I was invited by Mr. Suhaimi Ramly, the head coach of the training camps to deliver geometry lectures in the IMO 2016 training camps in January, March (pre-APMO camp) and April.
IMO 2016 Camp 2: January 12-16, 2016
Lecture Delivered: Trigonometric Tricks in Solving Geometry Problems
The basics and fundamentals are almost the same as posted on my Trigonometry page.
This lecture was given to the students who had not attended training camps in any of the previous years.
The problems with solutions included in the handouts were:
APMO 2013, Problem 1.
Let $ABC$ be an acute triangle with altitudes $AD$, $BE$, and $CF$, and let $O$ be the center of its circumcircle. Show that the segments $OA$, $OF$, $OB$, $OD$, $OC$, $OE$ dissect the triangle $ABC$ into three pairs of triangles that have equal areas.
APMO 2012, Problem 4.
Let $ ABC $ be an acute triangle. Denote by $ D $ the foot of the perpendicular line drawn from the point $ A $ to the side $ BC $, by $M$ the midpoint of $ BC $, and by $ H $ the orthocenter of $ ABC $. Let $ E $ be the point of intersection of the circumcircle $ \Gamma $ of the triangle $ ABC $ and the half line $ MH $, and $ F $ be the point of intersection (other than $E$) of the line $ ED $ and the circle $ \Gamma $. Prove that $ \tfrac{BF}{CF} = \tfrac{AB}{AC} $ must hold.
APMO 2013, Problem 5.
Let $ABCD$ be a quadrilateral inscribed in a circle $\omega$, and let $P$ be a point on the extension of $AC$ such that $PB$ and $PD$ are tangent to $\omega$. The tangent at $C$ intersects $PD$ at $Q$ and the line $AD$ at $R$. Let $E$ be the second point of intersection between $AQ$ and $\omega$. Prove that $B$, $E$, $R$ are collinear.
IMOSL 2011, G5.
Let $ABC$ be a triangle with incenter $I$ and circumcircle $\omega$. Let $D$ and $E$ be the second intersection points of $\omega$ with the lines $AI$ and $BI$, respectively. The chord $DE$ meets $AC$ at a point $F$, and $BC$ at a point $G$. Let $P$ be the intersection point of the line through $F$ parallel to $AD$ and the line through $G$ parallel to $BE$. Suppose that the tangents to $\omega$ at $A$ and at $B$ meet at a point $K$. Prove that the three lines $AE, BD,$ and $KP$ are either parallel or concurrent.
RIMO 2014, Day 1, Problem 2.
In a quadrilateral $ABCD$ with $\angle B=\angle D=90^{\circ}$, the extensions of $AB$ and $DC$ meet at $E$; and the extensions of $AD$ and $BC$ meet at $F$. A line through $B$ parallel to $CD$ intersects the circumcircle $\omega$ of the triangle $ABF$ at $G$ distinct from $B$; the line $EG$ intersects $\omega$ at $P$ distinct from $G$; and the line $AP$ intersects $CE$ at $M$. Prove that $M$ is the midpoint of $CE$.
IMOSL 2007, G2.
Given an isosceles triangle $ABC$ with $AB=AC$. The midpoint of side $BC$ is denoted by $M$. Let $X$ be a variable point on the shorter arc $MA$ of the circumcircle of triangle $ABM$. Let $T$ be the point in the angle domain $BMA$, for which $\angle TMX= 90^{\circ}$ and $TX=BX$. Prove that $\angle MTB-\angle CTM$ does not depend on $X$.
Problem solving session conducted: Trigonometric Tricks in Solving Geometry Problems
The problem solving session was themed on the same as that given in my lecture of trigonometry, except that this was targetted to a more experience group of students.
As such, the solution to the following difficult problem was demonstrated to the students:
IMO 2014, Problem 3.
Convex quadrilateral $ABCD$ has $\angle ABC = \angle CDA = 90^{\circ}$.
Point $H$ is the foot of the perpendicular from $A$ to $BD$.
Points $S$ and $T$ lie on sides $AB$ and $AD$, respectively,
such that $H$ lies inside triangle $SCT$ and
\[ \angle CHS - \angle CSB = 90^{\circ}, \quad \angle THC - \angle DTC = 90^{\circ}. \]
Prove that line $BD$ is tangent to the circumcircle of triangle $TSH$.
IMO 2016 Camp 3: March 4-8, 2016
Lectures Delivered: Geometry Big Guns
During that camp, Justin Lim (IMO 2014 gold medallist, currently an undergraduate at MIT) and I collectively covered a few advanced level tricks in geometry.
They are detailed below (together with the problems solvable using these theorems):
Harmonic Bundles
IMOSL 2004, G8.
