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?$
Problem
Source: IMO ShortList 1990, Problem 9 (HUN 3)
Tags: geometry, incenter, trigonometry, angle, midpoint, IMO Shortlist
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.
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