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Application of Algebra and Calculus in Euclidean Geometry

Huy Nguyen, Trung Le

Le Hong Phong High School - Ho Chi Minh City

Glossary

Abbreviations
MO Abbrev. for Mathematical Olympiad, the highest-level math contest for high school students, and preceded by the name of the region where the contest takes place - for example, VMO (Vietnam), USAMO (America), IMO (International).
TST Abbrev. for Team Selection Test, a test given to top performers in a national olympiad to select the nation’s representative team for an international olympiad.
Mathematical notions
\(a,b,c,p\) Unless otherwise specified, \(a,b,c\) denote the lengths of segment \(BC, CA, AB\) of \(\triangle ABC\) respectively, and \(p\) is one-half of its parameter, i.e., \(p = \frac{1}{2}(a+b+c)\).
\( (XYZ) \) For any three non-collinear points \(X, Y, Z\), denote the circumcircle of \(\triangle XYZ\).
\( (I;r) \) and \( (I) \) Denote the circle with center \(I\) and radius \(r\). When it is clear from the context that there is only one circle with center \(I\), the parameter \(r\) can be omitted.
\(\overline{AB}\) The algebraic length of segment \(AB\). If \(\overrightarrow{AB}\) points in the positive direction, \(\overline{AB} = |\overrightarrow{AB}|\); otherwise, \(\overline{AB} = -|\overrightarrow{AB}|\).
\( (ABCD) \) For any four collinear points \((A,B,C,D)\), denote the cross-ratio $$(ABCD) = \frac{\overline{AC}}{\overline{AD}} : \frac{\overline{BC}}{\overline{BD}}.$$ When \( (ABCD) = -1 \), \(A\) and \(B\) are harmonic conjugates of each other with respect to \(C\) and \(D\).
\( (AB, CD) \) The oriented angle starting at \(AB\) and ending at \(CD\).

Introduction

We begin with an analysis of a simple geometry problem, proposed by Dr. Trinh Le, deputy delegate of the Vietnam IMO team.

In $\abc$ consider a point $D$ on segment $BC$. Let $I$ and $J$ be the midpoints of $DB$ and $DC$ respectively. The perpendicular bisectors of $DB$ and $DC$ intersect $AB$ at $M$ and $AC$ at $N$ respectively. Let $O$ be the circumcenter of $\triangle ABC$. Prove that $AMON$ is a cyclic quadrilateral.

The following six solutions were devised by six students from the Ho Chi Minh City VMO team. Let us first take a look at their approaches and analyze the pros and cons of each.

Proof 1

Let $T$ be the midpoint of $BC$. Consider the Cartesian coordinate system $Txy$ where $T$ is the origin, $Tx$ is $BC$ and $Ty$ is the perpendicular bisector of $BC$. Denote the following coordinate points: $B(-1,0), C(1,0), D(d, 0)$ and $A(a,b)$. Let $U$ and $V$ be the midpoints of $AB$ and $AC$ respectively, and let $K$ be the projection of $O$ on $MN$. Furthermore, for any line $l$, denote $\vec{n_l}$ as its normal vector. We can then compute the followings:
  1. $ (AB): y=\dfrac{b(x+1)}{a+1}, \quad (AC): y=\dfrac{b(x-1)}{a-1}.$
  2. $U \lf \dfrac{a-1}{2}, \dfrac{b}{2} \ri, \ V \lf \dfrac{a+1}{2}, \dfrac{b}{2} \ri, M \lf \dfrac{d-1}{2}, \dfrac{b(d+1)}{2(a+1)} \ri , \ N \lf \dfrac{d+1}{2}, \dfrac{b(d-1)}{2(a-1)} \ri.$
  3. $x_O = x_T=0,\ \vto{OU} \parallel \vto{n_{AB}} \Rightarrow O \lf 0; \dfrac{a^2+b^2-1}{2ab} \ri.$
  4. $(MN): \dfrac{b(a-d)}{a^2-1} \lf x -\dfrac{d-1}{2} \ri + y - \dfrac{b(d+1)}{2(a+1)}=0.$
  5. $\vto{OK} \parallel \vto{n_{MN}} \Rightarrow \dfrac{x_K}{y_K - \frac{a^2+b^2-1}{2b}}=\dfrac{b(a-d)}{a^2-1}.$
Observe that $y_U = y_v = \frac{b}{2}$, so it turns out the line $UV$ is $(UV): y = \frac{b}{2}$. Furthermore, the point $K$ is uniquely determined from (4) and (5), and we notice that $\lf x_K = \frac{d-a}{2}, y_K = \frac{b}{2} \ri$ satisfies both. Hence $K \lf \frac{d-a}{2}, \frac{b}{2} \ri$ and therefore $K \in UV$. By the converse Simson's theorem $AMON$ is cyclic. $\qedsymbol$

