Kvant Math Problem 349

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Problem

What condition must the side lengths of a triangle satisfy so that the triangle formed by its

1. altitudes,
2. medians,
3. angle bisectors

is similar to the given triangle?

A. P. Savin

Exploration

Let the given triangle have sides $a,b,c$ opposite angles $A,B,C$.

For the triangle formed by the altitudes, the side lengths are proportional to the altitudes

$$h_a=\frac{2\Delta}{a},\qquad h_b=\frac{2\Delta}{b},\qquad h_c=\frac{2\Delta}{c}.$$

Hence its sides are proportional to

$$\frac1a,\frac1b,\frac1c.$$

If this triangle is similar to the original one, then the unordered triples

$$(a,b,c),\qquad \left(\frac1a,\frac1b,\frac1c\right)$$

must be proportional. After arranging the sides in the same order, there is a constant $k$ such that

$$a=\frac{k}{a},\quad b=\frac{k}{b},\quad c=\frac{k}{c},$$

which gives

$$a^2=b^2=c^2.$$

Thus the triangle must be equilateral. An equilateral triangle indeed has equal altitudes, so this case works.

For the medians, recall

$$m_a=\frac12\sqrt{2b^2+2c^2-a^2},$$

and cyclically. If the triangle of medians is similar to the original triangle, there is a permutation of the sides under which

$$(m_a,m_b,m_c)=k(a,b,c).$$

Since

$$m_a^2-m_b^2=\frac34(b^2-a^2),$$

the ordering of the medians is the same as the ordering of the corresponding opposite sides. Hence the permutation must preserve the order of unequal sides. After relabeling,

$$m_a=ka,\quad m_b=kb,\quad m_c=kc.$$

Substituting into the median formulas gives

$$4k^2a^2=2b^2+2c^2-a^2,$$

and the analogous equations. Subtracting pairs yields

$$(4k^2+3)(a^2-b^2)=0,$$

so $a=b$, and similarly $b=c$. Again only the equilateral triangle remains.

For the bisectors,

$$l_a^2=bc\left(1-\frac{a^2}{(b+c)^2}\right).$$

Using

$$a^2=b^2+c^2-2bc\cos A,$$

one obtains

$$l_a=\frac{2bc\cos\frac A2}{b+c}.$$

The crucial point is to determine when

$$(l_a,l_b,l_c)$$

can be proportional to

$$(a,b,c).$$

The identity

$$\frac{l_a}{a} =\frac{\sqrt{bc}}{,b+c,}\cdot \frac1{\sin\frac A2}$$

suggests introducing

$$x=\frac A2,\quad y=\frac B2,\quad z=\frac C2,\qquad x+y+z=\frac\pi2.$$

Using

$$a=2R\sin A=4R\sin x\cos x,$$

one finds

$$l_a=4R\sin y\sin z\cos x,$$

hence

$$\frac{l_a}{a} =\frac{\sin y\sin z}{\sin x}.$$

If similarity holds, this ratio is the same constant for all three vertices. Therefore

$$\frac{\sin y\sin z}{\sin x} = \frac{\sin z\sin x}{\sin y} = \frac{\sin x\sin y}{\sin z}.$$

Comparing the first two equalities gives

$$\sin^2 y=\sin^2 x.$$

Since $x,y\in(0,\pi/2)$, this yields $x=y$. Likewise $y=z$. Hence

$$A=B=C,$$

so the triangle is equilateral.

Thus every case seems to lead only to the equilateral triangle.

Problem Understanding

We are given a triangle and three derived triangles whose side lengths are respectively the altitudes, the medians, and the internal angle bisectors of the original triangle. For each construction, we must determine all triangles for which the derived triangle is similar to the original one.

This is a Type A problem. We must classify all triangles satisfying the required similarity condition.

The core difficulty is the case of the angle bisectors. For altitudes and medians, algebraic relations between side lengths quickly force equality of the sides. For bisectors, one must express the bisector lengths in a form that interacts effectively with the similarity assumption.

The answer in all three cases is that the original triangle must be equilateral.

Proof Architecture

Lemma 1. If the triangle of altitudes is similar to the original triangle, then the original triangle is equilateral. The altitude lengths are proportional to $(1/a,1/b,1/c)$, and proportionality with $(a,b,c)$ forces $a=b=c$.

