As in title: the problem is to construct triangle given difference of sides $b$ and $c$, then in-circle radius $r$, and height $h_{b}$.

The problem is from a set of problems exercising various formulas for distances between points such as foots of in-circle and ex-circle etc. One of such formulas claims that distance between foots of in-circle and ex-circle to side $a$ of given triangle (let's call these points $P$ and $P_{a}$) is equal to $b-c$, and that middle of side $a$ is also middle of segment between these two foots. I'm pretty sure that this formula is to be used in given construction, so I'd start by drawing line segment $PP_{a}$ of size $b-c$, then constructing perpendicular to this segment in point $P$, and finding in-circle center $S$ at distance $r$ along this perpendicular from point $P$. Then I can draw in-circle, however the problem is now how to utilize height $h_{b}$...

  • $\begingroup$ As an alternative to the answer provided by David below, the construction described above could be continued as follows: Find midpoint $A'$ of segment $PP_{a}$, that is also the midpoint of side $a$, and then draw a circle of radius $h/2$ using $A'$ as center. Then, tangent line common to this circle and in-circle will be line containing side $b$ of triangle, and the intersection of this line with line containing segment $PP_{a}$ will be vertex $C$. Vertex $B$ is reflection of $C$ around $A'$, and vertex $A$ is intersection of in-circle tangent through vertex $B$ and line containing side $b$. $\endgroup$
    – Crni
    Commented Feb 17, 2016 at 8:10

2 Answers 2


Let $A'$ be the midpoint of $BC$. We will use the fact that, as you say, the tangent from $A'$ to the incircle has length $(b-c)/2$.

First draw a line $l$ that will be $AC$. (Don't place $A$ and $C$ yet.) Now select a point $I$ at a distance $r$ from $l$ to be the incentre. Draw the incircle, which is tangent to $l$ at a point $Q$.

The power of $A'$ with respect to the incircle must be $(b-c)^2/4$. The set of all points with this property is a circle concentric to the incircle. It can be constructed by selecting one point on $l$ at a distance $(b-c)/2$ from $Q$, and drawing the circle centred at $I$ through that point.

But $A'$ must also be located on a line parallel to $l$ at a distance $h_b/2$ from it, on the same side as the incircle. Construct $A'$ as either intersection of this line with the previously constructed circle.

Now line $BC$ can be constructed as a tangent to the incircle passing through $A'$ (two possibilities). Vertex $C$ is the intersection of this line with $l$, and $B$ is the reflection through $A'$ of $C$. Finally, $A$ can be obtained by drawing the other tangent to the incircle from $B$ and taking the intersection with $l$.


There may me more elegant solutions, but this works:

YO need to determine the length of side b. This is determined in two steps: First, find the lengths of the other sides in terms of b; then set up an equation for b. We are given b-c = Δ, so c is simply given as:

c = b – Δ [1]

To determine a, note that the area of the triangle is given by one half the base times the height and also by one half the perimeter times the inradius. To see how the latter is derived, consider a triangle in which the inscribed circle has been constructed. Draw line segments form the center of this circle to the vertices of the triangle. This divides the area of the original triangle into three components that are each one half of one side times the inradius. Thus we have:

a= Δ + b
((h_b)/r-2) [2]

Note that (h_b)/r must be greater than 2 because of the triangle inequality. Attempting to solve the system with (h_b)/r < 2 would still give a triangle, but it would be turned “inside out” with the “inscribed circle” lying outside the triangle; the circle would be tangent to side a and the extensions of sides b and c rather than all three sides proper.

Now equate the area as given by one half the base times the height with the area as given by Heron’s Formula:

2b*(h_b)= [(a+b+c)(-a+b+c)(a-b+c)*(a+b-c)]^(1/2) [3]

Substitute [1] and [2] into [3], square both sides, clear fractions and divide by a common factor of b2 that appears (b is of course nonzero). This leads to a quadratic equation (which becomes linear in the specific case hb/r = 4:

(4r-(h_b))*((h_b)-2r)^2*b^2 +(12r-4(h_b))*((h_b)-2r)rΔ*b -4*r^2*[((h_b)-2r)*Δ^2+(h_b)*r^2]=0 [4]

Equation [4] has a unique positive root when 2 < (h_b)/r < 4, and this will give a unique triangle meeting the given values of Δ, r and (h_b) . For (h_b)/r > 4, we are not guaranteed a solution and with (h_b)/r > 4 there may be two roots.

  • $\begingroup$ Thanks for your reply, it indeed solves the problem in case when input values $b-c$, $r$ and $h_{b}$ given as numbers. This problem is, however, a geometric construction, with input values given just as line segments of unknown length, and the construction is to be conducted using ruler and compass only. $\endgroup$
    – Crni
    Commented Feb 17, 2016 at 6:57
  • $\begingroup$ Because we ended up with a quadratic equation, we can translatr it to a construction. To see how look up:en.m.wikipedia.org/wiki/Carlyle_circle $\endgroup$ Commented Feb 17, 2016 at 11:09
  • $\begingroup$ Oh, now I see... And we can use similar triangles to construct all of these products that appear in the coefficients of equations (as well as to conduct division by coefficient of $b^2$). This is quite interesting and general approach; in particular, it helps in the discussion part (no solution, or multiple solutions, etc.) of the construction. However, let me also note that for all this to work we'd still need segment of length 1 given, along with input data mentioned in the question - isn't that right? $\endgroup$
    – Crni
    Commented Feb 17, 2016 at 17:02

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