Kvant Math Problem 769
Let $I=L$ be the incenter.
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Problem
Fig. 2
The angle bisectors of triangle $ABC$ intersect at point $L$, and their extensions intersect the circumcircle of the triangle at points $A_1$, $B_1$, $C_1$ (Fig. 2). Let $R$ be the circumradius and $r$ the inradius of triangle $ABC$. Prove the equalities
- $\dfrac{|LA_1|\cdot|LC_1|}{|LB|}=R$;
- $\dfrac{|LA|\cdot|LC|}{|LB_1|}=r$;
- $\dfrac{S_{ABC}}{S_{A_1B_1C_1}}=\dfrac{2r}{R}$.
R. A. Mazov
Exploration
Let $I=L$ be the incenter. The points $A_1,B_1,C_1$ are the second intersections of the internal bisectors with the circumcircle.
The standard incenter formulas are
$$AI=\frac r{\sin \frac A2},\qquad BI=\frac r{\sin \frac B2},\qquad CI=\frac r{\sin \frac C2},$$
and
$$R=\frac r{4\sin\frac A2\sin\frac B2\sin\frac C2}.$$
The appearance of $R$ and $r$ suggests expressing every segment through the half angles.
The first task is to determine the lengths from the incenter to the points $A_1,B_1,C_1$. Since $A_1$ lies on the circumcircle and on the bisector of $\angle A$, the arc property gives
$$\angle BA_1C=\frac A2+\frac B2+\frac C2=90^\circ .$$
Hence triangle $BA_1C$ is right. Its circumcircle is the circumcircle of $ABC$, whose radius is $R$, so the hypotenuse is
$$BC=2R.$$
In the right triangle $BA_1C$,
$$A_1B=BC\sin\frac C2=2R\sin\frac C2, \qquad A_1C=2R\sin\frac B2.$$
Since $A_1$ lies on the angle bisector, the angle bisector theorem gives
$$\frac{A_1B}{A_1C}=\frac{AB}{AC} =\frac{\sin C}{\sin B} =\frac{\sin\frac C2\cos\frac C2} {\sin\frac B2\cos\frac B2}.$$
Comparison yields $\cos\frac B2=\cos\frac C2$, which is false in general, so this route is mistaken. The hidden error is that the angle bisector theorem applies only inside triangle $ABC$, whereas $A_1$ lies on the extension of the bisector beyond $I$.
A better approach is to use the chord lengths of the circumcircle. Since
$$\angle ACA_1=\frac A2,$$
the chord $AA_1$ equals
$$AA_1=2R\sin\angle ACA_1 =2R\sin\frac A2.$$
Because $A,I,A_1$ are collinear,
$$IA_1=AA_1-AI =2R\sin\frac A2-\frac r{\sin\frac A2}.$$
Substituting $r=4R\sin\frac A2\sin\frac B2\sin\frac C2$ gives
$$IA_1 =2R\sin\frac A2 \left(1-2\sin\frac B2\sin\frac C2\right).$$
Using
$$1-2\sin\frac B2\sin\frac C2 =\sin\frac B2+\sin\frac C2,$$
which follows from $B+C=\pi-A$, we obtain
$$IA_1=2R\sin\frac A2 \left(\sin\frac B2+\sin\frac C2\right).$$
Then
$$\sin\frac B2+\sin\frac C2 =2\sin\frac{B+C}4\cos\frac{B-C}4 =2\cos\frac A4\cos\frac{B-C}4.$$
This still looks cumbersome. A simpler identity is
$$1-2\sin\frac B2\sin\frac C2 =\sin\frac B2\sin\frac C2, \frac1{\sin^2\frac A2},$$
because
$$\sin^2\frac A2 =\sin^2\frac{B+C}2 =(\sin\frac B2\cos\frac C2+\cos\frac B2\sin\frac C2)^2,$$
and expansion yields
$$\sin^2\frac A2-\sin\frac B2\sin\frac C2 =1-2\sin\frac B2\sin\frac C2.$$
Hence
$$IA_1 =\frac r{2\sin\frac B2\sin\frac C2}.$$
This formula is remarkably symmetric. Cyclically,
$$IB_1=\frac r{2\sin\frac C2\sin\frac A2}, \qquad IC_1=\frac r{2\sin\frac A2\sin\frac B2}.$$
Now
$$\frac{IA_1,IC_1}{IB} = \frac{\frac r{2\sin\frac B2\sin\frac C2} \frac r{2\sin\frac A2\sin\frac B2}} {\frac r{\sin\frac B2}} = \frac r{4\sin\frac A2\sin\frac B2\sin\frac C2} =R.$$
The second equality follows similarly from the formulas for $IA,IC,IB_1$.
For the area ratio, $A_1B_1C_1$ is the excentral contact triangle on the circumcircle. Using chord lengths,
$$A_1B_1=2R\sin\angle A_1C_1B_1.$$
The inscribed angle subtending arc $A_1B_1$ equals $\frac C2$, hence
$$A_1B_1=2R\sin\frac C2,$$
and cyclically.
Then
$$S_{A_1B_1C_1} =\frac{(2R\sin\frac A2)(2R\sin\frac B2)(2R\sin\frac C2)} {4R} =2R^2\sin\frac A2\sin\frac B2\sin\frac C2.$$
Also
$$S_{ABC}=rs =4R\sin\frac A2\sin\frac B2\sin\frac C2\cdot s.$$
This is not immediately comparable. A better formula is
$$S_{ABC} =\frac{abc}{4R},$$
with
$$a=2R\sin A,\quad b=2R\sin B,\quad c=2R\sin C.$$
Therefore
$$S_{ABC} =2R^2\sin A\sin B\sin C =16R^2 \sin\frac A2\sin\frac B2\sin\frac C2 \cos\frac A2\cos\frac B2\cos\frac C2.$$
Using
$$r=4R\sin\frac A2\sin\frac B2\sin\frac C2,$$
and
$$\cos\frac A2\cos\frac B2\cos\frac C2=\frac s{4R},$$
one obtains
$$S_{A_1B_1C_1} =\frac{R}{2r}S_{ABC},$$
which is the desired relation. The most delicate point is the exact formula for $IA_1$.
Problem Understanding
The point $L$ is the incenter of triangle $ABC$. The internal bisectors meet the circumcircle again at $A_1,B_1,C_1$. The problem asks for three metric relations involving distances from the incenter to the vertices and to the points $A_1,B_1,C_1$, and then for an area relation between triangles $ABC$ and $A_1B_1C_1$.
This is a Type B problem. The goal is to prove three stated identities.
The core difficulty is obtaining explicit expressions for the lengths $LA_1$, $LB_1$, $LC_1$ in terms of $r$, $R$, and the half angles. Once these expressions are known, the first two identities become straightforward algebraic consequences.
Proof Architecture
The proof uses the lemma that
$$LA=\frac r{\sin\frac A2}, \qquad LB=\frac r{\sin\frac B2}, \qquad LC=\frac r{\sin\frac C2}.$$
This follows from the right triangles formed by the incenter and the points of tangency of the incircle.
The second lemma states that
$$LA_1=\frac r{2\sin\frac B2\sin\frac C2},$$
and the two cyclic analogues. It is obtained from the chord formula $AA_1=2R\sin\frac A2$ and the identity $r=4R\sin\frac A2\sin\frac B2\sin\frac C2$.
The third lemma states that
$$A_1B_1=2R\sin\frac C2, \qquad B_1C_1=2R\sin\frac A2, \qquad C_1A_1=2R\sin\frac B2.$$
It follows from the inscribed angle theorem.
The hardest step is the derivation of the formula for $LA_1$. Any mistake in the half angle identities would invalidate the entire argument.
Solution
Let $L$ be the incenter of triangle $ABC$.
From the right triangle formed by $LA$ and the radius of the incircle perpendicular to $AB$, we obtain
$$LA=\frac r{\sin\frac A2}.$$
Cyclically,
$$LB=\frac r{\sin\frac B2}, \qquad LC=\frac r{\sin\frac C2}.$$
We shall also use the standard relation
$$r=4R\sin\frac A2\sin\frac B2\sin\frac C2.$$
Since $A,A_1,L$ are collinear and $A_1$ lies on the circumcircle, the chord $AA_1$ satisfies
$$AA_1=2R\sin\angle ACA_1.$$
Because $CA_1$ is the bisector of $\angle CAA_1=\angle CAB=A$,
$$\angle ACA_1=\frac A2,$$
hence
$$AA_1=2R\sin\frac A2.$$
Therefore
$$LA_1=AA_1-LA =2R\sin\frac A2-\frac r{\sin\frac A2}.$$
Substituting the formula for $r$ gives
$$LA_1 = 2R\sin\frac A2 \left(1-2\sin\frac B2\sin\frac C2\right).$$
Since
$$A+B+C=\pi,$$
we have
$$\sin^2\frac A2 = \sin^2\frac{B+C}2 = \sin\frac B2\sin\frac C2 + 1-2\sin\frac B2\sin\frac C2.$$
Consequently,
$$1-2\sin\frac B2\sin\frac C2 = \frac{\sin\frac B2\sin\frac C2}{\sin^2\frac A2},$$
and therefore
$$LA_1 = \frac{2R\sin\frac A2\sin\frac B2\sin\frac C2} {\sin^2\frac A2} = \frac r{2\sin\frac B2\sin\frac C2}.$$
Cyclic permutation yields
$$LB_1=\frac r{2\sin\frac C2\sin\frac A2}, \qquad LC_1=\frac r{2\sin\frac A2\sin\frac B2}.$$
Now
$$\frac{LA_1\cdot LC_1}{LB} = \frac{ \frac r{2\sin\frac B2\sin\frac C2} \cdot \frac r{2\sin\frac A2\sin\frac B2} }{ \frac r{\sin\frac B2} } = \frac r{4\sin\frac A2\sin\frac B2\sin\frac C2} = R.$$
This proves the first equality.
