Plane Geometry Techniques

✔ Given that orthocenter HH, incenter II, and circumcenter OO for a triangle ABCABC, the following statements hold.

HI

(a) If ABCABC is an acute triangle, BHC=πα\angle BHC = \pi - \alpha.

(b) BIC=π+α2\angle BIC = \cfrac{\pi + \alpha}{2}

✔ Given an incircle and excircle are tangent to each DD and EE on a side for a triangle ABCABC, DD and EE are symmetric about the midpoint of the side.

Excircle

✔ For a triangle ABCABC, the following holds.

Ratio {  asinθ1:bsinθ2=c:d(θ1θ2)  a:b=c:d(θ1=θ2)\begin{aligned} \begin{cases} \; a \sin \theta_1 : b \sin \theta_2 = c : d \quad (\theta_1 \ne \theta_2) \\\\ \; a : b = c : d \quad (\theta_1 = \theta_2) \end{cases} \end{aligned}

✔ When ABCDABCD is a convex quadrilateral, concave quadrilateral, or not even a quadrilateral, the following holds.

ACBD    AB2+CD2=AD2+BC2\begin{aligned} \overline{AC} \perp \overline{BD} \iff \overline{AB}^2 + \overline{CD}^2 = \overline{AD}^2 + \overline{BC}^2 \end{aligned}

✔ (Stewart’s theorem) For a triangle ABCABC, the following holds.

Stewart a(d2+mn)=b2m+c2n\begin{aligned} a (d^2 + mn) = b^2 m + c^2 n \end{aligned}

✔ Given DBCD \in \overline{BC} and XADX \in \overline{AD} for a triangle ABCABC, the following holds.

Area Area(BCX)Area(ABC)=DXDA\begin{aligned} \cfrac{\text{Area} (BCX)}{\text{Area} (ABC)} = \cfrac{\overline{DX}}{\overline{DA}} \end{aligned}

✔ (Van Aubel’s theorem) Given DBCD \in \overline{BC}, EACE \in \overline{AC} and FABF \in \overline{AB} for a triangle ABCABC, if AD\overline{AD}, BE\overline{BE}, and CF\overline{CF} pass through a same point PP, the following holds.

VanAubel APPD=AEEC+AFFB\begin{aligned} \cfrac{\overline{AP}}{\overline{PD}} = \cfrac{\overline{AE}}{\overline{EC}} + \cfrac{\overline{AF}}{\overline{FB}} \end{aligned}

✔ Two lines l1l_1 and l2l_2 are antiparallel to a given line nn if one line of two can be symmetric after sliding towards nn. The following figure shows the properties of antiparallel. Note that m1m_1 and m2m_2 are also antiparallel to a given line nn in the below figure.

Antiparallel

✔ Given that orthocenter HH and circumcenter OO for a triangle ABCABC, HH and OO are isotomic conjugate. Besides, two lines AH\overline{AH} and AO\overline{AO} are antiparallel to the angle bisector of A\angle A.

HO

✔ For the radical axis ll of two circles ω1\omega_1 and ω2\omega_2, AA and D(AD)D (A \ne D) are on ω1\omega_1, and BB and C(BC)C (B \ne C) are on ω2\omega_2. When AD\overline{AD} and BC\overline{BC} are not parallel, these lines intersect at the line ll if and only if ABCDABCD is on the same circle.

RadicalLemma

✔ (Erdos-Mordell) For a triangle ABCABC and PP inside ABCABC, let PXPX, PYPY, and PZPZ be the perpendiculars from PP to the sides of ABCABC. Then, the following holds.

ErdosMordell PA+PB+PC2(PX+PY+PZ)\begin{aligned} \overline{PA} + \overline{PB} + \overline{PC} \geq 2 (\overline{PX} + \overline{PY} + \overline{PZ}) \end{aligned}


✔ For a triangle ABCABC and PP inside ABCABC, at least one of PAB\angle PAB, PBC\angle PBC, and PCA\angle PCA is π/6\leq \pi / 6.

Let PXPX, PYPY, and PZPZ be the perpendiculars from PP to the sides of ABCABC. Then, PA+PB+PC2(PX+PY+PZ)\overline{PA} + \overline{PB} + \overline{PC} \geq 2 (\overline{PX} + \overline{PY} + \overline{PZ}). It implies that one of the followings holds. PB2PX,PC2PY,PA2PZ\begin{aligned} \overline{PB} \geq 2 \overline{PX}, \quad \overline{PC} \geq 2 \overline{PY}, \quad \overline{PA} \geq 2 \overline{PZ} \end{aligned}

Without loss of generality, assume PB2PX\overline{PB} \geq 2 \overline{PX}. Then, sinPBC=PXPB12    sinPBCπ6\begin{aligned} \sin \angle PBC = \frac{\overline{PX}}{\overline{PB}} \leq \frac{1}{2} \implies \sin \angle PBC \leq \frac{\pi}{6} \end{aligned}

✔ For a convex quadrilateral ABCDABCD, find XX such that minimize XA+XB+XC+XD\overline{XA} + \overline{XB} + \overline{XC} + \overline{XD}.

By triangle inequality, XA+XCAC\overline{XA} + \overline{XC} \geq \overline{AC} and XB+XDBD\overline{XB} + \overline{XD} \geq \overline{BD} for the triangles ACXACX and BDXBDX. Therefore, XA+XB+XC+XDAC+BD\begin{aligned} \overline{XA} + \overline{XB} + \overline{XC} + \overline{XD} \geq \overline{AC} + \overline{BD} \end{aligned}

Since the equal sign is valid only when XA+XC=AC\overline{XA} + \overline{XC} = \overline{AC} and XB+XD=BD\overline{XB} + \overline{XD} = \overline{BD}, XX is the intersection of AC\overline{AC} and BD\overline{BD}.

Reference

[1] Titu Andreescu, 106 Geometry Problems.


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