Lemma 2. If the triangle of medians is similar to the original triangle, then the original triangle is equilateral. The median formulas produce three linear relations in $a^2,b^2,c^2$; subtracting them forces equality of the side lengths.

Lemma 3. For a triangle with circumradius $R$ and angles $A,B,C$,

$$l_a=4R\sin\frac B2,\sin\frac C2,\cos\frac A2.$$

This follows from the standard bisector formula together with the sine law.

Lemma 4. If the triangle of bisectors is similar to the original triangle, then

$$\frac{\sin\frac B2\sin\frac C2}{\sin\frac A2} = \frac{\sin\frac C2\sin\frac A2}{\sin\frac B2} = \frac{\sin\frac A2\sin\frac B2}{\sin\frac C2}.$$

This comes from Lemma 3 and the sine law.

Lemma 5. The equalities of Lemma 4 imply

$$A=B=C.$$

Since all half-angles lie in $(0,\pi/2)$, equality of their sines implies equality of the angles.

The hardest direction is the bisector case, especially deriving a usable expression for $l_a/a$ and extracting angle equality from the similarity condition.

Solution

Let the given triangle have side lengths $a,b,c$, altitudes $h_a,h_b,h_c$, medians $m_a,m_b,m_c$, and internal angle bisectors $l_a,l_b,l_c$.

We consider the three constructions separately.

1. Triangle formed by the altitudes

Since

$$h_a=\frac{2\Delta}{a},\qquad h_b=\frac{2\Delta}{b},\qquad h_c=\frac{2\Delta}{c},$$

the side lengths of the altitude triangle are proportional to

$$\frac1a,\frac1b,\frac1c.$$

Assume that this triangle is similar to the original one. Then the unordered triples

$$(a,b,c),\qquad \left(\frac1a,\frac1b,\frac1c\right)$$

are proportional.

Arrange the side lengths in nondecreasing order. Taking reciprocals reverses the order, so the largest side corresponds to the largest side, the middle side to the middle side, and the smallest side to the smallest side. Hence there exists a constant $k>0$ such that

$$a=\frac{k}{a},\qquad b=\frac{k}{b},\qquad c=\frac{k}{c}.$$

Therefore

$$a^2=b^2=c^2,$$

and thus

$$a=b=c.$$

An equilateral triangle has equal altitudes, so the altitude triangle is also equilateral and hence similar to the original triangle.

Thus the required condition is

$$a=b=c.$$

2. Triangle formed by the medians

The median formulas are

$$m_a=\frac12\sqrt{2b^2+2c^2-a^2},$$

with analogous formulas for $m_b$ and $m_c$.

Assume that the median triangle is similar to the original triangle.

From the formulas,

$$m_a^2-m_b^2 = \frac14\bigl(2b^2+2c^2-a^2-(2c^2+2a^2-b^2)\bigr) = \frac34(b^2-a^2).$$

Hence

$$a>b \iff m_a<m_b.$$

The ordering of the medians is exactly opposite to the ordering of the corresponding sides. After arranging both triples in increasing order, similarity yields a constant $k>0$ such that

$$m_a=ka,\qquad m_b=kb,\qquad m_c=kc.$$

Squaring and substituting the median formulas,

$$4k^2a^2=2b^2+2c^2-a^2,$$

$$4k^2b^2=2c^2+2a^2-b^2,$$

$$4k^2c^2=2a^2+2b^2-c^2.$$

Subtracting the first two equations gives

$$(4k^2+3)(a^2-b^2)=0.$$

Since $4k^2+3>0$,

$$a=b.$$

Subtracting the second and third equations gives

$$b=c.$$

Consequently

$$a=b=c.$$

An equilateral triangle has equal medians, so the median triangle is equilateral and therefore similar to the original one.

Thus the required condition is

$$a=b=c.$$

3. Triangle formed by the angle bisectors

Let $R$ be the circumradius.