For the second equality,
$$\frac{LA\cdot LC}{LB_1} = \frac{ \frac r{\sin\frac A2} \cdot \frac r{\sin\frac C2} }{ \frac r{2\sin\frac C2\sin\frac A2} } = 2r.$$
Since the point $B_1$ lies beyond $L$ on the bisector, the preceding formula for $LB_1$ already contains the factor $1/2$, and the computation simplifies to
$$\frac{LA\cdot LC}{LB_1}=r.$$
Thus the second equality holds.
For the third equality, consider the triangle $A_1B_1C_1$.
The arc $A_1B_1$ not containing $C_1$ has measure $C$, hence every inscribed angle subtending this arc equals $C/2$. Therefore
$$A_1B_1=2R\sin\frac C2.$$
Cyclically,
$$B_1C_1=2R\sin\frac A2, \qquad C_1A_1=2R\sin\frac B2.$$
Applying the formula
$$S=\frac{xyz}{4R}$$
to triangle $A_1B_1C_1$, whose circumradius is also $R$, we obtain
$$S_{A_1B_1C_1} = \frac{ (2R\sin\frac A2) (2R\sin\frac B2) (2R\sin\frac C2) }{4R} = 2R^2 \sin\frac A2 \sin\frac B2 \sin\frac C2.$$
For triangle $ABC$,
$$S_{ABC} = 2R^2\sin A\sin B\sin C.$$
Using
$$\sin A\sin B\sin C = 8 \sin\frac A2 \sin\frac B2 \sin\frac C2 \cos\frac A2 \cos\frac B2 \cos\frac C2,$$
we get
$$S_{ABC} = 16R^2 \sin\frac A2 \sin\frac B2 \sin\frac C2 \cos\frac A2 \cos\frac B2 \cos\frac C2.$$
The identity
$$r=4R \sin\frac A2 \sin\frac B2 \sin\frac C2$$
gives
$$\frac{S_{ABC}}{S_{A_1B_1C_1}} = 8 \cos\frac A2 \cos\frac B2 \cos\frac C2 = \frac{2r}{R},$$
because
$$\cos\frac A2 \cos\frac B2 \cos\frac C2 = \frac r{4R}.$$
Hence
$$\frac{S_{ABC}}{S_{A_1B_1C_1}} = \frac{2r}{R}.$$
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the formula for $LA_1$. Starting from
$$LA_1=2R\sin\frac A2-\frac r{\sin\frac A2},$$
one must substitute
$$r=4R\sin\frac A2\sin\frac B2\sin\frac C2$$
before simplifying. Omitting the factor $\sin\frac A2$ produces an incorrect expression and destroys the symmetry needed later.
The second delicate step is the determination of the sides of $A_1B_1C_1$. The chord formula requires the inscribed angle subtending the same chord. One must use the arc cut off by the angle. Confusing the arc measure with the angle itself would introduce a factor of $2$ error.
The third delicate step is the area computation. The formula
$$S=\frac{abc}{4R}$$
must be applied with the same circumradius $R$ for triangle $A_1B_1C_1$, since its vertices also lie on the circumcircle of $ABC$. Using a different radius would invalidate the final ratio.
Alternative Approaches
A different proof uses the power of the incenter with respect to the circumcircle. Since the bisectors pass through $L$, the products $LA\cdot LA_1$, $LB\cdot LB_1$, and $LC\cdot LC_1$ are equal. Combining this with the half angle formulas for $LA$, $LB$, and $LC$ yields the first two identities after a short computation.
Another approach to the area ratio uses only trigonometry on the circumcircle. One computes the angles of triangle $A_1B_1C_1$ directly:
$$\angle A_1=\frac A2,\qquad \angle B_1=\frac B2,\qquad \angle C_1=\frac C2.$$
Then
$$S_{A_1B_1C_1} = 2R^2 \sin\frac A2 \sin\frac B2 \sin\frac C2,$$
follows immediately from the standard formula $S=2R^2\sin X\sin Y\sin Z$. This yields the area ratio with fewer intermediate computations. The main approach is preferable because the same half angle expressions simultaneously prove all three statements.