Prove that there exists a unique point $M$ on the side $AD$ of a convex quadrilateral $ABCD$ such that \[\sqrt{S_{ABM}}+\sqrt{S_{CDM}} = \sqrt{S_{ABCD}}\]if and only if $AB\parallel CD$.
Problem
Source: Bulgaria National Olympiad 2023 Problem 4
Tags: geometry, area, convex quadrilateral
ACGNmath
12.04.2023 20:49
There's actually a surprisingly neat solution to this question. Square the equality to obtain \[S_{ABM}+S_{CDM}+2\sqrt{S_{ABM}S_{CDM}}=S_{ABCD}\]This is equivalent to \[2\sqrt{S_{ABM}S_{CDM}}=S_{BCM}\] Firstly, let us consider the case when $AB\parallel CD$. [asy][asy] size(7cm); pair A = dir(100); pair D = dir(200); pair C = dir(340); pair B = A+0.4*(C-D); draw(A--B--C--D--A--cycle); pair M = 0.6*A+0.4*D; pair X = foot(M,A,B); pair Y = foot(M,C,D); draw(A--X--Y,dashed); draw(B--M--C); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$M$",M,dir(M)); dot("$X$",X,dir(X)); dot("$Y$",Y,dir(Y)); [/asy][/asy] Let $X$ and $Y$ be the feet of the perpendiculars from $M$ to $AB$ and $CD$ respectively. Denote $AB=a, CD=b$ and the heights $MX$ and $MY$ by $x$ and $y$ respectively. Then \[2\sqrt{S_{ABM}S_{CDM}}=2\sqrt{\frac{ax}{2}\frac{by}{2}}=\sqrt{abxy}\]\[S_{BMC}=S_{ABCD}-S_{ABM}-S_{CDM}=\frac{1}{2}(a+b)(x+y)-\frac{1}{2}ax-\frac{1}{2}by=\frac{1}{2}(ay+bx).\]Therefore, \[ay+bx=2\sqrt{abxy}\quad\Rightarrow\quad (\sqrt{ay}-\sqrt{bx})^2=0\]This is achieved when $\frac{x}{y}=\frac{a}{b}$, or when $\frac{AM}{DM}=\frac{AB}{CD}$. It is clear that $M$ is unique in this case. Now, suppose that $AB$ is not parallel to $CD$. Let $D'$ be the point on $AD$ such that $AB\parallel CD'$. In the case where $D'$ lies outside segment $AD$, then we know that \[2\sqrt{S_{ABM}S_{CDM}}<2\sqrt{S_{ABM}S_{CD'M}}=\sqrt{abxy}\leq \frac{1}{2}(ay+bx)=S_{BCM}.\]Therefore, there cannot exist $M$ on $AD$ such that equality holds. In the case where $D'$ lies within segment $AD$, then consider what happens to the quantity $2\sqrt{S_{ABM}S_{CDM}}-S_{BCM}$ as we animate $M$ from $A$ to $D$. At points $A$ and $D$, this quantity is negative. Take $M$ to be the point on $AD'$ which divides it into the ratio $AB/CD'$. Then we know that at this point, we have \[2\sqrt{S_{ABM}S_{CDM}}-S_{BCM}>2\sqrt{S_{ABM}S_{CD'M}}-S_{BCM}=0\]Therefore, using the intermediate value theorem, since this function is continuous, there exists two points $M_1$ and $M_2$ in segments $AM'$ and $DM'$ respectively satisfying the desired quantity is equal to zero. Therefore there is no unique $M$ in this case, and we are done.
