Van Khea's areal problem

 The following problem is a generalization of my previous problem conjectured by Van Khea. We will give a proof.

We will be using standard notation: $\lvert{BC}\rvert=a$, $\lvert{AC}\rvert=b$, $\lvert{AB}\rvert=c$; $\angle{BAC}=\alpha$, $\angle{ABC}=\beta$ and $\angle{BCA}=\gamma$. If $X$, $Y$ and $Z$ are the vertices of a triangle, we denote its area $[XYZ]$.

Problem. Let $ABC$ be a triangle and $P$ any point on the plane of $ABC$. Let $X$, $Y$ and $Z$ be arbitrary points on sides $BC$, $AC$ and $AB$, respectively. Let $D$ be the reflection of $P$ around $X$. Similarly, define $E$ and $F$. Denote $U$, $V$ and $W$ the midpoints of sides $BC$, $AC$ and $AB$, respectively. Let $D'$ be the reflection of $D$ around $U$. Similarly, define $E'$ and $F'$ (see figure below). Prove that

$$[DEF]+[D'E'F']=[ABC].$$

Proof. Denote $AZ=g$, $BZ=h$, $BX=j$, $CX=k$, $CY=l$ and $AY=m$. The area of ​​triangle $XYZ$ can be expressed as follows

$$[XYZ]=\frac12bc\sin{\alpha}-\frac12gm\sin{\alpha}-\frac12hj\sin{\beta}-\frac12kl\sin{\gamma}.$$

Since triangles $XYZ$ and $DEF$ are homothetic with scale factor $2$, it follows that $[DEF]=4[XYZ]$. Thus, we have

\[{[DEF]}=2bc\sin{\alpha}-2gm\sin{\alpha}-2hj\sin{\beta}-2kl\sin{\gamma}.\tag{1}\]

Dividing both sides of $(1)$ by $[ABC]$ we have

\[\frac{[DEF]}{[ABC]}=4\left[1-\left(\frac{gm}{bc}+\frac{hj}{ca}+\frac{kl}{ab}\right)\right].\tag{2}\]

Segments $UX$ and $PD'$ are homothetic with center at $D$ and scale factor $2$. It follows that

$$PD'=2(j-\frac12a)=2j-a.$$

Similarly, we get $PE'=b-2l$ and $PF'=c-2g$. Since $UX$ and $PD'$ are homothetic segments, then $UX$ and $PD'$ are parallel and so are $VY$ and $PE'$. Hence $\angle{D'PE'}=\gamma$. Similarly, $\angle{D'PF'}=\beta$. So the area of triangle $D'E'F'$ is given by the expression

$$\begin{aligned}{[D'E'F']}&=[D'PE']+[D'PF']-[E'F'P]\\&=\frac{(2j-a)(b-2l)\sin{\gamma}}{2}+\frac{(2j-a)(c-2g)\sin{\beta}}{2}-\frac{(b-2l)(c-2g)\sin{(\beta+\gamma)}}{2}.\end{aligned}$$

Taking into account that $\sin{(\beta+\gamma)}=\sin{(\pi-\alpha)}=\sin{\alpha}$ and dividing by $[ABC]$ we obtain

\[\frac{[D'E'F']}{[ABC]}=\frac{(2j-a)(b-2l)}{ab}+\frac{(2j-a)(c-2g)}{ca}-\frac{(b-2l)(c-2g)}{bc}.\tag{3}\]

Adding equations $(2)$ and $(3)$, expanding and factorizing,

$$\frac{[DEF]}{[ABC]}+\frac{[D'E'F']}{[ABC]}=\frac{ca(4l+b)-4ga(m+l-b)-4j(b(g+h-c)+lc)-4klc}{abc}.$$

But $b=m+l$, $c=g+h$ and $a=j+k$, so

$$\begin{aligned}\frac{[DEF]}{[ABC]}+\frac{[D'E'F']}{[ABC]}&=\frac{ca(4l+b)-4jlc-4klc}{abc}\\&=\frac{ca(4l+b)-4cl(j+k)}{abc}\\&=\frac{ca(4l+b)-4cla}{abc}\\&=1.\end{aligned}$$

Therefore, 

$$[DEF]+[D'E'F']=[ABC].$$

$\square$

Note: The point $P$ may cross the side lines of the triangle $ABC$ in points either interior or exterior to the sides. The reasoning in cases other than that considered above requires only minor adjustments.

Remark: This theorem remains valid if $P$ is an arbitrary point in three-dimensional space.

