David's distance

Let $M$ , $N$ be midpoints of $AD$ , $BC$ respectively.

So that, MN is a symmertry axis of rectangle $ABCD$ , infer that MN is perpendicular to $X_1X_2$ . $(1)$

We have $X_1X_2$ is a chord of circle that has center $M$ . $(2)$

From $(1)$ and $(2)$ , we have $X_1$ and $X_2$ are symmetric by $MN$ so $d^2=4NX_1^2$ . $(3)$

Apply Pythagorean theorem to right triangle $NX_1M$ , we have $NX_1^2=MX_1^2-MN^2=\left( \frac {AD}{2} \right)^2-AB^2=6^2-5^2=11$ . $(4)$

(note: $X_1M$ is a median of right triangle $X_1AD$ )

From $(3)$ and $(4)$ , we have $d^2=44$ .

We know that, angle subtended in a right angle. So let us consider a circle with centre on midpoint of $AD$ say $O$ Since $\angle AX_{1}D = \angle AX_{2}D$ We have to take diameter of this circle to be $AD$ . Draw a perpendicular from $O$ to $P$ . $OP=5$ Draw segments $OX_{1}$ and $OX_{2}$ which are radii with radius = $6$ We can prove that $\Delta X_{1}PO \cong \Delta X_{2}PO$ by RHS congruency. Hence, $X_{1}P=X_{2}O$ By considering rt. $\Delta OPX_{2}$ with hypotenuse $OX_{2}=6$ by pythagoras theorem we have $PX_{2} = \sqrt{11}$ , $d=2 \sqrt{11} \Rightarrow d^{2}=44$

let DX1 = AX2 = x, AX1 = DX2 = y, and CX1 = BX2 = z by Phytagoras Theorem : x^2 + y^2 = 144 area of rectangle ABCD = 1/2 * 12 * 5 = 30 1/2 * x * y = 30, then y = 60/x.

x^2 + (60/x)^2 = 144 x^4 - 144x^2 + 3600=0 ( x^2 - (72+12 √(11) ) * ( x^2 - (72-12√(11)) ) = 0. x^2 = 72+12 √(11) or x^2 = 72-12 √(11) the value of x must be positive, then x = √(72-12 √(11)) = √66 - √6, y = 72 +12 √(11) = √66 + √6 and z = √((√66 - √6)^2 - 5^2)) = 6 - √11

the value of d^2 = (12 - 2 * z)^2 = (12 - 2 * (6 - √11) )^2 = (2√11)^2 = 44

We have : $A(0,0),B(0,5),C(12,5),D(12,0)$ , We are trying to find some point $M(x,5)$ laying on the line BC and verifying that the lines $AM$ and $MD$ are perpendicular this means that : $\overrightarrow{AM}\cdot \overrightarrow{MD}=0.$

From the coordinates, we get : $\overrightarrow{AM} \left( \begin{matrix} x \\ 5 \end{matrix} \right)$ , and : $\overrightarrow{MD} \left( \begin{matrix} 12-x \\ -5 \end{matrix} \right)$ , Hence we get : $x(12-x)-25=0 \Longrightarrow x=6\pm \sqrt{11}.$ We conclude that there are two possibilities for M as stated in the problem : $X_1(6+\sqrt{11},5),X_2(6-\sqrt{11})$ , the distance between them is : $d= 2 \sqrt{11}$ ,

Finally, we have : $d^2=\left(2\sqrt{11}\right)^2=44$ .

Let $O$ be the midpoint of $A$ and $D$ . Let $\theta_1$ and $\theta_2$ be $\angle AOX_1$ and $\angle DOX_2$ , respectively.Let $M$ be the midpoint of $X_1$ and $X_2$ .

Because $\angle AX_1D = \angle AX_2D = 90^o$ we can say that $X_1 and X_2$ lies on a circle whose diameter is $AD$ , specifically it's center is $O$ . Let's call this circle $\gamma$ .

Since $ABCD$ is a parallelogram, then we can say that the perpendicular distance of $M$ to line segment $AD$ is $AB=5$ . Since the radius of the circle $\gamma = \frac{AD}{2} = 6$ , then the distance from $O$ to $X_1$ and $X_2$ is also $6$ .

Thus if we use the pythagorean relation to the right triangle $X_1MO$ , we can see that $X_1M = \sqrt{11}$ . Moreover, using the same line of argument $X_2M = \sqrt{11}$ . It is clear that $MO$ is perpendicular to $AD$ and $BC$ because $cos \theta_1 = cos \theta_2$ and $ABCD$ is a rectangle.

Therefore the distance, $d$ , from $X_1$ to $X_2$ is $2\sqrt{11}$ . Thus, $d^2 = (2\sqrt{11})^2 = 44$

Let x1 and x2 divide BC into 3 segments with each of the length is (6-d/2), d, and (6-d/2). Suppose that Ax1 is x and Ax2 is y X^2=25+(6-d/2)^2 Y^2=25+(6+d/2)^2 And also x^2+y^2=144 By subtitution method, we can get d^2=44

Given the rectangle $ABCD$ with $AB = CD = 5$ and $BC = AD = 12$ , and the two points $X_1$ and $X_2$ such that $∠AX_1D$ and $∠AX_2D$ are both equal to $90^∘$ . So we draw line segments from $A$ to $X_1$ , $X_1$ to $D$ , $A$ to $X_2$ , and $X_2$ to $D$ inside $ABCD$ .

