Track / Stage 6 / 377 of 400 #1377 of 1964

Problem 1377

National olympiad, first round

Geometry Difficulty 6.8 Find the answer

Let $AOB$ be a given angle less than $180^{\circ}$ and let $P$ be an interior point of the angular region determined by $\angle AOB$ . Show, with proof, how to construct, using only ruler and compass, a line segment $CD$ passing through $P$ such that $C$ lies on the way $OA$ and $D$ lies on the ray $OB$ , and $CP:PD=1:2$ .

The source for this one didn't record the answer, so there is nothing to check what you type against. Work it on paper and mark yourself against the solution below.

Official solution

1. **Draw a Parallel Line to $OA$ through $P$ **:
- Start by drawing the given angle $\angle AOB$ with rays $OA$ and $OB$ .
- Place the point $P$ inside the angular region determined by $\angle AOB$ .
- Draw a line through $P$ that is parallel to $OA$ . This can be done using a compass and straightedge by copying the angle $\angle AOP$ at point $P$ .

2. **Intersection with $OB$ **:
- Let the parallel line through $P$ intersect the ray $OB$ at point $K$ .

3. **Construct Point $D$ on $OB$ **:
- Measure the distance $OK$ using a compass.
- Mark a point $D$ on $OB$ such that $KD = OK$ . This ensures that $OK = KD$ .

4. **Join Points $D$ and $P$ **:
- Draw the line segment $DP$ .

5. **Intersection with $OA$ **:
- Extend the line segment $DP$ to intersect the ray $OA$ at point $C$ .

6. **Verify the Ratio $CP:PD = 1:2$ **:
- By construction, $OK = KD$ , and since $P$ lies on the parallel line through $K$ , we can use Thales' theorem.
- Thales' theorem states that if a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally.
- Here, $\triangle OKD$ is divided by the parallel line through $P$ , so $CP:PD = 1:2$ .

$\blacksquare$

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.