IMO Math

Point $P$ lies in the interior of a triangle $ABC$. Lines $AP$, $BP$, and $CP$ meet the opposite sides of triangle $ABC$ at $A$', $B'$, and $C'$ respectively. Let $P_A$ the midpoint of the segment joining the incenters of triangles $BPC'$ and $CPB'$, and define points $P_B$ and $P_C$ analogously. Show that if \[ AB'+BC'+CA'=AC'+BA'+CB' \]then points $P,P_A,P_B,$ and $P_C$ are concyclic. 

  Problem:  Let $ABC$ be a triangle with circumcenter $O$ and incenter $I$. The points $D,E$ and $F$ on the sides $BC,CA$ and $AB$ respectively are such that $BD+BF=CA$ and $CD+CE=AB$. The circumcircles of the triangles $BFD$ and $CDE$ intersect at $P \neq D$. Prove that $OP=OI$. Solution 1:  Let circle with radius $OI$ intersect $BI, CI$ at $S, T \ne I$ respectively. Let $H$ be the foot of $S$ to $AB$ and let $BI$ meet $(ABC)$ at $M \ne A.$ It's known that if any circle passing through $S,B$ intersects $BC,BA$ at $D',F'$ respectively then $BD'+BF' = 2BH.$ But$$BH = BS \cdot \cos \left(\frac{1}{2} \angle ABC \right) = IM \cdot \cos \left(\frac{1}{2} \angle ABC \right) = MA \cdot \cos \left(\frac{1}{2} \angle ABC \right) = \frac{1}{2}AC$$so in fact $BFSD$ cyclic, similarly $CDTE$ cyclic. Then$$\angle TPS = 360^\circ - \angle TPD - \angle SPD = \frac{1}{2} \angle ABC  + \frac{1}{2} \angle ACB = \angle TIS$$so $P$ lies on $(TIS)$ centered at $O,$ done. $\square$ Soluti

Let $ABC$ be an equilateral triangle, and point $P$ on its circumcircle. Let $PA$ and $BC$ intersect at $D$, $PB$ and $AC$ intersect at $E$, and $PC$ and $AB$ intersect at $F$. Prove that the area of $\triangle DEF$ is twice the area of $\triangle ABC$. Solution 1:  Note that $$\angle{DPF}=\angle{FPE}=\angle{EPD}=120^{\circ}.$$ Now, using the sin area formula, we get $$[DEF]=[DPF]+[FPE]+[EPD]=\frac{\sqrt{3}}{4}\left(DP \cdot FP+FP \cdot EP+EP \cdot DP\right).$$We also have that $$[ABC]=BC^2 \cdot \frac{\sqrt{3}}{4}.$$ So, it suffices to prove $$\frac{\sqrt{3}}{4}\left(DP \cdot FP+FP \cdot EP+EP \cdot DP\right)=2 \cdot \frac{BC^2\sqrt{3}}{4} \implies DP \cdot FP+FP \cdot EP+EP \cdot DP=2BC^2.$$By the Law of Cosines on $\triangle{BPC}$, we obtain $$BC^2=b^2+c^2-2 \cdot b \cdot c \cdot \cos{120^{\circ}}=b^2+c^2+bc,$$ and Ptolemy's Theorem on quadrilateral $ABPC$ gives$$AP \cdot BC=BP \cdot AC+CP \cdot AB \implies AP=b+c.$$From some angle chasing, we know that $\triangle{ACD} \sim \tri

Let $ ABCDE$ be a convex pentagon such that \[ \angle BAC = \angle CAD = \angle DAE\qquad \text{and}\qquad \angle ABC = \angle ACD = \angle ADE. \]The diagonals $BD$ and $CE$ meet at $P$. Prove that the line $AP$ bisects the side $CD$.

 Let $ABC$ be a triangle. Circle $\Gamma$ passes through $A$, meets segments $AB$ and $AC$ again at points $D$ and $E$ respectively, and intersects segment $BC$ at $F$ and $G$ such that $F$ lies between $B$ and $G$. The tangent to circle $BDF$ at $F$ and the tangent to circle $CEG$ at $G$ meet at point $T$. Suppose that points $A$ and $T$ are distinct. Prove that line $AT$ is parallel to $BC$.

