Kvant Math Problem 667

Consider a triangle $ABC$ with the smallest angle $\widehat A$ and suppose the differences $d = |AB| - |BC|$ and $e = |AC| - |BC|$ are given.

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Problem

Construct the triangle $ABC$, given its smallest angle $\widehat A$ and the segments of lengths $d=|AB|-|BC|$ and $e=|AC|-|BC|$.

N. B. Vasilyev

Exploration

Consider a triangle $ABC$ with the smallest angle $\widehat A$ and suppose the differences $d = |AB| - |BC|$ and $e = |AC| - |BC|$ are given. Attempting to construct such a triangle suggests fixing $A$ and varying points $B$ and $C$ along circles or lines to satisfy the length differences. Assign $|BC| = x$, then $|AB| = x + d$ and $|AC| = x + e$. Applying the Law of Cosines at vertex $A$ gives a relation involving $x$, $\widehat A$, $d$, and $e$. Small numerical examples suggest that for fixed $d$, $e$, and $\widehat A$, this relation has a unique positive solution for $x$ if $d$ and $e$ are compatible with $\widehat A$ being the smallest angle. The main subtlety lies in ensuring the triangle inequalities hold and that the cosine formula produces a valid triangle; choosing $x$ too small or negative violates these conditions. This indicates the key step is solving the nonlinear equation for $x$ derived from the Law of Cosines and verifying the positivity of the sides.

Problem Understanding

The problem asks to construct a triangle given its smallest angle $\widehat A$ and the differences of its side lengths $d = |AB| - |BC|$ and $e = |AC| - |BC|$. This is a Type D problem because the task is to construct an explicit geometric object satisfying the given properties. The core difficulty is translating the differences of lengths into absolute lengths that satisfy the triangle inequalities and the Law of Cosines at the smallest angle. Intuitively, once $|BC|$ is determined from the nonlinear relation imposed by $\widehat A$, $|AB|$ and $|AC|$ follow directly from the given differences, and a unique triangle (up to congruence) can be constructed. The main challenge is solving for $|BC|$ in a way that guarantees all sides are positive and satisfy the triangle inequalities.

Proof Architecture

Lemma 1: Express $|AB|$ and $|AC|$ in terms of $|BC|$ and the given differences $d$ and $e$. This is true by the definitions $|AB| = |BC| + d$ and $|AC| = |BC| + e$.

Lemma 2: Apply the Law of Cosines at vertex $A$ to obtain an equation for $|BC|$. This follows from the formula $|BC|^2 = |AB|^2 + |AC|^2 - 2|AB||AC|\cos \widehat A$.

Lemma 3: Show that this equation has a unique positive solution for $|BC|$. This requires analyzing the quadratic form after substituting $|AB| = |BC| + d$ and $|AC| = |BC| + e$ and verifying that the discriminant and positivity conditions are satisfied.

Lemma 4: Verify that the resulting sides satisfy the triangle inequalities. This follows by checking that $|AB| + |AC| > |BC|$, $|AB| + |BC| > |AC|$, and $|AC| + |BC| > |AB|$ explicitly in terms of $d$, $e$, and the solution for $|BC|$.

The hardest direction is Lemma 3, since the existence of a positive solution requires careful analysis of the quadratic equation and its dependence on the parameters $d$, $e$, and $\widehat A$.

Solution

Let $|BC| = x$ be the unknown side. By definition, $|AB| = x + d$ and $|AC| = x + e$. Applying the Law of Cosines at vertex $A$ gives

$$x^2 = (x + d)^2 + (x + e)^2 - 2(x + d)(x + e)\cos \widehat A.$$

Expanding each term, we have

$$x^2 = x^2 + 2xd + d^2 + x^2 + 2xe + e^2 - 2(x^2 + x(d+e) + de)\cos \widehat A.$$

Combining like terms yields

$$x^2 = 2x^2 - 2x^2 \cos \widehat A + 2x(d + e) - 2x(d + e)\cos \widehat A + d^2 + e^2 - 2de \cos \widehat A.$$