Given a cyclic quadrilateral $ABCD$, let $M$ be the midpoint of the side $CD$, and let $N$ be a point on the circumcircle of triangle $ABM$. Assume that the point $N$ is different from the point $M$ and satisfies $\frac{AN}{BN}=\frac{AM}{BM}$. Prove that the points $E$, $F$, $N$ are collinear, where $E=AC\cap BD$ and $F=BC\cap DA$.
Poles and Polars + Brokard's theorem
IMO 2012, Problem 5.
Let $ABC$ be a triangle with $\angle BCA=90^{\circ}$, and let $D$ be the foot of the altitude from $C$. Let $X$ be a point in the interior of the segment $CD$. Let $K$ be the point on the segment $AX$ such that $BK=BC$. Similarly, let $L$ be the point on the segment $BX$ such that $AL=AC$. Let $M$ be the point of intersection of $AL$ and $BK$.
Show that $MK=ML$.
JOM 2013, G7.
Given a triangle $ABC$, let $l$ be the median corresponding to the vertex $A$. Let $E,F$ be the feet of the perpendiculars from $B,C$ to $AC,AB$. Reflect the points $E,F$ across $l$ to the points $P,Q$. Let $AP, AQ$ intersect $BC$ at $X,Y$. Let $\Gamma_1, \Gamma_ 2$ be the circumcircles of $\triangle EQY,\triangle FPX.$
Prove that $A$ lies on the line connecting the centers of $\Gamma_1, \Gamma_2$.
APMO 2013, Problem 5.
Let $ABCD$ be a quadrilateral inscribed in a circle $\omega$, and let $P$ be a point on the extension of $AC$ such that $PB$ and $PD$ are tangent to $\omega$. The tangent at $C$ intersects $PD$ at $Q$ and the line $AD$ at $R$. Let $E$ be the second point of intersection between $AQ$ and $\omega$. Prove that $B$, $E$, $R$ are collinear.
Homothety, Monge's theorem and Monge d'Alembert's theorem
IMOSL 2007, G8.
Point $ P$ lies on side $ AB$ of a convex quadrilateral $ ABCD$. Let $ \omega$ be the incircle of triangle $ CPD$, and let $ I$ be its incenter. Suppose that $ \omega$ is tangent to the incircles of triangles $ APD$ and $ BPC$ at points $ K$ and $ L$, respectively. Let lines $ AC$ and $ BD$ meet at $ E$, and let lines $ AK$ and $ BL$ meet at $ F$. Prove that points $ E$, $ I$, and $ F$ are collinear.
RMM 2012, Problem 6.
Let $ABC$ be a triangle and let $I$ and $O$ denote its incentre and circumcentre respectively. Let $\omega_A$ be the circle through $B$ and $C$ which is tangent to the incircle of the triangle $ABC$; the circles $\omega_B$ and $\omega_C$ are defined similarly. The circles $\omega_B$ and $\omega_C$ meet at a point $A'$ distinct from $A$; the points $B'$ and $C'$ are defined similarly. Prove that the lines $AA',BB'$ and $CC'$ are concurrent at a point on the line $IO$.
The following problem combines the idea of all theorems above:
IMO 2008, Problem 6.
Let $ABCD$ be a convex quadrilateral with $ BA\neq BC$. Denote the incircles of triangles $ ABC$ and $ ADC$ by $ \omega_{1}$ and $ \omega_{2}$ respectively. Suppose that there exists a circle $ \omega$ tangent to ray $ BA$ beyond $ A$ and to the ray $ BC$ beyond $ C$, which is also tangent to the lines $ AD$ and $ CD$. Prove that the common external tangents to $ \omega_{1}$ and $\omega_{2}$ intersect on $\omega$.
This was a project I did at the 2012 International Mathematics Tournament of Towns Summer Conference that I attended together with Justin.
One theorem discussed during the session was the Wolstenholme's theorem,
which could be useful in solving problems in Section 4 of the attached hyperlink, and the following:
APMO 2006, Problem 3.
Let $p\ge5$ be a prime and let $r$ be the number of ways of placing $p$ checkers on a $p\times p$ checkerboard so that not all checkers are in the same row (but they may all be in the same column). Show that $r$ is divisible by $p^5$. Here, we assume that all the checkers are identical.
IMO 2016 Camp 4: April 8-12, 2016
Lectures Delivered: Three geometry theorems (Simson's theorem, Casey's theorem, and Sawayama Thebault's theorem)
For students who want to excel in Olympiad geometry and be able to solve the hardest contest geometry problems comfortably,
mastery of these theorems is essential.
Simson's theorem
Statement: Given a point $D$ and a triangle $ABC$, the feet of perpendiculars from $D$ to sides $BC, CA, AB$ are collinear iff $D$ lies on the circumcircle of triangle $ABC$.
Applications:
IMO 2003, Problem 4.
Let $ABCD$ be a cyclic quadrilateral. Let $P,Q,R$ be the feet of the perpendiculars from $D$ to the lines $BC, CA, AB$ respectively. Show that $PQ=QR$ if and only if the bisectors of $\angle ABC$ and $\angle ADC$ are concurrent with $AC$.
