Kvant Math Problem 58

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Verdicts: SKIP + SKIP
Solve time: 6m24s
Source on kvant.digital

Problem

Three lines intersecting at a single point are given in the plane. A point is marked on one of them. It is known that the lines are the angle bisectors of some triangle, and the marked point is one of its vertices. Construct this triangle.

E. B. Dynkin, S. A. Molchanov, A. L. Rosenthal, A. K. Tolpygo

Mathematical Problems. 3rd ed. – Moscow: Nauka, 1971.

Exploration

The three given lines are the three internal angle bisectors of a triangle. Since all internal angle bisectors meet at the incenter, their common intersection point is the incenter $I$ of the sought triangle.

The marked point $A$ lies on one of the bisectors. Hence the line through $A$ and $I$ is the bisector of angle $A$ of the triangle. The other two given lines must be the bisectors at the remaining vertices.

Suppose these two bisectors are denoted by $\ell_B$ and $\ell_C$. Let

$$\angle(\ell_B,\ell_C)=\theta .$$

For a triangle with incenter $I$,

$$\angle BIC = 90^\circ+\frac A2.$$

The angle between two intersecting lines is determined only up to supplement. Since $\ell_B$ and $\ell_C$ are the lines $IB$ and $IC$, the acute angle between them equals

$$180^\circ-\left(90^\circ+\frac A2\right) =90^\circ-\frac A2.$$

Thus

$$A=180^\circ-2\theta .$$

This already determines the angle at the marked vertex.

The next question is how to recover the sides through $A$. Since $AI$ is the bisector of angle $A$, each side through $A$ must form with $AI$ the angle

$$\frac A2=90^\circ-\theta.$$

Therefore the two sides through $A$ are uniquely obtained by drawing through $A$ the two lines making angle $90^\circ-\theta$ with $AI$.

A possible hidden error is whether these lines actually meet $\ell_B$ and $\ell_C$ in points whose bisectors are $\ell_B$ and $\ell_C$. This must be checked carefully. If one of the sides through $A$ meets $\ell_B$ at $B$ and the other meets $\ell_C$ at $C$, then the side through $A$ and $B$ makes angle $A/2$ with $BI$, and the side through $A$ and $C$ makes angle $A/2$ with $CI$. Since

$$\frac A2=90^\circ-\theta,$$

the angle between $AB$ and $\ell_B$ equals the angle between $AC$ and $\ell_C$, namely $A/2$. Hence $\ell_B$ and $\ell_C$ are indeed bisectors at $B$ and $C$. This is the crucial point.

Problem Understanding

We are given three concurrent lines and a point $A$ lying on one of them. It is known that the three lines are the internal angle bisectors of a triangle and that $A$ is one vertex of that triangle. The task is to construct the triangle.

This is a Type D problem.

The common intersection point of the three bisectors is the incenter $I$ of the triangle. Since the marked point lies on one of the bisectors, the line $AI$ is the bisector at the vertex $A$. The core difficulty is to recover the sides through $A$ from the positions of the other two bisectors.

The answer is a unique triangle. The angle between the two bisectors not containing $A$ determines the angle at $A$, and once that angle is known, the two sides through $A$ are determined by symmetry about the bisector $AI$.

Proof Architecture

The first lemma states that if $\ell_B$ and $\ell_C$ are the bisectors at vertices $B$ and $C$ and meet at the incenter $I$, then the acute angle between $\ell_B$ and $\ell_C$ equals $90^\circ-\frac A2$; this follows from the standard identity $\angle BIC=90^\circ+\frac A2$.

The second lemma states that the angle $A$ is uniquely determined by the given three bisectors through the formula $A=180^\circ-2\theta$, where $\theta$ is the acute angle between the two bisectors not passing through the marked vertex.

The third lemma states that the sides through $A$ are exactly the two lines through $A$ making angle $A/2$ with the bisector $AI$; this is the defining property of an angle bisector.

The final verification shows that if these two lines meet $\ell_B$ and $\ell_C$ at points $B$ and $C$, then $\ell_B$ and $\ell_C$ are indeed the bisectors of angles $B$ and $C$.

The most delicate lemma is the last one, because it must be proved that the constructed triangle has precisely the prescribed bisectors.

