Geometry - Inadequate Information #2

Relevant wiki: Apollonius's Theorem

This solution uses Apollonius' Theorem .

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Use it on $\triangle ABD$ , $\triangle BCD$ , and $\triangle ACD$ respectively.

$\overline{AB}^2+\overline{AD}^2=2\left(\overline{AN}^2+\overline{BN}^2\right)~\cdots\boxed{A}~$

$\overline{BC}^2+\overline{CD}^2=2\left(\overline{CN}^2+\overline{BN}^2\right)~\cdots\boxed{B}~$

$\overline{AN}^2+\overline{CN}^2=2\left(\overline{MN}^2+\overline{AM}^2\right)~\cdots\boxed{C}~$

Calculate $\boxed{A}+\boxed{B}+2\times\boxed{C}$ .

$\begin{aligned} \overline{AB}^2+\overline{BC}^2+\overline{CD}^2+\overline{DA}^2&=4\overline{AM}^2+4\overline{BN}^2+4\overline{MN}^2 \\ &=\overline{AC}^2+\overline{BD}^2+4\overline{MN}^2 \end{aligned}$

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Therefore, $\overline{BC}^2+\overline{DA}^2=112$ .

From the question, we know that $\overline{BC}+\overline{DA}=14$ .

$\therefore~\overline{BC}\cdot\overline{DA}=\boxed{42}$ .

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Now let's prove that this configuration of points actually exists.

From the information above, we can safely say that $\overline{BC}=7-\sqrt{7}$ and $\overline{DA}=7+\sqrt{7}.$

Since we only need to prove that at least one configuration exists, I'm going to prove only one.

Consider $\angle BCD=\dfrac{\pi}{2}.$

Then let $O,~P,~Q$ be the center of the circle, and the points where $\overline{BC},~\overline{CD}$ contact with the circle, respectively.

We're going to draw a tangent line $\overline{BX}$ and $\overline{DY}$ from point $B$ and point $D,$ such that $\overline{BX}=5$ and $\overline{DY}=7+\sqrt{7}.$

$r$ is the radius of the circle, and then let's just say $0<2<r<4<7-\sqrt{7},$ to make this easier.

$\overline{BX},~\overline{CY}$ meets with the circle at points $R$ and $S$ respectively.

Then note that $\overline{CP}=\overline{CQ}=r.$

From it we can see that $\overline{BP}=7-\sqrt{7}-r,~\overline{XR}=-2+\sqrt{7}+r,~\overline{DQ}=9-r,~\overline{YS}=-2+\sqrt{7}+r.$

Oh. Seems like $\overline{XR}=\overline{YS}.$ Then $\angle XOR=\angle YOS.$

Then let's just say $\angle POC=\alpha,~\angle BOR=\beta,~\angle YOS=\gamma,~\angle DOQ=\delta.$

Since we already know $\overline{OX}=\overline{OY,}$ for $X$ and $Y$ to become an identical point, we need $\alpha+\beta+\gamma+\delta=\pi.$

Then, $\tan(\alpha+\delta)=-\tan(\beta+\gamma).$

$\tan\alpha=\dfrac{r}{r},~\tan\beta=\dfrac{7-\sqrt{7}-r}{r},~\tan\gamma=\dfrac{-2+\sqrt{7}+r}{r},~\tan\delta=\dfrac{9-r}{r}.$

$\begin{aligned} \tan(\alpha+\delta)+\tan(\beta+\gamma) &=\dfrac{\dfrac{r}{r}+\dfrac{9-r}{r}}{1-\dfrac{9-r}{r}}+\dfrac{\dfrac{7-\sqrt{7}-r}{r}+\dfrac{-2+\sqrt{7}+r}{r}}{1-\dfrac{(7-\sqrt{7}-r)(-2+\sqrt{7}+r)}{r^2}} \\ & =\dfrac{9}{2r-9}+\dfrac{5r}{2r^2-(9-2\sqrt{7})r+(21-9\sqrt{7})} = 0 \end{aligned}$

Then $10r^2-45r+18r^2-9(9-2\sqrt{7})r+9(21-9\sqrt{7})=0.$

Let $f(r)=28r^2-45r-9(9-2\sqrt{7})r+9(21-9\sqrt{7}).$

$\begin{aligned} f(2) &=22-9(18-4\sqrt{7}+21-9\sqrt{7}) \\ &<22-9(39-35)=22-36<0 \\ \\ f(4) &=268-9(36-8\sqrt{7}+21-9\sqrt{7}) \\ &>268-9(57-44)=151>0 \end{aligned}$

Therefore $f(r)=0$ has a solution where $2<r<4,$ and therefore, the configuration of points $A,~B,~C,~D$ exists such that it would satisfy the given condition.

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Formula: (Euler's midpoint theorem)

$\overline{AB}^2+\overline{BC}^2+\overline{CD}^2+\overline{DA}^2=\overline{AC}^2+\overline{BD}^2+4\overline{MN}^2$

$Let ~a=AB,~~b=BC,~~c=CD,~~d=DA;~~~~~~~p=AC,~~q=BD;~~~~~~~x=MN.\\ Pitot ~Theorem:-~~~~~a+c=b+d.~~~~~~\implies~(b^2+2bd+d^2)=(a+c)^2=196.\\ Euler's ~Midpoint~ Theorem:-~~~~~a^2+b^2+c^2+d^2=p^2+q^2+4x^2=218.~~~~~~\implies~b^2+d^2=218-5^2-9^2=112 . \\ \therefore~196=b^2+2bd+d^2=(b^2+d^2+2bd)=112+2bd.\\ \therefore~BC*DA=bd=\frac 1 2*(196-112)=\Large~~~~\color{#D61F06}{42}.$
The author Mr. H.M. 유 has already shown the existence of such a quadrilateral.