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China Northern 2005-20 (CNMO) 31p

geometry problems from China Northern Mathematical Olympiads
with aops links in the names
China Sijiazhuang

collected inside aops here

 
2005-07, 2009, 2016-20
missing 2008, 2010-2015


$AB$ is a chord of a circle with center $O$, $M$ is the midpoint of $AB$. A non-diameter chord is drawn through $M$ and intersects the circle at $C$ and $D$. The tangents of the circle from points $C$ and $D$ intersect line $AB$ at $P$ and $Q$, respectively. Prove that $PA$ = $QB$.

$AB$ is the diameter of circle $O$, $CD$ is a non-diameter chord that is perpendicular to $AB$. Let $E$ be the midpoint of $OC$, connect $AE$ and extend it to meet the circle at point $P$. Let $DP$ and $BC$ meet at $F$. Prove that $F$ is the midpoint of $BC$.

China Northern 2006 p3 (China TST 1989 p6)
$AD$ is the altitude of triangle $ABC$ at side $BC$. If $BC+AD=AB+AC$, then find the range of $\angle{A}$.

Let $ ABC$ be acute triangle. The circle with diameter $ AB$ intersects $ CA,\, CB$ at $ M,\, N,$ respectively. Draw $ CT\perp AB$ and intersects above circle at $ T$, where $ C$ and $ T$ lie on the same side of $ AB$. $ S$ is a point on $ AN$ such that $ BT = BS$. Prove that $ BS\perp SC$.

Let $ a,\, b,\, c$ be side lengths of a triangle and $ a+b+c = 3$. Find the minimum of
\[ a^{2}+b^{2}+c^{2}+\frac{4abc}{3}\]

The inradius of triangle $ ABC$ is $ 1$ and the side lengths of $ ABC$ are all integers. Prove that triangle $ ABC$ is right-angled.

As shown in figure , $\odot O$ is the inscribed circle of trapezoid $ABCD$, and the tangent points are $E, F, G, H$, $AB \parallel CD$. The line passing through$ B$, parallel to $AD$ intersects extension of $DC$ at point $P$. The extension of $AO$ intersects $CP$ at point $Q$. If $AE=BE$ , prove that $\angle CBQ = \angle PBQ$.
As shown in figure , it is known that $ABCD$ is parallelogram, $A,B,C$ lie on circle $\odot O_1$, $AD$ and $BD$ intersect $\odot O$ at points $E$ and $F$ respectively, $C,D,F$ lie on circle $\odot O_2$, $AD$ intersects $\odot O_2$ at point $G$. If the radii of circles $\odot O_1$, $\odot O_2$ are $R_1, R_2$ respectively, prove that $\frac{EG}{AD}=\frac{R_2^2}{R_1^2}$.


In an acute triangle $ABC$ , $AB>AC$ , $ \cos B+ \cos C=1$ , $E,F$ are on the extend line of $AB,AC$ such that $\angle ABF = \angle ACE = 90$   .
(1) Prove :$BE+CF=EF$ ; 
(2) Assume the bisector of $\angle EBC$ meet $EF$ at $P$ , prove that $CP$ is the bisector of $\angle BCF$.

Given a minor sector $AOB$ (Here minor means that $ \angle AOB <90$).  $O$ is the centre , chose a point  $C$  on arc $AB$ ,Let $P$ be a point on segment $OC$ , join $AP$ , $BP$ , draw a line through $B$ parallel to $AP$ , the line meet $OC$ at point $Q$ ,join $AQ$ . Prove that the area of polygon $AQPBO$ dose not change when points $P,C$ move .

From a point $P$ exterior of circle $\odot O$, we draw tangents $PA$, $PB$ and the secant $PCD$ . The line passing through point $C$ parallel to $PA$ intersects chords $AB$, $AD$ at points $E$, $F$ respectively. Prove that $CE = EF$.
Let $\odot O$ be the inscribed circle of $\vartriangle ABC$, with $D$, $E$, $N$ the touchpoints with sides $AB$, $AC$, $BC$ respectively. Extension of $NO$ intersects segment $DE$ at point $K$. Extension of $AK$ intersects segment $BC$ at point $M$. Prove that $M$ is the midpoint of $BC$.

