Construct a triangle given an angle, the side opposite the angle and the median to the other side (researching the number of solutions is not required).
Problem
Source: 2003 Oral Moscow Geometry Olympiad grade 9 p1
Tags: geometry, construction
RagvaloD
27.10.2020 09:45
Let $AB$ is this side,$\alpha$ - this angle,$m_c$ - median and $M$ is midpoint of $AB$ Easy to build triangle $ABC$ with $\angle CAB=90, \angle CBA=90-\alpha$ Now build circle $\omega_1$ - circumcircle of $ABC$ and circle $\omega_2$ - circle with center $M$ and radius $m_a$. Let $D$ is point of intersection $\omega_1,\omega_2$ Then $ABD$ is triangle, which we want to build.
parmenides51
24.12.2023 17:27
another construction using centroid
Let BL be the Known median, M be the midpoint of BC=a and G be the centroid of ABC
- Construct segment BC=a
- Construct an angle.<BAC=<A
- Locus of A are two arcs of circles with center M and radii R such that they "see" segment BC under angle <A (one at each side of AB) without the endpoints B,C
-Because MG is homothetic MA with center M and ratio MG/MA=1/3, locus of G are two arcs.of circles. with center M and radius R/3, (homothetic to the previous arcs, such that one at each side of AB) without the endpoints B,C
- We draw the circle (B, BL/3) i.e. with center B and radius BL/3 wich intersects one of the arcs of the locus of G at a point, let it be G'
- Triangle BCG' has been constructed
- Finally we extended MG' beyond G'to segment G'A' =2MG'
- Triangle A' BC is the wanted one
Latex when I use my pc