Let $ABC$ be an acute triangle with orthocenter $H$, and let $W$ be a point on the side $BC$, lying strictly between $B$ and $C$. The points $M$ and $N$ are the feet of the altitudes from $B$ and $C$, respectively. Denote by $\omega_1$ is the circumcircle of $BWN$, and let $X$ be the point on $\omega_1$ such that $WX$ is a diameter of $\omega_1$. Analogously, denote by $\omega_2$ the circumcircle of triangle $CWM$, and let $Y$ be the point such that $WY$ is a diameter of $\omega_2$. Prove that $X,Y$ and $H$ are collinear.

Let $ABC$ be a triangle with circumcircle $\Omega$ and incentre $I$. Let the line passing through $I$ and perpendicular to $CI$ intersect the segment $BC$ and the arc $BC$ (not containing $A$) of $\Omega$ at points $U$ and $V$ , respectively. Let the line passing through $U$ and parallel to $AI$ intersect $AV$ at $X$, and let the line passing through $V$ and parallel to $AI$ intersect $AB$ at $Y$ . Let $W$ and $Z$ be the midpoints of $AX$ and $BC$, respectively. Prove that if the points $I, X,$ and $Y$ are collinear, then the points $I, W ,$ and $Z$ are also collinear.

As I said in an earlier post, I will also post problems that are not from the IMO but I believe that they are of the same level. Let $ABC$ be a triangle with $AB<AC$ and $I$ be your incenter. Let $M$ and $N$ be the midpoints of the sides $BC$ and $AC$, respectively. If the lines $AI$ and $IN$ are perpendicular, prove that the line $AI$ is tangent to the circumcircle of $\triangle IMC$.

 In an acute-angled triangle $ABC$ on the sides $AB$, $BC$, $AC$ the points $H$, $L$, $K$ so that $CH \perp AB$, $HL \parallel AC$, $HK \parallel BC$. Let $P$ and $Q$ feet of altitudes of a triangle $HBL$, drawn from the vertices $H$ and $B$ respectively. Prove that the feet of the altitudes of the triangle $AKH$, drawn from the vertices $A$ and $H$ lie on the line $PQ$.

Today (March 18th) is my birthday! I was born on March 18th, 2009.  Thanks to my parents for supporting me in my journey so much :) I probably would've given up a long time back if it wasn't for them.  Now it's time for 2009 IMO SL #18, which is 2009 IMO SL #G3.  Let $ABC$ be a triangle. The incircle of $ABC$ touches the sides $AB$ and $AC$ at the points $Z$ and $Y$, respectively. Let $G$ be the point where the lines $BY$ and $CZ$ meet, and let $R$ and $S$ be points such that the two quadrilaterals $BCYR$ and $BCSZ$ are parallelogram. Prove that $GR=GS$. Let the incircle touch $\overline{BC}$ at $X$ and let the $A$-excircle $\omega_A$ touch $\overline{BC}$ at $X'$ and $\overline{AC}$ at $Y'$. Denote $\omega_R$ and $\omega_S$ as the circles centered at $R$ and $S$ respectively, both with radius $0$. Notice that $BR=CY=CX=BX'$ and $YR=BC=AY'-AY=YY'$, so we have that $\overline{BY}$ is the radical axis of $\omega_A$ and $\omega_R$. Similarly, $\overline{CZ}

I got several emails about how I haven't posted in a week, while I usually post during the first few days of the week. So don't worry, I'm still alive :D  Now let's get to the math.

Let $S$ be a convex quadrilateral $ABCD$ and $O$ a point inside it. The feet of the perpendiculars from $O$ to $AB, BC, CD, DA$ are $A_1, B_1, C_1, D_1$ respectively. The feet of the perpendiculars from $O$ to the sides of $S_i$, the quadrilateral $A_iB_iC_iD_i$, are $A_{i+1}B_{i+1}C_{i+1}D_{i+1}$, where $i = 1, 2, 3.$ Prove that $S_4$ is similar to S.

