# Trigonometry tricks

## Background Information

I discovered the trick of solving geometry problems using trigonometry in November 2011, when I started my preparation for IMO 2012 training camps. Due to many algebraic identities associated with geometry, trigonometry provides us with a hacky way to transform geometry into algebra. Thanks to trigonometry (combined with many other tools), my skills in geometry improved from not being able to solve geometry problems in the 'easy' category in IMO (Problems 1 or 4) to be able to get the 1 gold point on the difficult IMO 2014 Problem 3 (and later complete the solution after the contest).

## High-level concepts

Most geometry problems are about proving identities. Some of them, like determining the ratios between lengths of two lines or areas of two polygons, are more suggestive of using trigonometry. Others, as seen in the following, can also be transformed into its algebraic/trigonometric equivalent:

 Items to prove Trigonometric equivalent Two angles are the equal the sine/cosine of them are equal + they are on the same quadrant Three (or more) lines pass through the same point. The trigonometric version of Ceva's ratio is 1. Three (or more) points lie on the same line. The Menelaus' ratio is -1.

## Basics*

Note: Those name bolded in italics are unofficial names and are not guaranteed to be well-known. Please verify this in the contest should you decide to use it.

• Sine rule. For any triangle $ABC$, $\frac{AB}{AC}=\frac{\sin\angle ACB}{\sin\angle ABC}.$
• Cosine rule. For a triangle $ABC$, we have $BC^{2}=AB^{2}+CA^{2}-2\cdot AB\cdot CA\cdot\cos\angle BAC$.
Corollary. $\cos\angle BAC=\frac{AB^{2}+CA^{2}-BC^{2}}{2\cdot AB\cdot CA}.$
• Dissection of triangle area. Let $D$ be a point on $BC$ of $\triangle ABC$. Then $\frac {BD}{DC}=\frac{AB}{AC}\cdot\frac{\sin\angle BAD}{\sin\angle CAD}$.
Proof: It is (probably) well-known that $\frac {BD}{DC}=\frac{|\triangle ABD|}{|\triangle ADC|}$, (challenge: prove it yourself!) Now let $|\triangle ABC|$ be the area of $\triangle ABC$. Also let the perpendicular from points $D$ to lines $AB$ and $AC$ be $E$ and $F$ respectively. Then $\frac {BD}{DC}=\frac{|\triangle ABD|}{|\triangle ADC|}=\dfrac{\frac{1}{2}\cdot AB\cdot DE}{\frac{1}{2}\cdot AC\cdot DF}=\frac{AB\cdot (AD\cdot\sin\angle BAD)}{AC\cdot (AD\cdot\sin\angle CAD)}=\frac{AB}{AC}\cdot\frac{\sin\angle BAD}{\sin\angle CAD}.$
• Let $ABDC$ be a cyclic quadrilateral. Then $\frac {BD}{DC}=\frac{\sin\angle BAD}{\sin\angle CAD}$. In fact, chord length= diameter of the circle to which the chord belongs $\times$ sine of the angle subtended by the chord on the arc of the circle (or angle between the chord and the tangent to the circle at either of the chord's endpoint!) As long as ratio is concerned, the diameter of the circle is not so relevant in our solution compared to the sines.
• Let $D$ be in the angle domain of $\angle BAC$ of a triangle $BAC$ and let $AD$ intersect $BC$ at $E$. Then $\frac{BE}{EC}=\frac{AB}{AC}\cdot\frac{BD}{CD}\cdot\frac{\sin\angle ABD}{\sin\angle ACD}$. (See how it is equivalent to the trigo version of Ceva's theorem).
Corollary 1. Take $D$ as above. Then $\frac{\sin\angle BAD}{\sin\angle CAD}=\frac{BD}{CD}\cdot\frac{\sin\angle ABD}{\sin\angle ACD}.$
Corollary 2. Let $D$ and $E$ be in the angle domain of $\angle BAC$ of a triangle $BAC$. Then $\frac{\sin\angle BAD}{\sin\angle CAD}=\frac{\sin\angle BAE}{\sin\angle CAE}\iff\angle BAD=\angle BAE$ and $\angle CAD=\angle CAE$.
Corollary 3. Denote $D$ and $E$ the same way as we did in corollary 2. Then $A, D, E$ are collinear iff $\frac{BD\cdot\sin\angle ABD}{CD\cdot\sin\angle ACD}=\frac{BE\cdot\sin\angle ABE}{CE\cdot\sin\angle ACE}$.

## Good-To-Know

Trigonometric identities can be combined with the following identities in solving geometry problems:

• Menelaus' theorem: Given a triangle $ABC$ and let $D$, $E$, $F$ be on lines $BC, CA, AB$, respectively. Then $D,E,F$ are collinear if and only if $\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{FA}{FB}=-1$, where signed length is taken (by convention, the ratio $\frac{BD}{DC}$ is taken to be positive if $D$ lies on segment $BC$, and negative otherwise).
• Ceva's theorem: using the same notation above, $AD, BE, CF$ concurrent if and only if $\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{FA}{FB}=1$ (again, signed ratio is used).
• Ceva's theorem (trigonometric version) : using the same notation as above, $AD, BE, CF$ concurrent if and only if $\frac{\sin\angle BAD}{\sin\angle CAD}\cdot\frac{\sin\angle CBE}{\sin\angle EBA}\cdot\frac{\sin\angle FCA}{\sin\angle FCB}=1$ (try to imagine how would signed trigonometric ratio here looks like: it is rather important!).
• Properties of harmonic bundle: assume that $AD, BE, CF$ and let $DE$ intersect $AB$ at $F'$. Then $(F, F'; A, B)$ is harmonic. In other words $(CF, CF'; CA, CB)$ is harmonic pencil.