IMO 2007, Problem 2.
Consider five points $A,B,C,D$ and $E$ such that $ABCD$ is a parallelogram and $BCED$ is a cyclic quadrilateral. Let $\ell$ be a line passing through $A$. Suppose that $\ell$ intersects the interior of the segment $DC$ at $F$ and intersects line $BC$ at $G$. Suppose also that $EF=EG=EC$. Prove that $\ell$ is the bisector of angle $DAB$.
TOT Spring Fall 2008, Senior A-Level, Problem 7.
Each of three lines cuts chords of equal lengths in two given circles. The points of intersection of these lines form a triangle. Prove that its circumcircle passes through the midpoint of the
segment joining the centres of the circles.
Statement: Let $\omega_1$, $\omega_2$, $\omega_3$ and $\omega_4$ be four non-intersecting and mutually exclusive circles, and let $t_{ij}$ be the length of common exterior bitangent (i.e. segment connecting $A_iA_j$ s.t. $A_i$ on $\omega_i$, $A_j$ on $\omega_j$ and $A_iA_j$ tangent to both circles externally. ) Then there exists a circle tangent internally to all four circles (in the order of $\omega_1$, $\omega_2$, $\omega_3$ and $\omega_4$) if and only if:
$$t_{12}t_{34}+t_{23}t_{14}=t_{13}t_{24}.$$
Applications:
APMO 2006, Problem 4.
Let $A,B$ be two distinct points on a given circle $O$ and let $P$ be the midpoint of the line segment AB. Let $O_1$ be the circle tangent to the line $AB$ at $P$ and tangent to the circle $O$. Let $l$ be the tangent line, different from the line $AB$, to $O_1$ passing through $A$. Let $C$ be the intersection point, different from $A$, of $l$ and $O$. Let $Q$ be the midpoint of the line segment $BC$ and $O_2$ be the circle tangent to the line $BC$ at $Q$ and tangent to the line segment $AC$. Prove that the circle $O_2$ is tangent to the circle $O$.
IMO 2011, Problem 6.
Let $ABC$ be an acute triangle with circumcircle $\Gamma$. Let $\ell$ be a tangent line to $\Gamma$, and let $\ell_a, \ell_b$ and $\ell_c$ be the lines obtained by reflecting $\ell$ in the lines $BC$, $CA$ and $AB$, respectively. Show that the circumcircle of the triangle determined by the lines $\ell_a, \ell_b$ and $\ell_c$ is tangent to the circle $\Gamma$.
APMO 2014, Problem 5.
Circles $\omega$ and $\Omega$ meet at points $A$ and $B$. Let $M$ be the midpoint of the arc $AB$ of circle $\omega$ ($M$ lies inside $\Omega$). A chord $MP$ of circle $\omega$ intersects $\Omega$ at $Q$ ($Q$ lies inside $\omega$). Let $\ell_P$ be the tangent line to $\omega$ at $P$, and let $\ell_Q$ be the tangent line to $\Omega$ at $Q$. Prove that the circumcircle of the triangle formed by the lines $\ell_P$, $\ell_Q$ and $AB$ is tangent to $\Omega$.
Sawayama Thebault's theorem
Statement: Let $I$ be the incentre of $\triangle ABC$, $D$ a point on line $BC$. If a circle is tangent to the circumcircle of triagle $ABC$, to segment $DC$ at $E$ and segment $DA$ at $F$. Then $E,I,F$ are collinear.
This is another project I did at the Summer Conference mentioned above, partnered with Justin.
The project is to solve the following general problem:
There are $m$ cakes, each with weight 1.
We are to divide the cake into slices and distribute them to $n$ people such that everyone gets equal total weight of cake (that is, $\frac{m}{n}$).
Determine the maximum $k$ such that each slice has weight at least $k$
(basically, maximize the minimum possible slice).
Notice that in the official handout this quantity $k$ is denoted as $f(m,n)$.
Some auxillary results I discussed in the problem solving session were:
$f(m, km)=\frac{1}{k}$ for all positive integers $k, m$.
$f(km, m)=1$ for all positive inetegers $k, m$.
$f(m, n)\le\frac{m}{2n}$ for all $m, n$ such that $m$ does not divide $n$.
$f(m, n)\ge \frac{m}{3n}$ for all $m, n$ sastisfying $m\le n$.
When $\frac 23\le\frac mn\le\frac 34$, equality holds.
Remarkably, Justin and I submitted a solution to finding $f(31, 52)$ that was highly commended by the committee members (during the conference).
Have fun trying before looking at the answer!
Answer: $f(31, 52)=\frac{89}{364}$.
Acronyms Used
IMO: International Mathematical Olympiad
IMOSL: International Mathematical Olympiad, shortlisted problems
APMO: Asian Pacific Mathematical Olympiad
TOT: International Mathematics Tournament of Towns