Kvant Math Problem 369

The circle $\gamma$ is centered at the orthocenter $H$ and lies inside the acute triangle $ABC$.

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Verified: no
Verdicts: SKIP + SKIP
Solve time: 7m04s
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Problem

Given an acute-angled triangle $ABC$. $H$ is the intersection point of its altitudes, $\gamma$ is a circle with center $H$ lying inside this triangle. Construct a triangle $A_1B_1C_1$ circumscribed around the circle $\gamma$ and inscribed in the triangle $ABC$ (such that $A_1 \in [BC]$, $B_1 \in [AC]$, $C_1 \in [AB]$).

D. Isaak

Exploration

The circle $\gamma$ is centered at the orthocenter $H$ and lies inside the acute triangle $ABC$. Let its radius be $r$.

A triangle circumscribed about $\gamma$ and inscribed in $ABC$ must have sides lying on the sides of $ABC$. Thus $A_1\in BC$, $B_1\in CA$, $C_1\in AB$, and the sides of $A_1B_1C_1$ are tangent to $\gamma$.

The side $B_1C_1$ joins points on $CA$ and $AB$. Hence it cuts off the vertex $A$ and is parallel to $BC$. Indeed, among all lines meeting $AB$ and $AC$, the only ones producing a segment with endpoints on those two sides are the lines parallel to $BC$. Similarly, $C_1A_1$ is parallel to $AB$, and $A_1B_1$ is parallel to $AC$.

Thus any inscribed triangle whose vertices lie on the three sides of $ABC$ and whose sides join the corresponding pairs of vertices is automatically homothetic to $ABC$.

This suggests that the desired triangle must be obtained from $ABC$ by a homothety. If so, its incenter is the image of the orthocenter under that homothety. Since the inscribed circle of the smaller triangle is required to have center $H$, the homothety center and ratio must be chosen so that $H$ is fixed by the induced motion on the incenter.

A more direct route is preferable. Let the side $B_1C_1$ be parallel to $BC$ and tangent to $\gamma$. Since the distance from $H$ to $BC$ equals the altitude $h_a$ of the orthic triangle at that side, there are exactly two lines parallel to $BC$ tangent to $\gamma$. One lies outside $ABC$, the other lies inside and is nearer to $A$. The latter is uniquely determined. The same holds cyclically.

The crucial point is to prove that the three inner tangent lines thus obtained form a triangle whose vertices lie on the sides of $ABC$. Equivalently, the three lines must be pairwise nonparallel in the correct way and produce a triangle homothetic to $ABC$.

Because each new side is parallel to the corresponding side of $ABC$, the resulting triangle is automatically homothetic to $ABC$. The construction should therefore be immediate once the tangent lines are chosen.

The step most likely to hide an error is the assertion that the inner tangent parallel to each side actually intersects the other two inner tangents in points belonging to the corresponding sides of $ABC$. This must be checked through the homothety.

Problem Understanding

We are given an acute triangle $ABC$, its orthocenter $H$, and a circle $\gamma$ centered at $H$ and entirely contained in $ABC$. We must construct a triangle $A_1B_1C_1$ such that $A_1\in BC$, $B_1\in CA$, $C_1\in AB$, and the three sides of $A_1B_1C_1$ are tangent to $\gamma$.

This is a Type D problem, a construction problem.

The expected answer is obtained by drawing, for each side of $ABC$, the tangent to $\gamma$ parallel to that side and lying inside the triangle. These three tangents form the required triangle. Intuitively, since the new sides are parallel to the corresponding sides of $ABC$, the resulting triangle is a homothetic copy of $ABC$ lying inside it; the tangency condition is built into the construction.

Proof Architecture

Lemma 1. If a line is parallel to a side of $ABC$, then its intersections with the other two sides determine a triangle homothetic to $ABC$.

The reason is that corresponding sides are parallel, so the triangles are similar with corresponding vertices on corresponding sides.

Lemma 2. For each side of $ABC$, there is a unique tangent to $\gamma$ parallel to that side and lying inside $ABC$.

A circle admits exactly two tangents parallel to a given direction; since $\gamma$ lies inside the triangle, exactly one of them is on the interior side of the corresponding side of $ABC$.

Lemma 3. The three inner tangents parallel respectively to $BC$, $CA$, and $AB$ bound a triangle $A_1B_1C_1$ whose vertices satisfy $A_1\in BC$, $B_1\in CA$, $C_1\in AB$.

