AA_2, BB_2, CC_1 are concurrent, internal bisectors, midpoints

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AA_2, BB_2, CC_1 are concurrent, internal bisectors, midpoints

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Source: 2015 British FST2 p1

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#1 h Jan 14, 2021, 8:34 PM • 4 Y

Y by HWenslawski, Mango247, Mango247, Mango247

An acute-angled triangle $\vartriangle ABC$ is given, and $A_1,B_1, C_1$ are the midpoints of sides $BC,CA, AB$ respectively. The internal angle bisector of $\angle AC_1C$ meets $AC$ at $L$ , and the internal angle bisector of $\angle CC_1B$ meets $BC$ at $K$ . The line $LK$ intersects $B_1C_1$ at $A_2$ , and $A_1C_1$ at $B_2$ . Prove that the lines $AA_2, BB_2, CC_1$ are concurrent.

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L567

#2 h Jan 15, 2021, 5:54 AM • 1 Y

Y by HWenslawski

Let $AA_2 \cap BC = X$ , $BB_2 \cap AC = Y$ . Let $a, b, c$ be the lengths of the three sides and $m_c$ be the length of $CC_1$ . Using angle bisector theorem, we get that $AL = \frac{cb}{c + 2m_c}, LC = \frac{2m_cb}{c+2m_c}$ . Since $LB_1 = LC - B_1C$ , $LB_1 = b \frac{2m_c - c}{2m_c + c}$ . Also, observe that $LK || AB$ . So,

$\frac{BX}{XC} = \frac{C_1A_2}{A_2B_1} = \frac{AL}{LB_1} = \frac{c}{2m_c - c}$

Similarly, we get that $\frac{CY}{YA} = \frac{2m_c - c}{c}$ . Since $\frac{AC_1}{C_1B} = 1$ , by ceva's theorem, the three lines concur

@below Thanks for pointing out the mistakes!, ive fixed them now

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#3 h Jan 15, 2021, 8:00 AM

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Dear L567, thanks for your nice solution. However there are a few mistakes.

L567 wrote:

Let $AA_1 \cap BC = X$ , $BB_1 \cap AC = Y$

Let $AA_2 \cap BC = X$ , $BB_2 \cap AC = Y$

L567 wrote:

Since $LB = LC - BC$ , $LB = b \frac{2m_c - c}{2m_c + c}$

Since $LB_1 = LC - B_1C$ , $LB_1 = b \frac{2m_c - c}{2m_c + c}$

L567 wrote:

$\frac{BX}{XC} = \frac{C_1A_2}{A_2B} = \frac{AL}{LB}$

$\frac{BX}{XC} = \frac{C_1A_2}{A_2B_1} = \frac{AL}{LB_1}$

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oolite

#4 h Mar 21, 2022, 9:40 PM

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L567 wrote:

Also, observe that $LK || AB$

I'm impressed that this is so easy that you can just "observe" it!

It took me a while. Here's what I ended up doing:

Say that the line through $L$ parallel to $AB$ crosses $C_1C$ at $X$ and crosses $C_1K$ at $K^\prime$ . We need to show that $K^\prime=K$ . Note that $\angle C_1LX=\angle LC_1A=\angle XC_1L$ then angle-chase further in $\triangle K^\prime C_1L$ to find $LX=K^\prime X$ . But since $C_1C$ is a median, this means $K^\prime$ lies on $BC$ , i.e. $K^\prime = BC\cap C_1K$ which equals $K$ by definition.

Is there a neater way to prove that $KL\parallel AB$ ?

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#5 h Mar 22, 2022, 3:19 AM • 1 Y

Y by oolite

Oops, its not "obvious", I should have elaborated, $\frac{CK}{KB} = \frac{CC_1}{C_1B} = \frac{CC_1}{C_1A} = \frac{BL}{LA}$ so $KL || AB$

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