Proof 2

Let $N' \ne A$ be the intersection point of $(AMO)$ and $AC$. We prove $AN'=AN$, which would imply $N' \equiv N$. Denote $DB = x$, then $$ \begin{equation} BM=\frac{x}{2\cos B}, \ AM=c-\dfrac{x}{2\cos B}, \ AN=b-\dfrac{a-x}{2\cos C}. \label{am-an} \end{equation} $$ Since $AMON'$ is a cyclic quadrilateral, by Ptolemy's theorem: $$ \begin{equation} AO \cdot MN' =AM \cdot ON' + AN' \cdot OM. \label{ptolemy} \end{equation} $$ Further observe that $\angle{OMN'} = \angle{OAC}=90^0-B$ and $\angle{ON'M} =\angle{OAB}=90^0-C.$ Hence, in $\triangle OMN'$, applying the sine Rule, we have $$\dfrac{ON'}{OM}=\dfrac{\cos B}{\cos C}.$$ Multiplying both sides of $(\ref{ptolemy})$ by $\frac{\cos C}{OM}$ we have $$(AO\cos C) \cdot \frac{MN'}{OM} = (AM \cos C) \cdot \frac{ON'}{OM} + AN'\cos C.$$ which yields $$\begin{align*} AN'\cos C & = (AO \cos C) \cdot \frac{\sin \angle MON'}{\sin \angle ON'M} - (AM \cos C) \cdot \frac{\cos B}{cos C} \\ & = \lf \frac{a}{2\sin A} \cdot \cos C \cdot \frac{\sin A}{\cos C} \ri - \lf c - \frac{x}{2\cos B} \ri \cos B \\ & = \frac{a + x}{2} - c\cos B. \end{align*}$$ From $(\ref{am-an})$ we also have $$ AN \cos C=b \cos C - \dfrac{a-x}{2},$$ hence $AN' = AN$ would be equivalent to $$\frac{a + x}{2} - c\cos B = b \cos C - \dfrac{a-x}{2},$$ which can be rewritten as $a = b\cos C + c\cos B$, which is true. Thus $AN' = AN$ and therefore $N' \equiv N$. $\qedsymbol$

Proof 3

$(AMN)$ intersects the diameter $AK$ of $(ABC)$ at $O' \ne A$. We will prove that $AK=2AO'$, which implies $O' \equiv O$.
Observe that $\triangle O'MN \sim \triangle KBC$ so $$ \dfrac{OM}{KB}=\dfrac{MN}{BC}=\dfrac{O'N}{KC}=k, $$ which means that $$\begin{equation} MN=kBC,\ O'M=kKB, \ O'N=kKC. \label{mn} \end{equation}$$ Applying Ptolemy's theorem to cyclic quadrilaterals $AMO'N$ and $ABKC$ we have $$\begin{align} AO' \cdot MN & = AM \cdot O'N + AN \cdot O'M. \label{amo'n} \\ AK \cdot BC & = AB \cdot KC + AC \cdot KB. \label{ak} \end{align}$$ Dividing both sides of (\ref{amo'n}) by $k$ we have $$\begin{equation} AO \cdot BC = AM \cdot KC + AN \cdot KB. \label{ao} \end{equation}$$ From Solution 2, with $DB = x$, we also have $$AM=c-\dfrac{x}{2\cos B}, AN =b-\dfrac{a-x}{2\cos C}.$$ Further note that that $KB=2R\cos C$ and $KC=2R\cos B$. Looking at (\ref{ak}) and (\ref{ao}), to show $AK=2AO'$ we need to prove that $$AB \cdot KC + AC \cdot KB = 2(AM \cdot KC + AN \cdot KB),$$ which is equivalent to $$c\cos B + b\cos C= 2\cos B \lf c-\dfrac{x}{2\cos B} \ri+ 2\cos C \lf b-\dfrac{a-x}{2\cos C} \ri,$$ which reduces to $a=b\cos C + c\cos B$, which is true. Hence $O \equiv O' \in (AMN)$ and therefore $AMON$ is cyclic. $\qedsymbol$