## Pitfalls and notes

Although trigonometric bashing gives a shortcut in converting a problem to its easier equivalent form, it is not everything. As shown in some examples below, we need some other observations (possibly using other theorems) to turn a geometry problem into its 'ratio-friendly' form. At other times, trigonometric bashing needs to be combined with other identities in proving even a single auxillary result of a problem. Thus it is necessary to equip yourself with other geometry knowledge, too.

## Examples

Note: since this page is mainly about trigonometry, proofs of auxillary results that does not involve trigonometry are omitted (but quoted directly), if they can be eaily verified. Do not do this in any mathematical contest.

APMO 2017, Problem 2. Let $ABC$ be a triangle with $AB < AC$. Let $D$ be the intersection point of the internal bisector of angle $BAC$ and the circumcircle of $ABC$. Let $Z$ be the intersection point of the perpendicular bisector of $AC$ with the external bisector of angle $\angle{BAC}$. Prove that the midpoint of the segment $AB$ lies on the circumcircle of triangle $ADZ$.

IMO 2016, Problem 1. Triangle $BCF$ has a right angle at $B$. Let $A$ be the point on line $CF$ such that $FA=FB$ and $F$ lies between $A$ and $C$. Point $D$ is chosen so that $DA=DC$ and $AC$ is the bisector of $\angle{DAB}$. Point $E$ is chosen so that $EA=ED$ and $AD$ is the bisector of $\angle{EAC}$. Let $M$ be the midpoint of $CF$. Let $X$ be the point such that $AMXE$ is a parallelogram. Prove that $BD,FX$ and $ME$ are concurrent.

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 $TSH$.

IMOSL 2014, G4. Consider a fixed circle $\Gamma$ with three fixed points $A, B,$ and $C$ on it. Also, let us fix a real number $\lambda \in(0,1)$. For a variable point $P \not\in\{A, B, C\}$ on $\Gamma$, let $M$ be the point on the segment $CP$ such that $CM =\lambda\cdot CP$ . Let $Q$ be the second point of intersection of the circumcircles of the triangles $AMP$ and $BMC$. Prove that as $P$ varies, the point $Q$ lies on a fixed circle.

IMOSL 2014, G7. 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.

IMO 2013, Problem 3. Let the excircle of triangle $ABC$ opposite the vertex $A$ be tangent to the side $BC$ at the point $A_1$. Define the points $B_1$ on $CA$ and $C_1$ on $AB$ analogously, using the excircles opposite $B$ and $C$, respectively. Suppose that the circumcentre of triangle $A_1B_1C_1$ lies on the circumcircle of triangle $ABC$. Prove that triangle $ABC$ is right-angled.

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.

IMOSL 2012, G2. Let $ABCD$ be a cyclic quadrilateral whose diagonals $AC$ and $BD$ meet at $E$. The extensions of the sides $AD$ and $BC$ beyond $A$ and $B$ meet at $F$. Let $G$ be the point such that $ECGD$ is a parallelogram, and let $H$ be the image of $E$ under reflection in $AD$. Prove that $D,H,F,G$ are concyclic.

IMOSL 2012, G3. In an acute triangle $ABC$ the points $D,E$ and $F$ are the feet of the altitudes through $A,B$ and $C$ respectively. The incenters of the triangles $AEF$ and $BDF$ are $I_1$ and $I_2$ respectively; the circumcenters of the triangles $ACI_1$ and $BCI_2$ are $O_1$ and $O_2$ respectively. Prove that $I_1I_2$ and $O_1O_2$ are parallel.

IMOSL 2011, G7. Let $ABCDEF$ be a convex hexagon all of whose sides are tangent to a circle $\omega$ with centre $O$. Suppose that the circumcircle of triangle $ACE$ is concentric with $\omega$. Let $J$ be the foot of the perpendicular from $B$ to $CD$. Suppose that the perpendicular from $B$ to $DF$ intersects the line $EO$ at a point $K$. Let $L$ be the foot of the perpendicular from $K$ to $DE$. Prove that $DJ=DL$.

IMO 2009, Problem 4. Let $ABC$ be a triangle with $AB = AC$ . The angle bisectors of $\angle C AB$ and $\angle AB C$ meet the sides $B C$ and $C A$ at $D$ and $E$ , respectively. Let $K$ be the incentre of triangle $ADC$. Suppose that $\angle B E K = 45^\circ$ . Find all possible values of $\angle C AB$ .

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 $X$.

*These are cited directly from my notes written for the Malaysian IMO training camp in January 2016.