Let us focus now on trapezoid $ABX_2D$ , which is so because $AD || BX_2$ . Within $ABX_2D$ are two triangles, $ABX_2$ and $AX_2D$ , which are both right triangles because they contain the right angles $∠ABX_2$ and $∠AX_2D$ , respectively. Also, $X_1$ is in the segment $BX_2$ .

Let $x$ be the length of segment $AX_2$ , $y$ be that of segment $DX_2$ , $z$ , that of segment $BX_1$ , and $w = z + d$ ( $d$ is the length of segment $X_1X_2$ ; $w$ = length of segment $BX_2$ ). Furthermore, the side $X_2$ is common between the triangles $ABX_2$ and $AX_2D$ . From these information and the Pythagorean Theorem, $AB^2 + BX_2^2 = AX_2^2$ and $AX_2^2 + DX_2^2 = AD^2$ , or equivalently,

$1. 25 + w^2 = x^2$

$2. x^2 + y^2 = 144$

By substitution, we have $25 + w^2 + y^2 = 144$ , or

$3. w^2 + y^2 = 119$ .

Let us now take in triangle $DCX_2$ , which is also a right triangle because of right angle $∠DCX_2$ . Furthermore, segment $DX_2$ is the hypotenuse of $DCX_2$ . So letting $a$ be the length of side $CX_2$ ,

$4. a^2 + 25 = y^2$

Another substitution gives us:

$5. w^2 + a^2 = 94$

Another thing about $a$ is that it is also equal to $12 - w$ . Solving for $w$ in terms of $a$ ( $w = 12 - a$ ), and substituting this into $5$ :

$6. (12 - a)^2 + a^2 = 94$

Simplifying, and multiplying both sides of the equation by $\frac {1}{2}$ :

$7. a^2 - 12a + 25 = 0$

Using the Quadratic Formula to find $a$ , $a = CX_2 = 6 \pm \sqrt{11}$ .

This is one part. The second one comes when we consider trapezoid $DCX_1A$ with triangles $AX_1D$ and $DCX_1$ , and triangle $ABX_1$ , and the variables are assigned to corresponding segments on both polygons. Using the similar line of argument as the one we had before, we have $BX_1 = 6 \pm \sqrt{11}$ .

Finally, it can be verified that $BX_1 + X_1X_2 + CX_2 = BC$ . Thus, the lengths of $BX_1$ and $CX_2$ must not exceed that of $BC$ , which is 12. Out of the two possible values for $BX_1$ and $CX_2$ , the only acceptable value is $6 - \sqrt{11}$ . Therefore, $BX_1$ = $CX_2$ = $6 - \sqrt{11}$ , leaving $X_1X_2 = 2\sqrt{11}$ , which is the exact value of $d$ . The required answer is $d^2 = (2\sqrt{11})^2 = 44$ .

using to-ong triangular theorem

let's assume that $DX = x$

in the right angle triangle $ADX$ :

$AX^2 = 5^2 + x^2$

and in the right angle triangle $XBC$

$XB^2 = 5^2 + (12-x)^2$

and in the right angle triangle $ABX$

$AX^2 = 12^2 - XB^2$

$AX^2 = 12^2 -( 5^2 + (12-x)^2 )$

since $AX = AX$ :D we can say that

$12^2 - ( 5^2 + (12-x)^2 ) = 5^2 + x^2$

we can solve this equation easily and we will get that

$x = 6 - \sqrt{11}$ ... which is $DX1$

$x = 6 + \sqrt{11}$ ... which is $DX2$

we subtract them to get $X1X2 = 2 \sqrt{11}$

we square to get $\boxed{44}$

$AX_{1}D$ and $AX_{2}D$ lie in the same circle with circumcenter as the midpoint of AD, so the circumradius comes out to be $\frac {AD}{2}=6$ . Draw perpendiculars $X_{1}E$ and $X_{2}F$ to $AD$ . $X_{1}E = X_{2}E = AB = 5$ . In triangle $EX_{1}O$ , $EX_{1} = 5$ , $X_{1}O = 6$ . By applying pythagorean theorem on right triangle $EX_{1}O$ , we obtain $EO = \sqrt{11}$ . By symmetry we get $EO = OF = \sqrt{11}$

$\Rightarrow EF = 2 \sqrt{11}$

Since $EX_{1}X_{2}F$ is a rectangle.

$\Rightarrow X_{1}X_{2} = EF = 2 \sqrt{11}$

$\Rightarrow X_{1}X^2_{2} = 44$

i just draw it on paper,with my 90 degree ruler, and i got d is 6,6 so d^2=43,56 which mean 43 or 44 i answer 44 for the first time and it true :D