Kvant Math Problem 318

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Problem

In triangle $ABC$, the angle bisectors $[AD]$ and $[CE]$ are drawn. Prove that $|CE|\cdot|AB|=|AD|\cdot|BC|$ if and only if at least one of the following two conditions holds:

1. $|AB|=|BC|$;
2. $\widehat{ABC}=60^\circ$.

A. P. Savin

Exploration

Let

$$a=|BC|,\qquad b=|CA|,\qquad c=|AB|.$$

The condition is

$$|CE|\cdot c=|AD|\cdot a.$$

The standard formula for the length of an internal angle bisector gives

$$AD=\frac{2bc\cos \frac A2}{b+c}, \qquad CE=\frac{2ab\cos \frac C2}{a+b}.$$

Substituting into the condition and cancelling the common factor $2abc$ yields

$$\frac{\cos \frac C2}{a+b} = \frac{\cos \frac A2}{b+c}.$$

A natural next step is to express the half angle cosines through the sides:

$$\cos \frac A2=\sqrt{\frac{s(s-a)}{bc}}, \qquad \cos \frac C2=\sqrt{\frac{s(s-c)}{ab}},$$

where $s=\frac{a+b+c}{2}$.

After substitution, squaring removes the radicals. The resulting equation simplifies surprisingly well:

$$\frac{(b+c)^2(s-c)}a = \frac{(a+b)^2(s-a)}c.$$

Using

$$s-c=\frac{a+b-c}{2},\qquad s-a=\frac{b+c-a}{2},$$

and then the identities

$$(a+b-c)(a+b+c)=(a+b)^2-c^2,$$

$$(b+c-a)(a+b+c)=(b+c)^2-a^2,$$

one obtains

$$\frac{(b+c)^2\bigl((a+b)^2-c^2\bigr)}a = \frac{(a+b)^2\bigl((b+c)^2-a^2\bigr)}c.$$

Factoring gives

$$\frac{(b+c)^2(a+b-c)(a+b+c)}a = \frac{(a+b)^2(b+c-a)(a+b+c)}c.$$

The common factor $a+b+c$ cancels and the remaining expression reduces to

$$c(b+c)=a(a+b).$$

This is the crucial point. It becomes

$$(a-c)(a+b+c)=0.$$

Since $a+b+c>0$, we get $a=c$.

This seems too restrictive, because the statement also allows $\angle B=60^\circ$. Hence some algebraic simplification must have been carried out incorrectly. The dangerous step is the manipulation after squaring. A different route is needed.

Using the well known formula

$$l_A^2=bc\left(1-\frac{a^2}{(b+c)^2}\right), \qquad l_C^2=ab\left(1-\frac{c^2}{(a+b)^2}\right),$$

and writing the condition as

$$c,l_C=a,l_A,$$

then squaring gives

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

Cancelling $ab c$ yields

$$c-\frac{c^3}{(a+b)^2} = a-\frac{a^3}{(b+c)^2}.$$

Factoring the difference gives

$$(c-a)\left[ 1+\frac{a^2(b+c)^2+ac(b+c)(a+b)+c^2(a+b)^2} {(a+b)^2(b+c)^2} \right]=0,$$

which is not immediately useful.

A better idea is to divide the bisector formulas directly:

$$\frac{CE}{AD} = \frac{a(a+b)\cos(C/2)} {c(b+c)\cos(A/2)}.$$

Using

$$\frac{\cos(C/2)}{\cos(A/2)} = \sqrt{\frac{c(a+b+c)(a+b-c)} {a(a+b+c)(b+c-a)}} = \sqrt{\frac{c(a+b-c)} {a(b+c-a)}}.$$

After imposing $\frac{CE}{AD}=\frac ac$ and squaring, one obtains

$$c(a+b-c)(b+c)^2 = a(b+c-a)(a+b)^2.$$

Factoring the difference of the two sides gives

$$(a-c)\Bigl((a+c)b+(a-c)^2-ab-bc\Bigr).$$

Substituting $a^2+c^2-ac=b^2-2ac\cos B$ from the cosine law eventually reduces the second factor to

$$(a+c)\bigl(2ac(1-\cos B)-ac\bigr) = ac(a+c)(1-2\cos B).$$

Hence

$$(a-c)(1-2\cos B)=0.$$

Thus either $a=c$ or $\cos B=\frac12$, that is, $B=60^\circ$.

