Interior bisector of angle A meets BC at L

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Interior bisector of angle A meets BC at L

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Source: IMO 1987, Day 1, Problem 2

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orl

#1 h Nov 11, 2005, 8:01 PM • 3 Y

Y by ahmedosama, Adventure10, Mango247

In an acute-angled triangle $ABC$ the interior bisector of angle $A$ meets $BC$ at $L$ and meets the circumcircle of $ABC$ again at $N$ . From $L$ perpendiculars are drawn to $AB$ and $AC$ , with feet $K$ and $M$ respectively. Prove that the quadrilateral $AKNM$ and the triangle $ABC$ have equal areas.

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what

#2 h Jun 11, 2006, 6:35 PM • 2 Y

Y by Adventure10, Mango247

I use standard notations for triangles.

Since $AL$ is the bisector of $\angle {BAC}$ , we have $LM=LK$ , so $KM$ is perpendicular to $AN$ , that is to say, $[AKNM]=\frac 12\cdot KM\cdot AN$ .

In the cyclic quadrilater $AKLM$ we obtain $KM=AL\cdot \sin {\alpha}$ .

But $ABL$ and $ANC$ are similar, because $\angle{BAL}=\angle{NAC}=\frac{\alpha}2$ and $\angle{ABL}=\angle{ANC}=\beta$ and so $AL\cdot AN=b\cdot c$ .

Now substituting we have $[AKNM]=\frac 12\cdot KM\cdot AN=\frac 12\cdot AL\cdot \sin {\alpha}\cdot AN=\frac 12\cdot b\cdot c\cdot \sin {\alpha}=[ABC]$

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matematikolimpiyati

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matematikolimpiyati

#3 h Oct 28, 2013, 10:29 PM • 2 Y

Y by Adventure10, Mango247

$\angle BAN = \angle NAC = \angle NBC = \angle BCN$ .
Let the circumcircle of cyclic quadrilateral $AKLM$ meet $BC$ at $P$ , again.
$\angle KPL = \angle KAL = \angle CBN \Longrightarrow KP \parallel BN$ and $\angle LAM = \angle MPC = \angle PCN \Longrightarrow PM \parallel CN$ .
By area relation of parallel lines, we get $[BKP]=[KNP]$ and $[CMP]=[PMN]$ . So, $\begin{array}{rcl}[ABC] &=& [AKPM] + [BKP] + [CMP] \\ &=& [AKPM] + [KNP] + [PMN] \\ &=& [AKNM].\end{array}$

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#4 h Nov 2, 2013, 3:39 PM • 2 Y

Y by Adventure10, Mango247

Let $P, Q$ be the projections of $N$ onto $AB, AC$ respectively. Since $PN\parallel KL, NQ\parallel ML$ we infer that $[KLN]=[KLP],\ [LMN]=[LMQ]$ , hence $[AKNM]=[APL]+[AQL]$ . With $[BLP]=[CLQ]$ ( from $\triangle BNP\cong\triangle CQN\implies BP=CQ$ ) we are done, because, if $AB>AC$ , $[AKNM]= [APL]+[AQL] = [APL]+[BPL]+[AQL]-[CLQ]$ $=[ABL]+[ALC]=[ABC]$ ; if $AB<AC$ the proof is similar.

Best regards,
sunken rock

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#7 h Aug 7, 2015, 7:02 PM • 1 Y

Y by Adventure10

I don't think anyone has done it this way, so:
Solution
I have a question though, is working backwards like I did in my solution too "informal" for an olympiad proof?

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pad

#8 h Jun 21, 2018, 8:49 PM • 3 Y

Y by yayups, Adventure10, Mango247

Diagram
Solution

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9782 posts

jayme

#9 h Jun 23, 2018, 2:04 PM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,
again the Reim's theorem avoid angles chasing...

Sincerely
Jean-Louis

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PROF65

#10 h Jun 23, 2018, 6:15 PM • 2 Y

Y by Adventure10, Mango247

since $AL$ is $A$ -bisector then $LK=LM\implies ALK \equiv ALM\implies AK=AM\implies KM \perp AN$ besides $KM=\sin A .AL$ so $2[AKNM]=AN.KM=AN.AL\sin A=AB.AC\sin A=[ABC]$

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#12 h Jan 29, 2021, 5:30 AM • 1 Y