See also

Another Identity on Triangle Areas Arising from Reflection

Areal property of the circumcircle mid-arc triangle

 The following theorem is a property of the circumcircle mid-arc triangle which appears to be unknown.

We will be using standard notation: $\lvert{BC}\rvert=a$, $\lvert{AC}\rvert=b$, $\lvert{AB}\rvert=c$; $\angle{BAC}=\alpha$, $\angle{ABC}=\beta$ and $\angle{BCA}=\gamma$; $s$ for the semiperimeter; $R$ for the circumradius and $r$ for the inradius. If $X$, $Y$ and $Z$ are the vertices of a triangle, we denote its area $[XYZ]$.

Problem. Let $ABC$ be a triangle and $I$ its Incenter. Denote $\omega$ the circumcircle of $ABC$. Let $AI$ intersect $\omega$ again at $D$. Define $E$ and $F$ cyclically. Let $U$, $V$ and $W$ be the midpoints of $DE$, $EF$ and $FD$, respectively. Let $A'$ be the reflection of $A$ with respect to $V$. Define $B'$ and $C'$ cyclically (see figure 1). Then

$$[DEF]=[ABC]+[A'B'C'].\tag{1}$$

Figure 1

Lemma 1. If $ABC$ is a triangle and $DEF$ is its circumcircle mid-arc triangle, then

$$\boxed{\frac{[ABC]}{[DEF]}=\frac{2r}{R}.}\tag{2}$$

Proof. By property of inscribed angles, $\angle{BAF}=\angle{ACF}=\frac12\gamma$ and $\angle{CAE}=\frac12\beta$. It follows that

$$\begin{aligned}&\angle{EAF}=\alpha+\frac12\beta+\frac12\gamma\\&=\pi-\beta-\gamma+\frac12\beta+\frac12\gamma\\&=\pi-\frac12(\beta+\gamma).\end{aligned}$$

By the law of sines, 

$$\frac{EF}{\sin{(\pi-\frac12(\beta+\gamma))}}=\frac{EF}{\sin{\frac12(\beta+\gamma)}}=\frac{AF}{\sin{\frac12\gamma}}.\tag{3}$$

Since $\angle{AFC}=\beta$ and, again, by the law of sines, 

$$\frac{AF}{\sin{\frac12\gamma}}=\frac{b}{\sin{\beta}}.\tag{4}$$

From $(3)$ and $(4)$ we get

$$EF=\frac{b\sin{\frac12(\beta+\gamma)}}{\sin{\beta}}=2R\sin{\frac12(\pi-\alpha)}=2R\cos{\frac12\alpha}.\tag{5}$$

Analogously, we can find that

$$DF=2R\cos{\frac12\beta}\qquad ED=2R\cos{\frac12\gamma}.\tag{6}$$

Note that $\angle{DFC}=\frac12\alpha$ and $\angle{EFC}=\frac12\beta$, so $\angle{DFE}=\frac12(\alpha+\beta)$. Then the area of $DEF$ is given by
$$[DEF]=\frac{EF\cdot{DF}\sin{\frac12(\alpha+\beta)}}{2}=2R^2\cos{\frac12\alpha}\cos{\frac12\beta}\cos{\frac12\gamma}.\tag{7}$$
Elsewhere we have proved that $s=4R\cos{\frac12\alpha}\cos{\frac12\beta}\cos{\frac12\gamma}$ (for a proof, see section 3, c), here). As the area of $ABC$ can also be written as $[ABC]=rs$, we can rewrite it like this

$$[ABC]=4Rr\cos{\frac12\alpha}\cos{\frac12\beta}\cos{\frac12\gamma}.\tag{8}$$

Dividing $(8)$ by $(7)$ we get

$$\frac{[ABC]}{[DEF]}=\frac{2r}{R}.$$

$\square$

Lemma 2. Let $ABC$ be an triangle and $H$ its orthocenter. Let $l_a$, $l_b$ and $l_c$ be the perpendicular bisectors of sides $BC$, $AC$ and $AB$, respectively. Denote $H_1$, $H_2$ and $H_3$ the reflections of $H$ around $l_a$, $l_b$ and $l_c$, respectively. Then

 $$\boxed{\frac{[H_1H_2H_3]}{[ABC]}=\frac{(a^2-b^2)(a^2-c^2)}{b^2c^2}+\frac{(a^2-c^2)(b^2-c^2)}{a^2b^2}-\frac{(a^2-b^2)(b^2-c^2)}{a^2c^2}.}\tag{9}$$

