendicular to each other, equal in length, and bisect each other, the vertices $B$ and $D$ lie first on the perpendicular bisector of $A C$ and second on the circle with radius $\frac{A C}{2}$ centered at the midpoint of $A C$.

Therefore, a square $A B C D$ meets the conditions of the problem only if it can be obtained through the following construction:

1. We draw a segment $A C$ of length $A C=10.0 \, \text{cm}$ and construct its perpendicular bisector.
2. We draw a circle around the midpoint $M$ of the segment $A C$ with radius $\frac{1}{2} A C$.
3. If the circle intersects the perpendicular bisector at two points, let these points be $B$ and $D$. $A B C D$ is the desired square.

Proof: By construction, $B D \perp A C$. Furthermore, by construction, $A M = C M = B M = D M$. Consequently, by the congruence theorem (side-angle-side), the triangles $A M B, B M C, C M D$, and $D M A$ are congruent to each other. Therefore, $A B = B C = C D = D A$.

Since these triangles are also right-angled and isosceles with the vertex at $M$, their base angles are each $45^{\circ}$. Since each of the angles $\angle A B C, \angle B C D, \angle C D A, \angle D A B$ is equal to the sum of two of these base angles, each of them is $90^{\circ}$. Therefore, $A B C D$ is a square, and it has the prescribed diagonal length.

b) Let a rectangle $E F G H$ be inscribed in the square $A B C D$ as specified in the problem statement. Let $F G$ intersect the diagonal $A C$ at $P$ and $H E$ intersect the diagonal $A C$ at $Q$. Since the diagonal of a square bisects the interior angles as a line of symmetry, $\angle C A B = 45^{\circ}$.

Because $E F \parallel A C$, it follows that $\angle C A B = \angle F E B = 45^{\circ}$ as corresponding angles on intersecting parallel lines, and because $\angle F E H = 90^{\circ}$, it follows that $\angle H E A = 45^{\circ}$. Thus, the triangle $A E Q$ is isosceles due to the equal base angles, and it follows that $A Q = Q E$.

Similarly, $A Q = H Q$, so $E H = 2 A Q$. Correspondingly, $F G = 2 C P$. Since $E H = F G$, it follows that $A Q = C P$. Finally, because $E F \parallel Q P$ and $E Q \parallel F P$, it also follows that $E F = Q P$.

For the perimeter $u$ of the rectangle $E F G H$, we have:

$$
u = 2(E F + E H) = 2(Q P + A Q + C P) = 2 A C
$$

The perimeter of any such rectangle $E F G H$ is thus $20.0 \, \text{cm}$.

Solutions from the 3rd Round 1977 taken from [5]

### 4.20 18th Olympiad 1978

### 4.20.1 1st Round 1978, Class 7