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The largest square which fits into a circle is ABCD and EFGH is a square with G and H on the line CD and E and F on the circumference of the circle. Show that AB = 5EF. Similarly the largest equilateral triangle which fits into a circle is LMN and PQR is an equilateral triangle with P and Q on the line LM and R on the circumference of the circle. Show that LM = 3PQ

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Two Ladders

Two ladders are propped up against facing walls. The end of the first ladder is 10 metres above the foot of the first wall. The end of the second ladder is 5 metres above the foot of the second wall. At what height do the ladders cross?


Stage: 4 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

Thanks to Andrei of School 205 Bucharest for this well explained solution.

  • First, I drew a line $HG$, perpendicular on $P$, and a line through H, parallel with $PR$.
  • On it, I took a segment $HE$ of the same length with $HO$.
  • Then I finished drawing the square $HEFO$, drawing $EF$ perpendicular on $PR$. The line PE intersects $QR$ in $A$.
  • From $A$ I drew a parallel to $PR$ (let the intersection point with $PQ$ be $D$), and a perpendicular to $PR$ (let the intersection point with $PR$ be $B$).
  • This way I found 3 vertices of the rectangle $ABCD$, and I finished the rectangle finding the vertex $C$ on $PR$, so that $CD$ is perpendicular to $PR$.

The rectangle $ABCD$ is a square because:- Triangles $PEF$ and $PAB$ are similar, they are both right-angled triangles, with a common angle. The similarity ratio is: $${{PE}\over{PA}} = {{EF} \over {AB}}$$ Triangles $PEH$ and $PAD$ are similar, because angles $PEH$ and $PAD$ are equal and $HPE$ is a common angle. The similarity ratio $$\frac{PE}{PA} = \frac{HE}{AD}$$ From $(1)$ and $(2)$ $$ \frac{AB}{EF} = \frac{AD}{HE}$$ As $HE=EF$ (sides of a square) Therefore $ AB =AD$ $ ABCD$ is a rectangle with two adjacent sides equal

Therefore $ ABCD$ is a square.
Now, the construction of the inscribed square must be done in the following steps:

  • Choose a point $H$ on side $PQ$, near $P$
  • Draw line $HG$, perpendicular on $PR$
  • Take the distance $HG$ as the compass distance, and draw a circle arc, with centre $G$. $F$ is the point of intersection of this arc with $PR$.
  • Construct two circle arcs with centres $F$ and $H$ (with the same radius as before). Their intersection is point $E$, and $EFGH$ is a square.
  • Draw line $PE$. Let the intersection point of this line with side $QR$ be $A$.
  • Draw from $A$ parallel to $EF$ and $HE$. Their intersections with $PR$ and $QP$ are points $B$ and $D$ respectively.
  • Draw from $D$ a parallel to $AB$. Its intersection with $PR$ is $C$.
  • $ABCD$ is the square to be found.

The choice of point $P$, so that $E$ is interior to the triangle $PQR$ is not a restrictive condition, the construction is the same if $E$ is exterior to triangle $PQR$.