Proof. Assume first that $ABC$ is acute. From figure 2, note that 

$$HH_3=2(\frac{c}{2}-AC_1).\tag{10}$$

Figure 2

Let $CH$ intersect $AB$ in $C_1$. Applying the law of sines in the triangle $ACC_1$, we have that

$$b=\frac{AC_1}{\sin{(\frac{\pi}{2}-\alpha)}}=\frac{AC_1}{\cos{\alpha}}.$$

So, $AC_1=b\cos{\alpha}$. Substituting in $(10)$ and then substituting from the law of cosines $\cos{\alpha}=\frac{b^2+c^2-a^2}{2bc}$ we have

$$HH_3=2(\frac{c}{2}-b\cos{\alpha})=c-2b\cos{\alpha}=\frac{a^2-b^2}{c}.$$

Similarly, we get that $HH_1=\frac{b^2-c^2}{a}$ and $HH_2=\frac{a^2-c^2}{b}$. Now, note that 

$$[H_1H_2H_3]=[HH_2H_3]+[HH_1H_2]-[HH_1H_3].\tag{11}$$

As $HH_3$ and $HH_2$ are perpendicular to $l_c$ and $l_a$, respectively, then $\angle{H_2HH_3}=\alpha$ and similarly $\angle{H_2HH1}=\gamma$. Thus, equation $(11)$ can be written as follows

$$[H_1H_2H_3]=\frac{(a^2-b^2)(a^2-c^2)\sin{\alpha}}{2bc}+\frac{(a^2-c^2)(b^2-c^2)\sin{\gamma}}{2ab}-\frac{(a^2-b^2)(b^2-c^2)\sin{(\alpha+\gamma)}}{2ac}.\tag{12}$$

Dividing both sides of equation $(12)$ by $[ABC]=\frac12bc\sin{\alpha}=\frac12ab\sin{\gamma}=\frac12ac\sin{\beta}$ and taking into account that $\sin{(\alpha+\gamma)}=\sin{(\pi-\beta)}=\sin{\beta}$.

$$\frac{[H_1H_2H_3]}{[ABC]}=\frac{(a^2-b^2)(a^2-c^2)}{b^2c^2}+\frac{(a^2-c^2)(b^2-c^2)}{a^2b^2}-\frac{(a^2-b^2)(b^2-c^2)}{a^2c^2}.$$

$\square$

Remark. For simplicity, we have written the proof under the assumption that $ABC$ is acute, which is the configuration used in the argument above. The right and obtuse cases can be handled with the same identities after the usual minor adjustments.

Note that the terms on the right hand side of the equation $(9)$ are the products of the Mollweide's formulas. Substituting from Mollweide's formulas and applying the identity of sine of double angle, we can write the equation $(9)$ as follows

$$\boxed{\frac{[H_1H_2H_3]}{[ABC]}=\frac{\sin{(\alpha-\beta)}\sin{(\alpha-\gamma)}}{\sin{\gamma}\sin{\beta}}+\frac{\sin{(\alpha-\gamma)}\sin{(\beta-\gamma)}}{\sin{\alpha}\sin{\beta}}-\frac{\sin{(\alpha-\beta)}\sin{(\beta-\gamma)}}{\sin{\alpha}\sin{\gamma}}.}\tag{13}$$

Back to the main problem

It is well-known that the orthocenter of the circumcircle mid-arc triangle is the incenter of the reference triangle. So $IB\perp{FD}$. Moreover, because of properties of inscribed angles, $$\angle{IDF}=\angle{FCB}=\angle{FDB}=\frac12\gamma$$ 

and $$\angle{IFD}=\angle{CAD}=\angle{DAB}=\angle{DFB}=\frac12\alpha.$$ 

Hence, by $ASA$, $DFI\cong{DFB}$. As a consequence, $I$ is the reflection of $B$ around $FD$. As $B'$ is the reflection of $B$ around $W$, it follows that the perpendicular bisector of $FD$ must also bisect $IB'$, meaning that $B'$ is the reflection of the orthocenter of $DEF$, $I$, around the perpendicular bisector of $DF$. Similarly, we conclude that $A'$ and $C'$ are the reflections of $I$ around the perpendicular bisectors of $EF$ and $DE$, respectively. Now our goal will be to show that $\frac{[A'B'C']}{[DEF]}=1-\frac{2r}{R}$. Note that $\angle{EFD}=\frac12(\alpha+\beta)$, $\angle{EDF}=\frac12(\beta+\gamma)$ and  $\angle{DEF}=\frac12(\alpha+\gamma)$. Applying formula $(13)$ to triangles $A'B'C'$ and $DEF$ and substituting angles we have

