Find the perimeter of the quadrilateral $KLMN$.

Question

In the figure, determine the perimeter of the quadrilateral $KLMN$.(The answer provided to the problem is : $8\sqrt2$)

I could find the radii of the circumference but I can't find the sides of the quadrilateral

$\triangle ENB =: sin 45^o = \frac{4-r_2}{4}\implies \boxed{r_2 = 4-2\sqrt2}$ $\triangle PKF: (4+r_1)^2 -(4-r_1)^2 =B^2 \implies OB = 4\sqrt{r_1}\\ \therefore \triangle KOE: (4-r_1)^2=r_1^2+(4\sqrt r_1 - 4)^2 \implies\boxed{ r_1 = \frac{16}{9} = r_3}$ $\triangle LNE: (4+r_4)^2=(2\sqrt2)^2+(8-r_4-2\sqrt2)^2 \implies \boxed{r_4 = \frac{40-{16\sqrt2}}{17}}$

$MN = KN$

$ML=KL$

Answer 1

This may not be the most concise approach but I think it's feasible. I didn't obtain $8\sqrt{2}$ as the final answer, though.

• You have $\sin\measuredangle{BEN}$, $\cos\measuredangle{BEN}$, $\sin\measuredangle{OEK}$, and $\cos\measuredangle{OEK}$, so you can figure out $\cos\measuredangle{KEN}$, as $\measuredangle{KEN}=\pi-\measuredangle{BEN}-\measuredangle{OEK}$. With $KE$ and $EN$, applying the law of cosines gives you $KN$.
• It's not difficult to calculate $OB$, and with $OK$, one can easily figure out $\sin\measuredangle{OBK}$ and $\cos\measuredangle{OBK}$, and thus $\cos\measuredangle{KBL}$ as $\measuredangle{KBL}=\frac{\pi}{4}-\measuredangle{OBK}$. It's also straightforward to calculate $BK$ and $BL$, so applying the law of cosines gives you $KL$.

Since $KN=MN$ and $KL=ML$, the perimeter of quadrilateral $KLMN$ is $2(KN+KL)$.

Answer 2

Hint: Considering the figure you can find following measures using the radii you found:

1- $BE=BG-r_4$

2- $LC$ from measures BC, BL and $\angle CBL=45^o$

3- $CP=PB$ and $\angle LCP$ then $LP$.

4- show $PW=r_4$

5- find $\angle PCW$ then $\angle PCW$.

6- $\angle LCW=\angle PCW+\angle LCP$.

7- Find $MC=ML$ from LC and $\angle LCW$

8- Find $\angle KLM$, $LM=LK$ so you can find $KM$

9- Now you have measures of LN and KM and LM=LK, you can find $ MN=NK$.

Answer 3

One way to go in such situations is to use inversion w.r.t. to a chosen point that makes the "most complicated circles" to become lines. I will denote by $*$ the inversion in the corner denoted (my notations!) below by $O$, the origin of the axes in the picture, and with power $8^2=64$:

Some words on the picture, so that it become less intimidating. The axes will be used to have a simpler way to address points, i will not make huge cartesian computations. On the $Ox$ axis there are the points $A_1,A_2,A_3,A_4$ placed with step four units. So for instance $OA_2=8$, so w.r.t. inversion star the point $A_2$ is fixed, $A_2=A_2^*$. Also, since $OA_1\cdot OA_4=4\cdot 16=\left(\frac 12\cdot8\right)\left(2\cdot8\right)=8^2$ the points $A_1$ and $A_4$ correspond w.r.t. star, $A_1^*=A_4$, $A_4^*=A_1$.

The points $B_1,B_2,B_3,B_4$ on the $Oy$ axis correspond to $A_1,A_2,A_3,A_4$ by symmetry w.r.t. the angle bisector of the shown first quadrant.

The piece of paper from the question corresponds to the $8\times 8$ square $OA_2S^*B_2$, here $S^*=(8,8)$. Let $S$ be the point $S=(4,4)$. Then the notation is compatible with the inversion, since $OS\cdot OS^*=4\sqrt2\cdot 8\sqrt 2=8^2$.

There is a violet quarter circle from $A_2$ to $B_2$, and the given picture lives inside the corresponding quarter disk. Insid this region there are:

• the pink circle is $(N)$, centered in $N$, and there is not much to say any longer about $N$, it is simply $N=OSS^*\cap A_1B_1$. The picture above show its transform, but we are not needing this information.
• the blue circle is $(M)$, centered in $M$. It is tangent to $Ox$, the red half-circle $\overset\frown{OSA_2}$, and the green half-circle $\overset\frown{OSB_2}$. It lives inside the half-disk with arc $\overset\frown{OSA_2}$ and outside the other half-disk.
• the orange circle is $(L)$, centered in $L$. It is tangent to the violet quarter circle $\overset\frown{A_2B_2}$, to the red half-circle $\overset\frown{OSA_2}$, and to the green half-circle $\overset\frown{OSB_2}$. It lives inside the quarter-disk with radius $8$, and outside the half-disks with radius $4$.
• there is a reflected blue circle $(K)$, centered in $K$, but we do not really need it. The point $M$ covers the information we also need for $K$.

---

So far we have only identified the given data. Let us transform by inversion.