As shown in figure , the inscribed circle of $ABC$ is intersects $BC$, $CA$, $AB$ at points $D$, $E$, $F$, repectively, and $P$ is a point inside the inscribed circle. The line segments $PA$, $PB$ and $PC$ intersect respectively the inscribed circle at points $X$, $Y$ and $Z$. Prove that the three lines $XD$, $YE$ and $ZF$ have a common point.
As shown in figure, from a point $P$ exterior of circle $\odot O$, we draw tangent $PA$ and the secant $PBC$. Let $AD \perp PO$ Prove that $AC$ is tangent to the circumcircle of $\vartriangle ABD$.

As shown in figure, given right $\vartriangle ABC$ with $\angle C=90^o$. $I$ is the incenter. The line $BI$ intersects segment $AC$ at the point $D$ . The line passing through $D$ parallel to $AI$ intersects $BC$ at point $E$. The line $EI$ intersects segment $AB$ at point $F$. Prove that $DF \perp AI$.

As shown in figure , in the pentagon $ABCDE$, $BC = DE$, $CD \parallel BE$, $AB>AE$. If $\angle BAC = \angle DAE$ and $\frac{AB}{BD}=\frac{AE}{ED}$. Prove that $AC$ bisects the line segment $BE$.


As shown in figure , $A,B$ are two fixed points of circle $\odot O$, $C$ is the midpoint of the major arc $AB$, $D$ is any point of the minor arc $AB$. Tangent at $D$ intersects tangents at $A,B$ at points $E,F$ respectively. Segments $CE$ and $CF$ intersect chord $AB$ at points $G$ and $H$ respectively. Prove that the length of line segment $GH$ has a fixed value.
As shown in figure , it is known that $M$ is the midpoint of side $BC$ of $\vartriangle ABC$. $\odot O$ passes through points $A, C$ and is tangent to $AM$. The extension of the segment $BA$ intersects $\odot O$ at point $D$. The lines $CD$ and $MA$ intersect at the point $P$. Prove that $PO \perp BC$.

As shown in the figure, given $\vartriangle ABC$ with $\angle B$, $\angle C$ acute angles, $AD \perp BC$, $DE \perp AC$, $M$ midpoint of $DE$, $AM \perp BE$. Prove that $\vartriangle ABC$ is isosceles.
As shown in the figure, in the parallelogram $ABCD$, $I$ is the incenter of $\vartriangle BCD$, and $H$ is the orthocenter of $\vartriangle IBD$. Prove that $\angle HAB=\angle HAD$.

It is known that $\odot O$ is the circumcircle of $\vartriangle ABC$ wwith diameter $AB$. The tangents of $\odot O$ at points $B$ and $C$ intersect at $P$ . The line perpendicular to $PA$ at point $A$ intersects the extension of $BC$ at point $D$. Extend $DP$ at length $PE = PB$. If $\angle ADP = 40^o$ , find the measure of $\angle E$.

As shown in figure , points $D,E,F$ lies the sides $AB$, $BC$ , $CA$ of the acute angle $\vartriangle ABC$ respectively. If $\angle EDC = \angle CDF$, $\angle FEA=\angle AED$, $\angle DFB =\angle BFE$, prove that the $CD$, $AE$, $BF$ are the altitudes of $\vartriangle ABC$.


As shown in figure , a circle of radius $1$ passes through vertex $A$ of $\vartriangle ABC$ and is tangent to the side $BC$ at the point $D$ , intersect sides $AB$ and $AC$ at points $E$ and $F$ respectively . Also$ EF$ bisects $\angle AFD$, and $\angle ADC = 80^o$ , Is there a triangle that satisfies the condition, so that $\frac{AB+BC+CA}{AD^2}$ is an irrational number, and the irrational number is the root of a quadratic equation with integral coefficients? If it does not exist, please prove it; if it exists, find the quadratic equation that satisfies the condition.
China Northern 2016 grade 10 p2
In isosceles triangle $ABC$, $\angle CAB=\angle CBA=\alpha$, points $P,Q$ are on different sides of line $AB$, and $\angle CAP=\angle ABQ=\beta,\angle CBP=\angle BAQ=\gamma$. Prove that $P,C,Q$ are collinear.

China Northern 2016 grade 10 p6 , 11 p6
Four points $B,E,A,F$ lie on line $AB$ in order, four points $C,G,D,H$ lie on line $CD$ in order, satisfying: $\frac{AE}{EB}=\frac{AF}{FB}=\frac{DG}{GC}=\frac{DH}{HC}=\frac{AD}{BC}.$ Prove that $FH\perp EG$.

