Show that ∠AKM = ∠KPC - [UKRMO 2009 Grade 8]

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0 replies

jlacosta

Apr 2, 2025

0 replies

functional equation interesting

skellyrah 10

N 23 minutes ago by jasperE3

find all functions IR->IR such that $xf(x+yf(xy)) + f(f(x)) = f(xf(y))^2 + (x+1)f(x)$

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skellyrah

Apr 24, 2025

jasperE3

23 minutes ago

Line through orthocenter

juckter 14

N an hour ago by lpieleanu

Source: Mexico National Olympiad 2011 Problem 2

Let $ABC$ be an acute triangle and $\Gamma$ its circumcircle. Let $l$ be the line tangent to $\Gamma$ at $A$ . Let $D$ and $E$ be the intersections of the circumference with center $B$ and radius $AB$ with lines $l$ and $AC$ , respectively. Prove the orthocenter of $ABC$ lies on line $DE$ .

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juckter

Jun 22, 2014

lpieleanu

an hour ago

Geometry with parallel lines.

falantrng 33

N an hour ago by joshualiu315

Source: RMM 2018,D1 P1

Let $ABCD$ be a cyclic quadrilateral an let $P$ be a point on the side $AB.$ The diagonals $AC$ meets the segments $DP$ at $Q.$ The line through $P$ parallel to $CD$ mmets the extension of the side $CB$ beyond $B$ at $K.$ The line through $Q$ parallel to $BD$ meets the extension of the side $CB$ beyond $B$ at $L.$ Prove that the circumcircles of the triangles $BKP$ and $CLQ$ are tangent .

33 replies

falantrng

Feb 24, 2018

joshualiu315

an hour ago

subsets of {1,2,...,mn}

N.T.TUAN 9

N an hour ago by AshAuktober

Source: USA TST 2005, Problem 1

Let $n$ be an integer greater than $1$ . For a positive integer $m$ , let $S_{m}= \{ 1,2,\ldots, mn\}$ . Suppose that there exists a $2n$ -element set $T$ such that
(a) each element of $T$ is an $m$ -element subset of $S_{m}$ ;
(b) each pair of elements of $T$ shares at most one common element;
and
(c) each element of $S_{m}$ is contained in exactly two elements of $T$ .

Determine the maximum possible value of $m$ in terms of $n$ .

9 replies

N.T.TUAN

May 14, 2007

AshAuktober

an hour ago

No more topics!

Show that ∠AKM = ∠KPC - [UKRMO 2009 Grade 8]

Amir Hossein 2

N Mar 17, 2022 by Mahdi_Mashayekhi

In the triangle $ABC$ given that $\angle ABC = 120^\circ .$ The bisector of $\angle B$ meet $AC$ at $M$ and external bisector of $\angle BCA$ meet $AB$ at $P.$ Segments $MP$ and $BC$ intersects at $K$ . Prove that $\angle AKM = \angle KPC .$

2 replies

Amir Hossein

Jan 21, 2011

Mahdi_Mashayekhi

Mar 17, 2022

Show that ∠AKM = ∠KPC - [UKRMO 2009 Grade 8]

G H J

Reveal topic tags

geometry

angle bisector

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Amir Hossein

5452 posts

Amir Hossein

#1 h Jan 21, 2011, 11:46 AM • 3 Y

Y by son7, Adventure10, Mango247

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FantasyLover

1784 posts

FantasyLover

#2 h Jan 22, 2011, 1:55 AM • 2 Y

Y by Adventure10, Mango247

Solution

Lemma. In $\triangle ABC$ , internal angle bisector of $\angle ABC$ and external angle bisectors of $\angle BCA$ and $\angle CAB$ are concurrent.
Proof. It suffices to prove that $B$ -excenter $I_B$ lies on the angle bisector of $\angle B$ . Let $X$ and $Y$ be the projections of $I_B$ onto $BA$ and $BC$ , respectively. In $\triangle BXI_B$ and $\triangle BYI_B$ , $BX=BY$ and $I_BX=I_BY$ , implying that $\triangle BXI_B\cong \triangle BYI_B$ . Hence, $I_B$ lies on the angle bisector of $\angle B$ , and the lemma is proven. $\square$

According to above lemma, $P$ is the $M$ -excenter of $\triangle MBC$ , and therefore $K$ lies on the angle bisector of $\angle BMC$ .

By the lemma again, since $K$ is $A$ -excenter of $\triangle ABM$ , we have that $K$ lies on the angle bisector of $\angle BAC$ .

From here on, easy angle chasing shows that $\angle AKM=\angle KPC=30^{\circ}$ . We are done. $\blacksquare$

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Mahdi_Mashayekhi

694 posts

Mahdi_Mashayekhi

#3 h Mar 17, 2022, 4:40 PM

Y by

Note that $BP$ is external bisector of $\angle MBC$ so $P$ is excenter of $BMC$ . $\angle CPM = \angle 60 + \angle \frac{A}{2} - \angle CMP = \angle 30$ . we have $\frac{CM}{MA} . \frac{AP}{BP} . \frac{BK}{KC} = \frac{CB}{AB} . \frac{CA}{CB} . \frac{BK}{KC} = 1$ so $\frac{BK}{KC} = \frac{AB}{AC}$ so $AK$ is angle bisector of $A$ . Now we have $\angle AMK = \angle 30 + \angle \frac{A}{2} - \angle \frac{A}{2} = \angle 30$ .
We're Done.

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