• The axis $Ox$ is transformed into itself.
• The axis $Oy$ is transformed into itself.
• The violet quarter-circle $\overset\frown{A_2B_2}$ is transformed into itself.
• The red half-circle $\overset\frown{OSA_2}$ is transformed into the half-line $A_2S^*$. It is also marked with red.
• The green half-circle $\overset\frown{OSB_2}$ is transformed into the half-line $B_2S^*$. It is also marked with green.
• The blue circle $(M)$ is tangent to $Ox$, red half-circle, green half-circle. So it is mapped to a circle tangent to $Ox$, red line, green line. There is only one such circle, it fits inside the square $S^*A_2A_4?$, and its center $M'$ projects thus on the axes in $A_2$ and $B_1$. (Note that $M'\ne M^*$. For this reason we use a prime to denote the center of a transformed circle. We keep this convention in the sequel.) But $(M)$ and $(M')$ correspond not only through inversion, but also through a similarity centered in $O$. We know the factor of this similarity by considering projections on $Ox$, so let us use only one coordinate, the $x$-value for points on this line. Firstly, the projection of $M'$ is $A_2+4=8+4=12$. So the projection of $M$ is by inversion $8^2/12=16/3=12\cdot(4/9)$. So $M$ is $\frac 49 M'=\frac 49(12,4)=\left(\frac {16}3,\frac {16}9\right)$. The radius of $(M)$ is thus the second component, $16/9$, OP denotes it by $r_3$.
• The orange circle $(L)$ is mapped into a circle $(L')$ tangent to the violet arc, the red line, and the blue line. Let $D=D^*$ be the point of contact
  the violet arc with $(L)$ and $(L')$, so $D=D^*=4\sqrt 2(1,1)$, the only point the point on the angle bisector $OSS^*$ with $OD=8=OA_2=OB_2$.

Many computations on the common line $ONSLDL'S^*$ of many relevant points will follow, so let us introduce $$ e =\frac 1{\sqrt 2}(1,1)\ , $$ the unit vector on this line with orientation in the first quadrant. Then we can write simpler for instance $N=2\sqrt 2e$, $S=2N=4\sqrt 2e$, $D=8e=D^*$, $S^*=8\sqrt 2 e$.

Let now $x$ be the distance from $L'$ to the red vertical or/and green horizontal line/s. It is thus the radius of $(L')$, so also $x=LD$. And $x\sqrt 2$ is $L'S^*$. This gives $$(\sqrt 2+1)x=DL'+L'S^*=DS^*=OS^*-OD=8\sqrt 2-8=8(\sqrt 2-1)\ .$$ So $$x=\frac 1{\sqrt 2+1}8(\sqrt 2-1)=8(\sqrt 2-1)^2=8(3-2\sqrt 2)=24-16\sqrt2\ . $$ I used the last formula to construct the tangent point of $(L')$ with the green line, by drawing a corresponding parallel to the line trough $S^*$ and $(24,0)$. This is not needed in the computations, but was needed while drawing the picture. Now we have the formula for $L'$, it is on the angle bisector $OSS^*$, $L'=D+xe=(8+x)e=\dots$ $%(32-16\sqrt 2)e=16\sqrt 2(\sqrt 2-1)e$

The point $L'$ is not "compatible" with the inversion, it does not have a "good/relevant" tranform, the interesting point is rather the point $Z$ such that $DZ$ is a diameter of $(L')$. (Not shown in the picture.) It is $$ \begin{aligned} Z &=2L'-D=2(D+xe)-D=D+2xe=(8+2x)e=(56-32\sqrt 2)e \\ &=8(7-4\sqrt 2)e\ ,\\ &\qquad\text{ so by inversion of power $8^2$:} \\ Z^*&= \frac8{7-4\sqrt 2}e=\frac 8{49-32}(7+4\sqrt 2)e=\frac 8{17}(7+4\sqrt 2)e\ , \\ L &=\frac 12(D+Z^*)=4e+\frac 4{17}(7+4\sqrt 2)e =\frac 4{17}(17+7+4\sqrt 2)e \\ &=\frac {16}{17}(6+\sqrt 2)e=\frac {16}{17}(3\sqrt 2+1)\cdot(1,1) \ ,\\ \operatorname{diameter}(\ (L)\ ) &=Z^*D=\left(8-\frac 8{17}(7+4\sqrt 2)\right)=\frac 8{17}(17-7-4\sqrt 2) \\ &=\frac {16}{17}(5-2\sqrt 2)\ , \end{aligned} $$ which is in OP notations exactly $2r_4$. (This is a the good point to upvote the question, everything was correctly calculated.)

---

---

It remains to find the sides of the quadrilaterals. I see only a simple, direct way to do that, we project on the axes, $NM$ and $ML$, use the known lengths of the projections and Pythagoras to conclude: $$ \begin{aligned} NM^2 &= \left(\frac {16}3-2\right)^2+ \left(\frac {16}9-2\right)^2 = \frac {100}9 +\frac4{81}=\frac{904}{9^2}\ ,\\ NM &=\frac{\sqrt{904}}9\approx 3.34073252852731\dots\ , \\[2mm] ML^2 &= \left(\frac{16}{17}(1+3\sqrt 2)-\frac {16}3\right)^2+ \left(\frac{16}{17}(1+3\sqrt 2)-\frac {16}9\right)^2 \\ &= \left(\frac{16}{17\cdot 9}\right)^2\cdot \Big[\ \Big( 9(1+3\sqrt 2)-3\cdot 17 \Big)^2 \Big( 9(1+3\sqrt 2)-17 \Big)^2 \ \Big] \\ &= \left(\frac{16}{17\cdot 9}\right)^2\cdot \Big[\ 4744-2700\sqrt 2 \ \Big] \ ,\\ ML &= \frac{32}{17\cdot 9}\sqrt{1186-675\sqrt 2}\approx 3.1816009679134\dots \ , \end{aligned} $$ and the computed values correspond to those shown by geogebra, here we profit from the fact that we have an exact drawing:

We do not have a beautiful result... So i will stop here.