Y by teomihai

We use conventional notation, setting $BC=a, AC=b$ , and $AB=c$ . Also, let $\angle BAC=\theta$ and $AL=d$ .
Note trivially that $[ABC]=\frac{1}{2}bc\sin\theta$ . We note that by the Angle Bisector Theorem, we have $BL=\frac{ab}{b+c}$ and $LC=\frac{ac}{b+c}$ . Now by Power of a point, note that $LM=\frac{a^2bc}{d(b+c)^2}$ . We now compute $[AKNM]=2[AKN]=2\left(\frac{1}{2}d^2\sin\frac{\theta}{2}\cos\frac{\theta}{2}+\frac{a^2bc}{2(b+c)^2}\sin\frac{\theta}{2}\cos\frac{\theta}{2}\right)=\frac{1}{2}d^2\sin\theta+\frac{a^2bc}{2(b+c)^2}\sin\theta$ .
It is well-known that $d^2=\frac{bc}{(b+c)^2}((b+c)^2-a^2)$ so we have that $[AKNM]=\frac{bc}{2(b+c)^2}((b+c)^2-a^2)\sin\theta+\frac{a^2bc}{2(b+c)^2}\sin\theta=\frac{bc}{2(b+c)^2}\sin\theta((b+c)^2-a^2+a^2)=\frac{1}{2}bc\sin\theta$ as desired.

This post has been edited 1 time. Last edited by Pleaseletmewin, Jan 29, 2021, 5:31 AM

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#13 h Oct 15, 2021, 3:32 AM

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WLOG, suppose $AB \le AC$ , and let $P, Q$ be the projections of $N$ onto $AB, AC$ respectively. By Thales', we know $APNQ$ is cyclic, so the Spiral Center Lemma implies $N$ is the Miquel Point of $PQCB$ , i.e. $NPB \overset{+}{\sim} NQC$ . Now, since $N$ is the midpoint of arc $BC$ , we have $NB = NC$ , so the two aforementioned triangles are actually congruent.

Since $AN$ is the perpendicular bisector of $KM$ , we know $\frac{[NLK] + [NLM]}{[AKLM]} = \frac{2 \cdot [NLK]}{2 \cdot [ALK]} = \frac{NL}{AL}.$ Now, observe that $\frac{[BLK] + [CLM]}{[AKLM]} = \frac{\frac{BK}{AK} \cdot [ALK] + \frac{CM}{AM} \cdot [ALM]}{2 \cdot [ALK]}$ $= \frac{BK + CM}{2 \cdot AK}.$
Because $NPB \cong NQC$ , we have $AB + AC = (AP - BP) + (AQ + CQ) = (AP + AQ) + (CQ - BP) = 2 \cdot AP$ so $\frac{BK + CM}{2 \cdot AK} = \frac{(AB - AK) + (AC - AM)}{2 \cdot AK} = \frac{AB + AC - 2 \cdot AK}{2 \cdot AK}$ $= \frac{2 \cdot AP - 2 \cdot AK}{2 \cdot AK} = \frac{KP}{AK} = \frac{NL}{AL}.$ Now, it's easy to see this implies $[NLK] + [NLM] = [BLK] + [CLM].$ Adding $[AKLM]$ to both sides finishes. $\blacksquare$

Remark: Embarrassingly, this problem took me over an hour because I tried to use Trig in my initial attempt at a length bash.

Anyways, this length bash should mostly be natural, as working backwards from $\frac{[BLK] + [CLM]}{[AKLM]} = \frac{BK + CM}{2 \cdot AK} \overset{?}{=} \frac{NL}{AL}$ is a fairly direct line of attack. The hardest step is substituting $BK + CM = AB + AC - 2 \cdot AK$ but the rest of my solution follows smoothly afterwards.

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#14 h Jun 17, 2022, 2:12 PM

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Let $X,Y$ be the projections of $N$ onto $AB,AC$ respectively. From the fact that $L,N$ lie on the A-bisector we know that $KL=ML$ and $XN = YN$ . Now
$[ABC] = [ACL] +[ABL] = \frac{(AB+AC)\cdot KL}{2},$ $[AKNM]=AK \cdot XN,$ so it is sufficient to prove that $\frac{(AB+AC)\cdot KL}{2\cdot AK \cdot XN} = 1$ . From Thales' theorem $\frac{KL}{XN} = \frac{AK}{AX}$ , so
$\frac{(AB+AC)\cdot KL}{2\cdot AK \cdot XN} = \frac{(AB+AC)\cdot AK}{2\cdot AK \cdot AX} = \frac{(AB+AC)}{2 \cdot AX} = \frac{(AX \pm BX+AY \mp CY)}{2 \cdot AX} = 1$ as desired.

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#15 h Jun 17, 2022, 7:14 PM

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Consider projections $P,Q$ of $N$ onto $AB,AC$ respectively. Throughout the solution all areas are oriented. $|NB|=|NC|,|NP|=|NQ|\implies NBP\cong NCQ$ $|BP|=|CQ|,|LK|=|LM|\implies \text{area} (LBP)=\text{area} (LCQ).$ Finally we deduce $\text{area} (AKNM)-\text{area} (ABC)=\text{area} (LBP)+\text{area} (LQC)=0\text{ } \blacksquare$

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