$$\frac{[A'B'C']}{[DEF]}=\frac{\sin{\frac12(\alpha-\gamma)}\sin{\frac12(\beta-\gamma)}}{\cos{\frac12\beta}\cos{\frac12\alpha}}+\frac{\sin{\frac12(\beta-\gamma)}\sin{\frac12(\beta-\alpha)}}{\cos{\frac12\gamma}\cos{\frac12\alpha}}-\frac{\sin{\frac12(\alpha-\gamma)}\sin{\frac12(\beta-\alpha)}}{\cos{\frac12\gamma}\cos{\frac12\beta}}. \tag{14}$$

Note that the right hand side of the equation $(14)$ are, again, products of the Mollweide's formulas. Thus, we have

$$\begin{aligned}\frac{[A'B'C']}{[DEF]}&=\frac{(a-c)(b-c)}{ab}+\frac{(b-c)(b-a)}{ac}-\frac{(a-c)(b-a)}{bc}\\&=1-\frac{c(a+b-c)}{ab}+1-\frac{b(a-b+c)}{ac}+1-\frac{a(-a+b+c)}{bc}.\end{aligned}\tag{15}$$

Substituting from the half-angle formulas, 

$$\begin{aligned}\frac{[A'B'C']}{[DEF]}&=3-\frac{2a\cos^2{\frac12\alpha}+2b\cos^2{\frac12\beta}+2c\cos^2{\frac12\gamma}}{s}\\&=1-\frac{a\cos{\alpha}+b\cos{\beta}+c\cos{\gamma}}{s}.\end{aligned}\tag{16}$$

Substituting from the law of sines, factorizing and applying double-angle identity for sine,

$$\begin{aligned}\frac{[A'B'C']}{[DEF]}&=1-\frac{R(\sin{2\alpha}+\sin{2\beta}+\sin{2\gamma})}{s}.\end{aligned}\tag{17}$$

But $\sin{2\alpha}+\sin{2\beta}+\sin{2\gamma}=4\sin{\alpha}\sin{\beta}\sin{\gamma}=\frac{abc}{2R^3}=\frac{2rs}{R^2}$, hence

$$\frac{[A'B'C']}{[DEF]}=1-\frac{R}{s}\left(\frac{2rs}{R^2}\right)=1-\frac{2r}{R}.\tag{18}$$

Finally, from $(2)$ and $(18)$,

$$[ABC]+[A'B'C']=\frac{2r}{R}[DEF]+\left(1-\frac{2r}{R}\right)[DEF]=[DEF].$$

$\square$

Some corollary inequalities:

• $\frac{(a^2-b^2)(a^2-c^2)}{b^2c^2}+\frac{(a^2-c^2)(b^2-c^2)}{a^2b^2}\geq\frac{(a^2-b^2)(b^2-c^2)}{a^2c^2}.$ (follows from $(9)$)
• $\frac{\sin{(\alpha-\beta)}\sin{(\alpha-\gamma)}}{\sin{\gamma}\sin{\beta}}+\frac{\sin{(\alpha-\gamma)}\sin{(\beta-\gamma)}}{\sin{\alpha}\sin{\beta}}\geq\frac{\sin{(\alpha-\beta)}\sin{(\beta-\gamma)}}{\sin{\alpha}\sin{\gamma}}.$ (follows from $(13)$)
• $\frac{c(a+b-c)}{ab}+\frac{b(a-b+c)}{ac}+\frac{a(-a+b+c)}{bc}\leq3.$ (follows from $(15)$) (this is IMO, 1964/2)
• $a\cos{\alpha}+b\cos{\beta}+c\cos{\gamma}\leq{s}.$ (follows from $(16)$)
• $\sin{2\alpha}+\sin{2\beta}+\sin{2\gamma}\leq\frac{s}{R}.$ (follows from $(17)$)
• $R\geq{2r}.$ (This is Euler's Inequality! It follows from (18))

Relation $(1)$ seems like a good companion for property 4 of the Garcia-reflection triangle. See M. Dalcín, S. N. Kiss Some Properties of the García Reflection Triangles 119--126.

Update. Van Khea has generalized my problem. I have given a proof of his nice generalization which is available at Van Khea's areal problem.