AoPS - Problem

Source: IMO ShortList 1990, Problem 9 (HUN 3)

Tags: geometry, incenter, trigonometry, angle, midpoint, IMO Shortlist

orl

15.08.2008 20:55

The incenter of the triangle $ABC$ is $K.$ The midpoint of $AB$ is $C_1$ and that of $AC$ is $B_1.$ The lines $C_1K$ and $AC$ meet at $B_2,$ the lines $B_1K$ and $AB$ at $C_2.$ If the areas of the triangles $AB_2C_2$ and $ABC$ are equal, what is the measure of angle $\angle CAB?$

The QuattoMaster 6000

15.08.2008 22:58

orl wrote: The incenter of the triangle $ABC$ is $K.$ The midpoint of $AB$ is $C_1$ and that of $AC$ is $B_1.$ The lines $C_1K$ and $AC$ meet at $B_2,$ the lines $B_1K$ and $AB$ at $C_2.$ If the areas of the triangles $AB_2C_2$ and $ABC$ are equal, what is the measure of angle $\angle CAB?$

Let the tangency points between the incircle with

$AB$ and

$AC$ respectively be

$D$ and

$E$ . Let the midpoints of

$CD$ and

$BE$ be

$Y$ and

$X$ respectively. Furthermore, let the semiperimeter of

$\triangle ABC$ be

$s$ ,

$AB = c$ ,

$BC = a$ , and

$AC = b$ . It is well-known that

$B_1$ ,

$X$ , and

$K$ are collinear as are

$C_1$ ,

$Y$ , and

$K$ . Notice that

$1=\frac {[AB_2C_2]}{[ABC]} = \frac {\frac {AC_2\cdot AB_2\sin \angle BAC}{2}}{\frac {AB\cdot AC\sin \angle BAC}{2}} = \frac {AC_2}{AB}\cdot \frac {AB_2}{AC}$ Apply Menelaus to

$\triangle AEB$ with transversal

$C_2XB_1$ , we have that

$\frac {AC_2}{C_2B}\cdot \frac {BX}{XE}\cdot \frac {EB_1}{AB_1} = 1$ . We have that

$BX = XE$ , so

$\frac {AC_2}{C_2B} = \frac {AB_1}{EB_1} = \frac {\frac {b}{2}}{\frac {b}{2} - (s - a)} = \frac {b}{b - (a + b + c - 2a)} = \frac {b}{a - c}$ which means that

$\frac {C_2B}{AC_2} = \frac {a - c}{b}$ , so

$\frac {AB}{AC_2} = 1 + \frac {C_2B}{AC_2} = \frac {a + b - c}{b}$ . Similarly,

$\frac {AC}{AB_2} = \frac {a + c - b}{c}$ , so

$1 = \frac {AC}{AB_2}\cdot \frac {AB}{AC_2} = \frac {a^2 - (b - c)^2}{bc} = \frac {a^2 - b^2 - c^2 + 2bc}{bc}$

$bc = a^2 - b^2 - c^2 + 2bc$

$a^2 = b^2 + c^2 - bc$

$\cos \angle BAC = \frac {b^2 + c^2 - a^2}{2bc} = \frac {1}{2} = \cos 60$ from which it follows that

$\angle BAC = 60$ .

sunken rock

11.01.2010 18:05

One can use Cristea's theorem ( the Theorem of Transversal ) to find $AB_2$ and $AC_2$ , getting the same value for the ratio $\frac{AB \cdot AC}{AB_2 \cdot AC_2}$ as to The QuattoMaster 6000. Best regards, sunken rock

lym

12.01.2010 18:02

My Method： Let $BK, CK$ intersect $AC, AB$ at $D, E$ respectively， $CC_3 \parallel B_1K, BB_3 \parallel C_1K,CC_3 \cap BB_3=F,FJ \parallel AB, FL \parallel AC$ Then $C_2, B_2$ are the midpoints of $AC_3, AB_3$ respectively， $K$ is the common midpoint of $AF, EJ, DL$ . $\triangle BEK \cong \triangle BGK$ $\frac{AC}{AC_3}=\frac{FC}{FC_3}=\frac{CJ}{2KJ}$ ，similarly， $\frac{AB}{AB_3}=\frac{BL}{2KL}$ ， $S(AB_3C_3)=4S(AB_2C_2), BK$ bisector $EG,GL \parallel BK$ So on the basis of this, we can get： $S(ABC)=S(AB_2C_2) \Longleftrightarrow 4S(ABC)=S(AB_3C_3) \Longleftrightarrow \frac{CJ}{KJ}=\frac{KL}{BL}$ $\Longleftrightarrow \frac{CG}{BG}=\frac{KL}{BL} \Longleftrightarrow GL \parallel CK \Longleftrightarrow CK$ bisector $DG \Longleftrightarrow \triangle CDK \cong \triangle CGK$ $\Longleftrightarrow \angle KEB+\angle KDC=\angle KGB+\angle KGC=180^{\circ} \Longleftrightarrow A, D, K, E$ are on a circle $\Longleftrightarrow \angle BAC+\angle BKC=180^{\circ} \Longleftrightarrow \angle BAC=60^{\circ}$ . Q. E. D.

Attachments: