Using a compass and a ruler, construct a triangle given a side, the opposite angle, and the radius of the inscribed circle.

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If the angle bisectors of angles $B$ and $C$ of triangle $A B C$ intersect at point $O$, then $\angle B O C=90^{\circ}+\frac{1}{2} \angle A$.

## Solution

If $O$ is the center of the inscribed circle of the desired triangle $A B C$, then in triangle $B O C$ the following are known: $B C=a$ (given side), the height drawn from vertex $O$ (given radius $r$), and $\angle B O C=90^{\circ}+\frac{1}{2} \angle A (\angle A=\alpha-$ given angle).

From this, the following construction follows. We construct an arc on the chord $B C$ that subtends an angle of $90^{\circ}+\frac{\alpha}{2}$. Then we draw a line parallel to the line $B C$ and at a distance equal to $r$. If this line intersects the constructed arc, then each point of intersection is the center of the inscribed circle of the desired triangle. If the tangents drawn from points $B$ and $C$ to the constructed circle intersect at point $A$, then $A$ is the third vertex of the desired triangle $A B C$.

![](https://cdn.mathpix.com/cropped/2024_05_06_cb0f0d605674d837f9abg-48.jpg?height=410&width=648&top_left_y=1290&top_left_x=708)

![](https://cdn.mathpix.com/cropped/2024_05_06_cb0f0d605674d837f9abg-48.jpg?height=100&width=1216&top_left_y=1758&top_left_x=17)

Find the geometric locus of the midpoints of segments connecting a given point, lying outside a given circle, with points on this circle.

## Hint

Consider a homothety with the center at the given point and a coefficient of $\frac{1}{2}$.

## Solution

Consider a homothety with the center at the given point $M$ and a coefficient of $\frac{1}{2}$. Under this homothety, the given circle will transform into a circle with a radius equal to half the radius of the given circle, and the center will be the midpoint of the segment $O M$, where $O$ is the center of the given circle.

Any point on this circle is the midpoint of a segment with endpoints at the given point $M$ and on the given circle. On the other hand, the midpoint of any such segment lies on the constructed circle.

![](https://cdn.mathpix.com/cropped/2024_05_06_cb0f0d605674d837f9abg-49.jpg?height=368&width=1112&top_left_y=-1&top_left_x=479)

## Answer

A circle.

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