Proof 4

Let $U,V$ be the midpoints of $AB,AC$ respectively and $R,S$ be the projections of $U,V$ onto $BC$ respectively. It is easy to see that $IR=JS$. Indeed, on axis $BC$, denote the direction of $\vto{BC}$ as the positive direction. We then have $$\begin{align*} & \ds{IR}=\ds{BR}-\ds{BI}=BU \cos (BR, BU) - \dfrac{x}{2} = \dfrac{c\cos B - x}{2}, \\ & \ds{JS}=\ds{CS}-\ds{CJ}=CV\cos (CS, CA) + \dfrac{a-x}{2}=\dfrac{-b\cos C +a-x}{2}. \end{align*}$$ Since $a=b\cos C + c\cos B$, $\ds{IR}=\ds{JS}$ so $IR=JS$.
Applying Thales's theorem for $UR \parallel MI$ and $VS \parallel NJ$ we have $$\dfrac{MU}{IR}=\dfrac{BU}{BR}=\dfrac{1}{\cos B}, \ \dfrac{NV}{SJ}=\dfrac{CN}{CJ}=\dfrac{1}{\cos C}.$$ Hence $$\dfrac{MU}{NV}=\dfrac{IR}{JS} \cdot \dfrac{\cos C}{\cos B}=\dfrac{\cos C}{\cos B} =\dfrac{OU}{OV}.$$ Observe that $\triangle OMU$ and $\triangle ONV$ are directionally similar. Let $f$ be the spiral similarity with center $O$, angle $(OU,OV)$ and ratio $k=\frac{OU}{OV}$, then $f(U)=V, f(M)=N$.
Hence $(MO,MU) \equiv (NO,NV) \ \pmod{\pi}$, so $AMON$ is a cyclic quadrilateral. $\qedsymbol$

Proof 5

Let $E$ be the reflection of $D$ across $MN$. We then have $$\angle{MEN}=\angle{MDN}=180^0-\angle{MDB}-\angle{NDC}=180^0-B-C=A.$$ Hence, $AENM$ is a cyclic quadrilateral, which implies $\angle{AME}=\angle{ANE}$. Note that $ME = MD = MB$ and $NE = ND = NC$, so $\triangle MBE$ and $\triangle NCE$ are isosceles. In other words, $\angle{AME} = 2\angle{MBE}$ and $\angle{ANE} = 2\angle{NCE}$. Therefore $\angle{MBE}=\angle{NCE}$, so $AECB$ is cylic and $E \in (ABC)$.
Observe that $OM$ and $ON$ are the perpendicular bisectors of $EB$ and $EC$ respectively. Let $X=OM \cap BE, \ Y = ON \cap CE$, then $EXOY$ is a cyclic quadrilateral.
Hence $$\angle{MON}=180^0-\angle{XEY}=180^0-A,$$ which means $AMON$ is cyclic. $\qedsymbol$