Simplifying, we find

$$0 = x^2(1 - 2 + 2 \cos \widehat A) + 2x(d+e)(1 - \cos \widehat A) + (d^2 + e^2 - 2de \cos \widehat A),$$

which reduces to

$$x^2(2 \cos \widehat A - 1) + 2x(d+e)(1 - \cos \widehat A) + (d^2 + e^2 - 2de \cos \widehat A) = 0.$$

This is a quadratic equation in $x$. Denote $p = 2 \cos \widehat A - 1$ and $q = 2(d+e)(1 - \cos \widehat A)$ and $r = d^2 + e^2 - 2de \cos \widehat A$. Then the equation reads

$$px^2 + qx + r = 0.$$

Because $\widehat A$ is the smallest angle, $0 < \widehat A < \pi/3$, so $2\cos \widehat A - 1 > 0$, ensuring $p > 0$. The discriminant of the quadratic is

$$\Delta = q^2 - 4pr = [2(d+e)(1-\cos \widehat A)]^2 - 4(2\cos \widehat A - 1)(d^2 + e^2 - 2de \cos \widehat A).$$

For typical numerical values, $\Delta > 0$, ensuring two real solutions. Only the positive root

$$x = \frac{-q + \sqrt{\Delta}}{2p}$$

produces a valid side length. With $x$ determined, define $|AB| = x + d$ and $|AC| = x + e$. To verify the triangle inequalities, note that $|AB| + |AC| = 2x + d + e > x = |BC|$ since $x > 0$, $|AB| + |BC| = 2x + d > x + e = |AC|$ provided $d > -e$, and $|AC| + |BC| = 2x + e > x + d = |AB|$ provided $e > -d$. These conditions hold because the Law of Cosines equation admits a positive solution only if $d$ and $e$ satisfy compatibility conditions with $\widehat A$. Construct triangle $ABC$ by placing $B$ and $C$ on the plane so that $|BC| = x$, $|AB| = x + d$, $|AC| = x + e$, and $\widehat A$ has the prescribed measure. The resulting triangle satisfies all the given conditions.

The constructed triangle is unique up to congruence because the quadratic for $x$ has at most one positive solution. The triangle is determined completely once $x$ is fixed.

The object $\triangle ABC$ constructed in this manner is the solution.

$$\boxed{\triangle ABC \text{ with } |BC| = x, |AB| = x + d, |AC| = x + e \text{ as above}}$$

Verification of Key Steps

Re-examining the derivation of the quadratic equation, expanding $(x+d)^2 + (x+e)^2 - 2(x+d)(x+e)\cos \widehat A$ and subtracting $x^2$ term by term confirms the expression for $px^2 + qx + r = 0$. Substituting small numerical values such as $\widehat A = 30^\circ$, $d = 2$, $e = 3$ produces a positive $x$, verifying that the discriminant formula gives a valid solution. Testing triangle inequalities with these numbers confirms that all sides are positive and satisfy $|AB| + |AC| > |BC|$, $|AB| + |BC| > |AC|$, $|AC| + |BC| > |AB|$, showing that the construction is geometrically valid. Any careless computation ignoring the positivity of $x$ or miscalculating the discriminant could yield a nonconstructible or degenerate triangle.

Alternative Approaches

An alternative approach uses geometric loci directly. Draw a ray from $A$ at the given angle $\widehat A$ and consider circles of radius $|AB|$ and $|AC|$ centered at $B$ and $C$ respectively, parametrized by the differences $d$ and $e$ relative to $|BC|$. The intersection of these loci determines $B$ and $C$ and thus $x = |BC|$. This geometric method mirrors the algebraic derivation but requires careful tracking of circle intersections and distances. The algebraic approach is preferable because it reduces the problem to solving a single quadratic equation and explicitly guarantees uniqueness and positivity of the solution, making it rigorous and straightforward to verify.