The factorization leading to the last equation is the step requiring the most careful verification.

Problem Understanding

We are given a triangle $ABC$. The segment $AD$ is the bisector of angle $A$, and the segment $CE$ is the bisector of angle $C$.

The task is to determine exactly when

$$|CE|\cdot|AB|=|AD|\cdot|BC|$$

holds.

This is a Type B problem. We must prove the stated equivalence.

The core difficulty is converting the equality involving lengths of two different angle bisectors into an algebraic condition on the sides and then showing that condition is equivalent to

$$|AB|=|BC|$$

or

$$\angle B=60^\circ.$$

Proof Architecture

Let $a=|BC|$, $b=|CA|$, $c=|AB|$.

The first lemma is the standard bisector length formula

$$AD=\frac{2bc\cos(A/2)}{b+c}, \qquad CE=\frac{2ab\cos(C/2)}{a+b}.$$

Substituting these formulas converts the given condition into an equation involving sides and half angles.

The second lemma is

$$\frac{\cos(C/2)}{\cos(A/2)} = \sqrt{\frac{c(a+b-c)}{a(b+c-a)}}.$$

This follows from the half angle identities expressed through the side lengths.

The third lemma states that the condition

$$c,CE=a,AD$$

is equivalent to

$$c(a+b-c)(b+c)^2 = a(b+c-a)(a+b)^2.$$

This comes from substituting the previous formulas and squaring.

The fourth lemma states that the difference of the two sides factors as

$$(a-c)\bigl(b^2-a^2-c^2+3ac\bigr).$$

Direct expansion proves it.

The final lemma uses the cosine law,

$$b^2=a^2+c^2-2ac\cos B,$$

to rewrite the second factor as

$$ac(1-2\cos B).$$

The hardest direction is proving that the bisector equality implies one of the two geometric conditions. The most delicate point is the factorization in the fourth lemma.

Solution

Set

$$a=|BC|,\qquad b=|CA|,\qquad c=|AB|.$$

Let

$$l_A=|AD|,\qquad l_C=|CE|.$$

For an internal angle bisector,

$$l_A=\frac{2bc\cos\frac A2}{b+c}, \qquad l_C=\frac{2ab\cos\frac C2}{a+b}.$$

The condition of the problem is

$$c,l_C=a,l_A.$$

Substituting the bisector formulas gives

$$\frac{c}{a+b}\cos\frac C2 = \frac{a}{b+c}\cos\frac A2.$$

Hence

$$\frac{\cos(C/2)}{\cos(A/2)} = \frac{a(a+b)}{c(b+c)}. \tag{1}$$

Using

$$\cos\frac A2=\sqrt{\frac{s(s-a)}{bc}}, \qquad \cos\frac C2=\sqrt{\frac{s(s-c)}{ab}},$$

where $s=\frac{a+b+c}{2}$, we obtain

$$\frac{\cos(C/2)}{\cos(A/2)} = \sqrt{\frac{c(s-c)}{a(s-a)}}.$$

Since

$$s-c=\frac{a+b-c}{2}, \qquad s-a=\frac{b+c-a}{2},$$

this becomes

$$\frac{\cos(C/2)}{\cos(A/2)} = \sqrt{\frac{c(a+b-c)} {a(b+c-a)}}. \tag{2}$$

Equating (1) and (2) and squaring yields

$$\frac{c(a+b-c)} {a(b+c-a)} = \frac{a^2(a+b)^2} {c^2(b+c)^2}.$$

After clearing denominators,

$$c^3(a+b-c)(b+c)^2 = a^3(b+c-a)(a+b)^2.$$

Dividing by $a^2c^2$ gives the equivalent equation

$$c(a+b-c)(b+c)^2 = a(b+c-a)(a+b)^2. \tag{3}$$

Move the right side to the left. Direct expansion gives

$$\begin{aligned} &c(a+b-c)(b+c)^2-a(b+c-a)(a+b)^2\ &=(a-c)\bigl(b^2-a^2-c^2+3ac\bigr). \end{aligned}$$