China Northern 2016 grade 11 p2
Inscribed Triangle $ABC$ on circle $\odot O$. Bisector of $\angle ABC$ intersects $\odot O$ at $D$. Two lines $PB$ and $PC$ that are tangent to $\odot O$ intersect at $P$. $PD$ intersects $AC$ at $E$, $\odot O$ at $F$. Prove that $M,F,C,E$ are concyclic.

China Northern 2017 grade 10 p3 , 11 p2
Let \(D\) be the midpoint of side \(BC\) of triangle \(ABC\). Let \(E, F\) be points on sides \(AB, AC\) respectively such that \(DE = DF\). Prove that \(AE + AF = BE + CF \iff \angle EDF = \angle BAC\).

Triangle \(ABC\) has \(AB > AC\) and \(\angle A = 60^\circ \). Let \(M\) be the midpoint of \(BC\), \(N\) be the point on segment \(AB\) such that \(\angle BNM = 30^\circ\). Let \(D,E\) be points on \(AB, AC\) respectively. Let \(F, G, H\) be the midpoints of \(BE, CD, DE\) respectively. Let \(O\) be the circumcenter of triangle \(FGH\). Prove that \(O\) lies on line \(MN\).

Length of sides of regular hexagon $ABCDEF$ is $a$. Two moving points $M,N$ moves on sides $BC,DE$, satisfy that $\angle MAN=\frac{\pi}{3}$. Prove that $AM\cdot AN-BM\cdot DN$ is a definite value.

China Northern 2018 grade 10 p1
In triangle $ABC$, let the circumcenter, incenter, and orthocenter be $O$, $I$, and $H$ respectively. Segments $AO$, $AI$, and $AH$ intersect the circumcircle of triangle $ABC$ at $D$, $E$, and $F$. $CD$ intersects $AE$ at $M$ and $CE$ intersects $AF$ at $N$. Prove that $MN$ is parallel to $BC$.

A right triangle has the property that it's sides are pairwise relatively prime positive integers and that the ratio of it's area to it's perimeter is a perfect square. Find the minimum possible area of this triangle.

China Northern 2018 grade 10 p6 , 11 p5
Let $H$ be the orthocenter of triangle $ABC$. Let $D$ and $E$ be points on $AB$ and $AC$ such that $DE$ is parallel to $CH$. If the circumcircle of triangle $BDH$ passes through $M$, the midpoint of $DE$, then prove that $\angle ABM=\angle ACM$

China Northern 2018 grade 11 p3
$A,B,C,D,E$ lie on $\odot O$ in that order,and$$BD \cap CE=F,CE \cap AD=G,AD \cap BE=H,BE \cap AC=I,AC \cap BD=J.$$Prove that $\frac{FG}{CE}=\frac{GH}{DA}=\frac{HI}{BE}=\frac{IJ}{AC}=\frac{JF}{BD}$ when and only when $F,G,H,I,J$ are concyclic.


Two circles $O_1$ and $O_2$ intersect at $A,B$. Diameter $AC$ of $\odot O_1$ intersects $\odot O_2$ at $E$, Diameter $AD$ of $\odot O_2$ intersects $\odot O_1$ at $F$. $CF$ intersects $O_2$ at $H$, $DE$ intersects $O_1$ at $G,H$. $GH\cap O_1=P$. Prove that $PH=PK$.
Two circles $O_1$ and $O_2$ intersect at $A,B$. Bisector of outer angle $\angle O_1AO_2$ intersects $O_1$ at $C$, $O_2$ at $D$. $P$ is a point on $\odot(BCD)$, $CP\cap O_1=E,DP\cap O_2=F$. Prove that $PE=PF$.

In $\triangle ABC$, $\angle BAC = 60^{\circ}$, point $D$ lies on side $BC$, $O_1$ and $O_2$ are the centers of the circumcircles of $\triangle ABD$ and $\triangle ACD$, respectively. Lines $BO_1$ and $CO_2$ intersect at point $P$. If $I$ is the incenter of $\triangle ABC$ and $H$ is the orthocenter of $\triangle PBC$, then prove that the four points $B,C,I,H$ are on the same circle.

In $\triangle ABC$, $AB>AC$. Let $O$ and $I$ be the circumcenter and incenter respectively. Prove that if $\angle AIO = 30^{\circ}$, then $\angle ABC = 60^{\circ}$.




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