Solution

Let the three given lines meet at the point $I$. Since they are the angle bisectors of some triangle, $I$ is the incenter of that triangle.

Let the marked point be $A$. The given line passing through $A$ will be denoted by $\ell_A$. Then $\ell_A=AI$ is the bisector of angle $A$.

Denote by $\ell_B$ and $\ell_C$ the other two given lines.

Let $\theta$ be the acute angle between $\ell_B$ and $\ell_C$.

Since $\ell_B$ and $\ell_C$ are the lines $IB$ and $IC$, the angle $\angle BIC$ is the obtuse angle formed by these lines. For every triangle,

$$\angle BIC=90^\circ+\frac A2.$$

Hence the acute angle between the same two lines equals

$$180^\circ-\left(90^\circ+\frac A2\right) =90^\circ-\frac A2.$$

Therefore

$$\theta=90^\circ-\frac A2,$$

and consequently

$$A=180^\circ-2\theta.$$

Thus the angle at the vertex $A$ is determined.

Construct through $A$ the two lines $m$ and $n$ forming with the bisector $\ell_A$ equal angles

$$\frac A2=90^\circ-\theta$$

on opposite sides of $\ell_A$.

Let

$$B=m\cap \ell_B,\qquad C=n\cap \ell_C.$$

The triangle $ABC$ is the required construction.

It remains to verify it.

By construction, the lines $AB$ and $AC$ are the lines $m$ and $n$, and each makes angle $A/2$ with $\ell_A$. Hence $\ell_A$ bisects angle $A$.

Since $\ell_B$ forms angle $\theta$ with $\ell_C$, and

$$\theta=90^\circ-\frac A2,$$

the angle between $\ell_B$ and $AB$ equals $A/2$. The angle between $\ell_B$ and $AI$ also equals $A/2$, because $AI$ is the bisector of angle $A$. Thus the rays $BA$ and $BC$ are symmetric with respect to $\ell_B$. Therefore $\ell_B$ bisects angle $B$.

The same argument applied to $\ell_C$ shows that $\ell_C$ bisects angle $C$.

Hence the three angle bisectors of the constructed triangle are exactly the three given lines. Since $A$ is the prescribed marked vertex, the construction satisfies all requirements.

The required triangle is

$$\boxed{\text{the triangle }ABC\text{ obtained by the construction above}.}$$

Verification of Key Steps

The identity

$$\angle BIC=90^\circ+\frac A2$$

may be derived directly. Since $BI$ and $CI$ are angle bisectors,

$$\angle IBC=\frac B2,\qquad \angle ICB=\frac C2.$$

In triangle $BIC$,

$$\angle BIC =180^\circ-\frac B2-\frac C2 =180^\circ-\frac{B+C}{2}.$$

Using $A+B+C=180^\circ$,

$$\angle BIC =180^\circ-\frac{180^\circ-A}{2} =90^\circ+\frac A2.$$

The passage from the angle between the lines $\ell_B$ and $\ell_C$ to the value of $A$ requires choosing the acute angle between the lines. If one mistakenly uses the obtuse angle, one obtains $A<0$, which is impossible. The acute angle is

$$\theta=180^\circ-\angle BIC,$$

hence $\theta=90^\circ-\frac A2$.

The verification that $\ell_B$ is the bisector at $B$ cannot be replaced by an appeal to symmetry alone. One must check that the angle between $AB$ and $\ell_B$ equals the angle between $\ell_B$ and $BC$. The first equals $A/2$ by construction. The second equals the angle between $\ell_B$ and $AI$, which is also $A/2$ because the acute angle between $\ell_B$ and $\ell_C$ is $\theta=90^\circ-A/2$. This establishes the bisector property rigorously.

Alternative Approaches

A different approach uses the incenter characterization by equal distances from the sides. After computing $A$ from the angle between the other two bisectors, one may construct the sides through $A$ as before. Then the remaining side can be obtained as the common tangent to the two angle sectors determined by the bisectors $\ell_B$ and $\ell_C$. This leads to the same triangle but introduces additional constructions involving perpendiculars from the incenter.

The main approach is preferable because it uses only the fundamental relation $\angle BIC=90^\circ+\frac A2$ and the defining property of an angle bisector. Every step is immediate from the geometry of the incenter, and the construction emerges directly from the given data.