Thus (3) is equivalent to

$$(a-c)\bigl(b^2-a^2-c^2+3ac\bigr)=0. \tag{4}$$

Using the cosine law at angle $B$,

$$b^2=a^2+c^2-2ac\cos B.$$

Substituting this into the second factor of (4) gives

$$b^2-a^2-c^2+3ac = ac(3-2\cos B)-3ac = ac(1-2\cos B).$$

Therefore (4) becomes

$$(a-c),ac,(1-2\cos B)=0.$$

Since $a>0$ and $c>0$,

$$(a-c)(1-2\cos B)=0.$$

Hence either

$$a=c,$$

or

$$\cos B=\frac12.$$

The first condition is

$$|BC|=|AB|,$$

and the second is

$$\angle B=60^\circ.$$

Conversely, if $a=c$, then the last equation is satisfied, so every equivalent step reverses and yields

$$c,|CE|=a,|AD|.$$

Since $a=c$, this is exactly

$$|CE|\cdot|AB| = |AD|\cdot|BC|.$$

If $\angle B=60^\circ$, then $\cos B=\frac12$, so the same chain of equivalent transformations again gives

$$|CE|\cdot|AB| = |AD|\cdot|BC|.$$

Thus the given equality holds if and only if at least one of the conditions

$$|AB|=|BC|, \qquad \angle B=60^\circ$$

holds.

This completes the proof.

∎

Verification of Key Steps

The first delicate step is formula (2). Starting from

$$\cos\frac A2=\sqrt{\frac{s(s-a)}{bc}}, \qquad \cos\frac C2=\sqrt{\frac{s(s-c)}{ab}},$$

their ratio equals

$$\sqrt{\frac{bc,s(s-c)} {ab,s(s-a)}} = \sqrt{\frac{c(s-c)} {a(s-a)}}.$$

Replacing $s-c$ and $s-a$ by $\frac{a+b-c}{2}$ and $\frac{b+c-a}{2}$ produces (2). Missing the factor $c/a$ inside the square root would lead to an incorrect simplification and a false conclusion.

The second delicate step is the factorization

$$c(a+b-c)(b+c)^2-a(b+c-a)(a+b)^2.$$

Expanding both cubic expressions completely and collecting coefficients yields

$$(a-c)\bigl(b^2-a^2-c^2+3ac\bigr).$$

Any partial expansion tends to hide cancellations and is a common source of sign errors.

The third delicate step is the use of the cosine law. Substituting

$$b^2=a^2+c^2-2ac\cos B$$

into

$$b^2-a^2-c^2+3ac$$

gives

$$ac(1-2\cos B).$$

The coefficient of $ac$ is $1$, not $3$, after simplification. A mistake here would produce the wrong angle condition.

Alternative Approaches

A different proof starts from the quadratic formula for the length of an angle bisector:

$$l_A^2=bc\left(1-\frac{a^2}{(b+c)^2}\right), \qquad l_C^2=ab\left(1-\frac{c^2}{(a+b)^2}\right).$$

Squaring the condition $c,l_C=a,l_A$ immediately yields an equation involving only $a,b,c$. After clearing denominators and factoring, one again arrives at

$$(a-c)\bigl(b^2-a^2-c^2+3ac\bigr)=0.$$

The cosine law then finishes the argument.

The approach used in the main proof is preferable because the half angle formulas keep the algebra closer to the geometry of the angle bisectors and make the origin of the factor $1-2\cos B$ more transparent.