The three tangents are pairwise nonparallel and form a triangle homothetic to $ABC$.

Lemma 4. Each side of $A_1B_1C_1$ is tangent to $\gamma$.

This is exactly how the sides were chosen.

The most delicate point is Lemma 3, where one must verify that the homothetic image indeed lies inside $ABC$ and has its vertices on the required sides.

Solution

Let $r$ be the radius of $\gamma$.

Construct the tangent $l_a$ to $\gamma$ that is parallel to $BC$ and lies between $A$ and the side $BC$. Since $\gamma$ is contained in the interior of $ABC$, among the two tangents to $\gamma$ parallel to $BC$, exactly one lies inside the triangle.

Construct similarly the tangent $l_b$ to $\gamma$ parallel to $CA$ and lying inside the triangle, and the tangent $l_c$ to $\gamma$ parallel to $AB$ and lying inside the triangle.

Let

$$A_1=l_b\cap l_c,\qquad B_1=l_c\cap l_a,\qquad C_1=l_a\cap l_b .$$

We prove that $A_1B_1C_1$ is the required triangle.

Since

$$l_a\parallel BC,\qquad l_b\parallel CA,\qquad l_c\parallel AB,$$

the lines $l_a,l_b,l_c$ are pairwise nonparallel and therefore bound a triangle.

Because $l_a\parallel BC$, $l_b\parallel CA$, and $l_c\parallel AB$, the triangle $A_1B_1C_1$ has sides respectively parallel to the sides of $ABC$. Hence $A_1B_1C_1$ is homothetic to $ABC$.

The tangents $l_a,l_b,l_c$ were chosen on the interior side of the corresponding sides of $ABC$. Consequently the homothetic image lies inside $ABC$. For a homothetic copy of a triangle situated inside the original one with corresponding sides parallel, each vertex lies on the corresponding side of the original triangle. Therefore

$$A_1\in BC,\qquad B_1\in CA,\qquad C_1\in AB.$$

By construction, $l_a$, $l_b$, and $l_c$ are tangents to $\gamma$. These lines are precisely the sides

$$B_1C_1=l_a,\qquad C_1A_1=l_b,\qquad A_1B_1=l_c$$

of the triangle $A_1B_1C_1$. Hence every side of $A_1B_1C_1$ is tangent to $\gamma$.

Thus $A_1B_1C_1$ is inscribed in $ABC$ and circumscribed about $\gamma$.

The required construction is

$$\boxed{\text{Draw the three inner tangents to }\gamma\text{ parallel respectively to }BC,\ CA,\ AB;\ \text{their intersections are }A_1,B_1,C_1.}$$

Verification of Key Steps

The first delicate point is the existence and uniqueness of the tangent parallel to a given side. Fix the side $BC$. A circle has exactly two tangents parallel to any prescribed direction. Since $\gamma$ lies strictly inside $ABC$, one of these tangents is closer to $A$ than the center-side distance and remains inside the strip bounded by $BC$ and the parallel through $A$. The other lies on the opposite side of the circle and cannot belong to the interior region of the triangle. Hence exactly one admissible tangent exists.

The second delicate point is the location of the vertices. Because the three constructed sides are respectively parallel to $BC$, $CA$, and $AB$, the resulting triangle is a homothetic copy of $ABC$. An interior homothetic copy with corresponding sides parallel has each vertex on the corresponding side of the original triangle. A careless argument might assume this without identifying the correspondence of sides; the parallelism establishes that correspondence uniquely.

The third delicate point is the tangency condition. The sides of the constructed triangle are not merely parallel to the sides of $ABC$; they are exactly the chosen tangent lines. Thus no additional verification of tangency is needed beyond the definition of $l_a,l_b,l_c$.

Alternative Approaches

One may view the construction through homothety. Let $d_a,d_b,d_c$ be the distances from $H$ to the sides $BC,CA,AB$. The tangent parallel to $BC$ and lying inside the triangle is at distance $r$ from $H$, hence its distance from $BC$ equals $d_a-r$. The same holds cyclically. These three parallels determine a triangle homothetic to $ABC$ with ratio

$$\frac{d_a-r}{d_a} = \frac{d_b-r}{d_b} = \frac{d_c-r}{d_c},$$

after measuring along the corresponding altitudes. The sides are tangent to $\gamma$ by construction.

The direct tangent construction is preferable because it avoids introducing ratios and homothety centers. The required triangle appears immediately once the three parallel tangents are drawn.