Proof 6

Let $E \ne D$ be the other intersection point of $(DMN)$ and $BC$.
We have $\angle{EMN}=\angle{EDN}=C, \ A = \angle{MDN}=\angle{MEN}$ so $\triangle EMN \sim \triangle ACB$.
Let $O'$ be the orthocenter of $\triangle EMN$. We observe that $BMO'E, AMO'N, CNO'E$ are cyclic quadrilaterals, hence $$\angle{O'BE}=\angle{O'ME}=\angle{O'NE}=\angle{O'CE},$$ so $O'C = O'B$. Similarly we have $O'A = O'B \ (=O'C)$ and thus $o'$ is the orthocenter of $\triangle ABC$. In other words, $O' \equiv O$.
We also have $\angle{MON}=180^0-\angle{MDN}=180^0-A$, which means $AMON$ is cyclic.
We can observe the relation among the six solutions given above. While all solutions involve the additional construction of some geometric objects (new point, new line, or new circle), the more sophisticated this construction is, the simpler the subsequent proof becomes.
  • Solution 1 is the most simple approach: by considering the problem in the context of Cartesian coordinates. It is then straightforward to calculate the coordinates of every point. The proof does apply a small twist, however, by invoking Simson's theorem to turn the problem of cyclicity into that of collinearity, thereby avoiding equations of circles which are usually fairly complicated.
  • Solution 2 and 3 "reverse" the problem by constructing the point $N' \ne A$ as the intersection of $AC$ and $(AMO)$. The motivation here is that, instead of having to deduce cyclicity, we assume cyclicity is already given. What remains is to show that both $N'$ and $N$ share a common unique property, which in turn implies that $N'$ is indeed $N$. This is a recommended approach when one has not figured out how to effectively use all of the information provided by the problem.
  • Solution 4 builds upon the realization that all properties of $M$ and $N$ depend only on the location of $D$ on $BC$, i.e., the length of the segment $DB$. Hence we try to convert all computations to those involving only the segments on $BC$. Thanks to this approach, there are much less algebraic work to do.
  • Solution 5 and 6 are two purely geometric solutions that one needs rich experience in geometry to come up with. In particular, the idea of solution 5 is that since $\angle{MDN}=A$, we create a reflection of $D$ in order to obtain a cyclic quadrilateral. In solution 6, we use the following lemma:
    Given $\triangle ABC$ with three points $X,Y,Z$ on $BC,CA,AB$ respectively such that $\triangle XYZ \sim \triangle ABC$. It follows that the orthocenter of $\triangle XYZ$ is also the circumcenter of $\triangle ABC$.

Background

We first highlight a number of theorems and solutions that are highly applicable. Note that there are a number of textbook formulas - Pythagorean's Theorem, Thales' theorem, Law of sines and Law of cosines - that are not introduced but also relevant here.

Useful equatilies.

Triangle lengths.

Given $\abc$ $(AB < AC)$ with altitude $AD$ and bisector $AF$. Let $E$ be the tangent point of $\abc$'s incircle and $BC$. We then see that $$BD=\dfrac{a^2+c^2-b^2}{2a}, \ BE = \dfrac{a+c-b}{2}, \ BF=\dfrac{ac}{b+c},$$ which in turn yields $$DE=BE-BD=\dfrac{(b-c)(b+c-a)}{2a}.$$ Furthermore, from Stewart's theorem, we have $$AD=\sqrt{AB \cdot AC - DB \cdot DC }= \dfrac{2\sqrt{bcp(p-a)}}{b+c}=\dfrac{2bc}{b+c} \cdot \cos \dfrac{A}{2}.$$

Trigonometric expressions.

In any $\triangle ABC$, we have $$\begin{aligned} & \cos \dfrac{A}{2}=\sqrt{\dfrac{p(p-a)}{bc}} , \ \tan \dfrac{A}{2} =\sqrt{\dfrac{(p-b)(p-c)}{p(p-a)}},\\ & \cot A =\dfrac{b^2+c^2-a^2}{4S} , \ \cot \dfrac{A}{2}=\dfrac{b+c-a}{2r}. \end{aligned}$$

Concurrency and collinearity.

In $\abc$ consider points $D \in BC, E \in CA, F \in AB$. Let $L$ be another point on $BC$.