Construct a triangle from 3 segments

Question

My father recalled the following problem for his 6th grade Geometry class in the school in Soviet Union. I struggled hopelessly, maybe someone here will have an idea.

Construct a triangle from a side, as well as a median and a bisector from the opposite angle towards that side.

My Idea - Updated With A Picture

Given the side (dark blue line), median (red line) and bisector (black line), we must construct the entire triangle from which these were taken.

Since the side is already given, we already have 2 vertices of the triangle, so we just need to construct the third vertex.

Take the side, call it $AB$ (it's dark blue in the picture) and construct its midpoint $M$. Centered on this midpoint, draw a circle with the radius of the size of the median (it's the red circle in the picture).

The final point we need to construct will need to lie on that circle. We just need to use the bisector criterion to fix this point, but I am not sure how to do that.

(On the drawing, one potential vertex is picked on the circle and the black line is placed to be its bisector.)

It also seems there are 2 solutions, one above the side and one below, so we can restrict ourselves to the upper semicircle and the solution will be unique.

Thank you very much for your help.

Answer 1

The left endpoint of the blue line is $A$, the right endpoint is $B$. The top vertex is $C$. The intersection point between the black line and the blue line is $N$. We are given the length for $AB=2L$, and $CN=d$, the radius of the red circle is $r$. Set up coordinate system with origin at the midpoint of $AB$, and call it $O$. So we have $A(-L,0), B(L,0), C(x_c, y_c)$. As long as we find the coordinate $(x_c, y_c)$, we are done.

$$x_c^2+ y_c^2=r^2\tag{1}$$

First, express the angle $\angle A=\arctan\frac{y_c}{x_c+L}, ~~\angle B=\arctan\frac{y_c}{L-x_c}, ~~~\angle C=\pi-\angle A-\angle B$

$$\angle CNB=\angle A+\frac{1}2\angle C$$

Slope for black line:

$$k=\tan \angle CNB$$

Straight line equation for black line:

$$y-y_c=k(x-x_c)$$

We get coordinate for point $N$

$$(x_c-\frac{y_c}{k},0)$$

Since we are given the length for $CN=d$, we get

$$d^2=\left(\frac{y_c}{k}\right)^2+y_c^2\tag{2}$$

Plug in the expression for $k$ and solve equation (1) and (2), we can find the coordinate for $(x_c, y_c)$.

Done.

Answer 2

Let $a$ be the reference side opposite angle $A$. Some formulas for median and bisector are known.

For median at A:

$$ 2 m_a= \sqrt{2 b^2+2 c^2-a^2} $$

Bisector length $b_a$

$$ b_a= \frac{2 bc }{b+c} \cos A/2 $$

$$ \cos A= \frac{b^2+c^2-a^2}{2 b c } =2 \cos^2 (A/2) -1 $$

Clubbing the second and first relations together $ (b,c) $ can be solved for in terms of $a$ with algebra which is a good exercise for the student.

A particular numerical example:

Given$ (a, m_a, b_a)=( 1,0.68, 0.52 ) $ the approx. computer solution for $(b,c)$ is (0.4196,1.1175).