Marinchoo
12.04.2023 22:35
Assume there exists a unique such point $M$. Let $AB=a$, $CD=b$, $\angle A = \alpha$, $\angle B = \beta$, $AD = x$, $AM = m$, $S_{ABCD}=S$. Note that:
\[S_{ABM} = \frac{AB\cdot AM \cdot \sin\angle BAM}{2} = \frac{am\sin\alpha}{2}\quad\text{ and }\quad S_{CDM} = \frac{DC \cdot DM \cdot \sin\angle CDM}{2} = \frac{b(x-m)\sin\beta}{2}\]\[\Longrightarrow\sqrt{S_{ABM}}+\sqrt{S_{CDM}} = \sqrt{S_{ABCD}}\iff 2\sqrt{S_{ABM}\cdot S_{CDM}} = S_{BCM}\]\[\iff 4S_{ABM}\cdot S_{CDM} = S^2_{BCM} = (S - S_{ABM} - S_{CDM})^2\]\[\iff 4\left(\frac{am\sin\alpha}{2}\right)\left(\frac{b(x-m)\sin\beta}{2}\right) = \left(S - \frac{am\sin\alpha}{2} - \frac{b(x-m)\sin\beta}{2}\right)^2 \quad (\star)\]\[\iff 4ab\sin\alpha\sin\beta m(x-m) = \left((2S-bx\sin\beta) - m(a\sin\alpha - b\sin\beta)\right)^2\]\[\iff 4ab\sin\alpha\sin\beta m(x-m) = m^2(a\sin\alpha - b\sin\beta)^2 - 2(2S-bx\sin\beta)(-a\sin\alpha + b\sin\beta)m+(2S-bx\sin\beta)^2\]\[\iff m^2(a\sin\alpha + b\sin\beta)^2 - 2m\left((2S-bx\sin\beta)(-a\sin\alpha + b\sin\beta)-2abx\sin\alpha\sin\beta\right)+(2S-bx\sin\beta)^2 = 0\]The last is quadratic in $m$ with the leading coefficient being strictly positive as $a\sin\alpha +b\sin\beta>0$ hence if there exists a root $m$, the equation must also have a second root $m'$. From Vieta's on the constant term it follows that $m'>0$ and one can also note that $m'<x$ as otherwise the LHS is negative and the RHS non-negative in $(\star)$. Therefore we must have $m' = m$ due to the uniqueness, i.e. the discriminant is zero:
\[\left((2S-bx\sin\beta)(-a\sin\alpha + b\sin\beta)-2abx\sin\alpha\sin\beta\right)^2 = (a\sin\alpha + b\sin\beta)^2(2S-bx\sin\beta)^2\]Clearly, $2S-bx\sin\beta = 2S-2S_{ACD} = 2S_{ABC}>0$, i.e. $(a\sin\alpha + b\sin\beta)(2S-bx\sin\beta)>0$, and also
\[(2S-bx\sin\beta)(-a\sin\alpha + b\sin\beta)-2abx\sin\alpha\sin\beta < (2S-bx\sin\beta)( - a\sin\alpha + b\sin\beta) < (2S-bx\sin\beta)(a\sin\alpha + b\sin\beta)\]implying that we can remove the second powers as follows:
\[\iff (2S-bx\sin\beta)(a\sin\alpha - b\sin\beta)+2abx\sin\alpha\sin\beta = (a\sin\alpha + b\sin\beta)(2S-bx\sin\beta)\]\[\iff 2abx\sin\alpha\sin\beta = 2b\sin\beta(2S-bx\sin\beta)\]\[\iff ax\sin\alpha + bx\sin\beta = 2S\iff S_{ABD}+S_{ADC} = S_{ABCD}\iff AB\parallel CD\]
dgrozev
13.04.2023 09:57
A more geometric solution, I think!
sami1618
10.07.2024 18:02
Here is a solution using affine transforms: Consider the transformation such that $A=(0,1)$, $D=(0,0)$, and $C=(1,0)$. Then $B$ is mapped to $(x,y)$ for positive $x$ and $y$ whose sum is greater then $1$ and $M$ is a variable point $M=(0,m)$ with $m\in (0,1)$. Then the equality becomes $$\sqrt{\frac{x(1-m)}{2}}+\sqrt{\frac{m}{2}}=\sqrt{\frac{x+y}{2}}$$Notice that at $m=0$ and at $m=1$ the LHS is smaller then the RHS. Thus for the equation to have exactly one solution the RHS must be the maximum of the LHS. Notice that the Cauchy-Shwartz inequality yields $$\frac{x+1}{2}=\left(\frac{x}{2}+\frac{1}{2}\right)\left((1-m)+m\right)\geq \left(\sqrt{\frac{x(1-m)}{2}}+\sqrt{\frac{m}{2}}\right)^2$$Notice that equality holds at $m=\frac{1}{x+1}$. Thus we must have $y=1$ which clearly works, finishing the problem.