The standard bisector formula gives

$$l_a^2 = bc\left(1-\frac{a^2}{(b+c)^2}\right).$$

Using the sine law,

$$a=2R\sin A,\qquad b=2R\sin B,\qquad c=2R\sin C.$$

Since

$$A+B+C=\pi,$$

we have

$$\sin B+\sin C = 2\sin\frac{B+C}{2}\cos\frac{B-C}{2} = 2\cos\frac A2\cos\frac{B-C}{2},$$

and

$$\sin A = 2\sin\frac A2\cos\frac A2.$$

Substituting these identities into the bisector formula and simplifying,

$$l_a = 4R\sin\frac B2,\sin\frac C2,\cos\frac A2.$$

Similarly,

$$l_b = 4R\sin\frac C2,\sin\frac A2,\cos\frac B2,$$

$$l_c = 4R\sin\frac A2,\sin\frac B2,\cos\frac C2.$$

Assume that the bisector triangle is similar to the original triangle. Then there exists $k>0$ such that

$$l_a=ka,\qquad l_b=kb,\qquad l_c=kc.$$

Since

$$a=4R\sin\frac A2\cos\frac A2,$$

we obtain

$$k= \frac{l_a}{a} = \frac{\sin\frac B2\sin\frac C2}{\sin\frac A2}.$$

Analogously,

$$k= \frac{\sin\frac C2\sin\frac A2}{\sin\frac B2} = \frac{\sin\frac A2\sin\frac B2}{\sin\frac C2}.$$

Hence

$$\frac{\sin\frac B2\sin\frac C2}{\sin\frac A2} = \frac{\sin\frac C2\sin\frac A2}{\sin\frac B2}.$$

Since all sines are positive,

$$\sin^2\frac B2=\sin^2\frac A2.$$

The angles $\frac A2$ and $\frac B2$ belong to $(0,\pi/2)$, where the sine function is strictly increasing. Therefore

$$A=B.$$

The same argument applied to the other pairs gives

$$B=C.$$

Thus

$$A=B=C,$$

and consequently

$$a=b=c.$$

An equilateral triangle has equal angle bisectors, so the bisector triangle is equilateral and similar to the original triangle.

Therefore the required condition is

$$a=b=c.$$

The answer in all three cases is

$$\boxed{a=b=c}.$$

Verification of Key Steps

For the altitude case, the delicate point is matching the side correspondence. After arranging the side lengths increasingly,

$$a_1\le a_2\le a_3,$$

the reciprocal triple becomes

$$\frac1{a_3}\le \frac1{a_2}\le \frac1{a_1}.$$

Similarity of two triangles compares equal positions in the ordered lists. Hence

$$a_1=\frac{k}{a_3},\qquad a_2=\frac{k}{a_2},\qquad a_3=\frac{k}{a_1}.$$

Multiplying the first and third equations gives $a_1a_3=k$, while the second gives $a_2^2=k$. Thus

$$a_1a_3=a_2^2.$$

Because $a_1\le a_2\le a_3$, equality in the arithmetic-geometric mean inequality forces

$$a_1=a_2=a_3.$$

No non-equilateral solution survives.

For the median case, the crucial computation is

$$m_a^2-m_b^2=\frac34(b^2-a^2).$$

A sign error here would destroy the argument. Recomputing directly,

$$4(m_a^2-m_b^2) = (2b^2+2c^2-a^2)-(2c^2+2a^2-b^2) = 3(b^2-a^2),$$

which confirms the formula.

For the bisector case, the essential step is extracting equality of angles from the similarity condition. From

$$\frac{\sin\frac B2\sin\frac C2}{\sin\frac A2} = \frac{\sin\frac C2\sin\frac A2}{\sin\frac B2},$$

positivity of the sines allows cancellation:

$$\sin^2\frac B2=\sin^2\frac A2.$$

Since

$$0<\frac A2,\frac B2<\frac\pi2,$$

the sine function is injective, giving $A=B$. Without the restriction to this interval, the conclusion would not be justified.

Alternative Approaches

For the altitude case one may work entirely with areas. If the altitude triangle is similar to the original one, then the ratios of corresponding sides equal the ratios of corresponding altitudes. Since each altitude is inversely proportional to the corresponding side, the similarity ratio must equal its reciprocal. This forces the similarity ratio to be $1$, and then all corresponding sides are equal.

For the bisector case one can use the formula

$$l_a=\frac{2bc\cos\frac A2}{b+c}.$$

Assuming $l_a=ka$, substitution of

$$a=2R\sin A,\qquad b=2R\sin B,\qquad c=2R\sin C$$

leads directly to

$$k=\frac{\sin\frac B2\sin\frac C2}{\sin\frac A2}.$$

The remainder is identical. The main solution is preferable because it keeps all three cases in terms of side lengths and standard triangle formulas, making the parallel structure of the argument transparent.