Let $\omega$ be the circumcircle of a triangle $ABC$. Denote by $M$ and $N$ the midpoints of the sides $AB$ and $AC$, respectively, and denote by $T$ the midpoint of the arc $BC$ of $\omega$ not containing $A$. The circumcircles of the triangles $AMT$ and $ANT$ intersect the perpendicular bisectors of $AC$ and $AB$ at points $X$ and $Y$, respectively; assume that $X$ and $Y$ lie inside the triangle $ABC$. The lines $MN$ and $XY$ intersect at $K$. Prove that $KA=KT$.

Let $ABC$ be an acute-angled triangle and let $D, E$, and $F$ be the feet of altitudes from $A, B$, and $C$ to sides $BC, CA$, and $AB$, respectively. Denote by $\omega_B$ and $\omega_C$ the incircles of triangles $BDF$ and $CDE$, and let these circles be tangent to segments $DF$ and $DE$ at $M$ and $N$, respectively. Let line $MN$ meet circles $\omega_B$ and $\omega_C$ again at $P \ne M$ and $Q \ne N$, respectively. Prove that $MP = NQ$.

Suppose medians $m_a$ and $m_b$ of a triangle are orthogonal. Prove that  a) Using medians of that triangle it is possible to construct a rectangular triangle.  b) The following inequality:\[5(a^2+b^2-c^2) \geq 8ab,\]is valid, where $a,b$ and $c$ are side length of the given triangle.

Let $ ABCD$ be a convex quadrilateral, and let $ R_A, R_B, R_C, R_D$ denote the circumradii of the triangles $ DAB, ABC, BCD, CDA,$ respectively. Prove that $R_A + R_C > R_B + R_D$ if and only if $ \angle A + \angle C > \angle B + \angle D.$

This problem was requested to be solved by a user. If you would like to request a problem too, please use the form on the left menu of the blog.  Triangle $ABC$ has side lengths $AB=4$, $BC=5$, and $CA=6$. Points $D$ and $E$ are on ray $AB$ with $AB<AD<AE$. The point $F \neq C$ is a point of intersection of the circumcircles of $\triangle ACD$ and $\triangle EBC$ satisfying $DF=2$ and $EF=7$. Then $BE$ can be expressed as $\tfrac{a+b\sqrt{c}}{d}$, where $a$, $b$, $c$, and $d$ are positive integers such that $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a+b+c+d$. 

Prove that there exists a positive constant $c$ such that the following statement is true: Consider an integer $n > 1$, and a set $\mathcal S$ of $n$ points in the plane such that the distance between any two different points in $\mathcal S$ is at least 1. It follows that there is a line $\ell$ separating $\mathcal S$ such that the distance from any point of $\mathcal S$ to $\ell$ is at least $cn^{-1/3}$. (A line $\ell$ separates a set of points S if some segment joining two points in $\mathcal S$ crosses $\ell$.) Suppose that among all projections of points in $\mathcal{S}$ onto some line $m$, the maximum possible distance between two consecutive projections is $\delta$. We need to prove $\delta \ge \Omega(n^{-1/3})$.  At this point, define $A,B$ as the two farthest points in $\mathcal{S}$. So, all points lie in the intersections of the circles centered at $A,B$ with radius $AB \geq 1$.  Choose chord $XY$ in circle $B$ where $XY \perp AB$ and $d(A, XY)=\frac{1}{2}$. Also, let $\mat

If $a,b,c$ are the sides of a triangle, prove that \[\frac{\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}+\frac{\sqrt{c+a-b}}{\sqrt{c}+\sqrt{a}-\sqrt{b}}+\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\leq 3 \]

Let $ABC$ be an acute triangle and let $M$ be the midpoint of $AC$. A circle $\omega$ passing through $B$ and $M$ meets the sides $AB$ and $BC$ at points $P$ and $Q$ respectively. Let $T$ be the point such that $BPTQ$ is a parallelogram. Suppose that $T$ lies on the circumcircle of $ABC$. Determine all possible values of $\frac{BT}{BM}$.