Find the angle x formed in the figure below

Question

Outside the square $ABCD$, a semi-circle of diameter $BC$ is constructed. On the arc $BC$, point $L$ is marked. If $AL=2(BL)$, calculate $m \sphericalangle ALC$. (Answer: $60^\circ$)

I made the drawing and got a solution by trigonometry. Does anyone have a solution using geometry?

$x= BL$, $AL = 2x$

$α,β$ : $\sphericalangle LBC, \sphericalangle ALC$

$\triangle CLB_{(ret.)}\\ \angle BCL = 90-\alpha \implies \angle ACL =135^o +\alpha\\ \angle ACB = 90-\beta \implies \angle BAL = \beta - \alpha$

$\begin{array} :\triangle ACL:\dfrac{l\sqrt{2}}{\sin\beta}=\dfrac{2x}{\sin(135^{\circ}-\alpha)}(I)\\\triangle LAB:\dfrac{2x}{\sin(90^{\circ}+\alpha)}=\dfrac{x}{\sin(\beta-\alpha)}=\dfrac{l}{\sin(90^{\circ}-\beta)}\end{array}$

$\dfrac{2x}{\sin(90^o+α)}=\dfrac{l}{\sin(90^o−β)} ⟺ 2x=\dfrac{l\cos α}{\cos β}\\ sen(135^o−α)=\frac{\sqrt2(\cos α+\sin α)}{2}\\ \text{From (I) : } \dfrac{l\sqrt2}{\sin β}=\dfrac{l\cos α}{\dfrac{\sqrt2}{2}(\cos α+\sin α)\cos β}\\ \therefore\ 1+\tan α=\tan β\ (∗)\ . \\ \dfrac{2x}{\sin(90^o+α)}=\dfrac{x}{\sin(β−α)} \\\qquad \implies 2(\sin β \cos α−\cos β \sin α)= \cos α\qquad(\div \cos\alpha \cos\beta) \\\qquad \implies\tan β−\tan α=\dfrac{1}{2\cos β}\\ \text{From $(∗)$ : }1=\dfrac{1}{2\cos β}\ ⟹\ \cos β=\dfrac{1}{2} \therefore \boxed{β=60^\circ } $

(Solution by ani_pascual)

Answer 1

Consider $E$, the midpoint of $CA$. $LE$ bisects $\angle CLB$(since $CLBE$ is cyclic).

Consider $B'$ such that $\triangle ALB'$ is right isosceles with $\angle ALB' =90^{\circ}$.

Then, $\triangle AEL \sim \triangle ABB'$, following from $\displaystyle \frac {AE}{AB}=\frac {AL}{AB'}=\frac {1}{\sqrt2}$ and $\angle EAL= \angle BAB'$.

$\implies \angle BB'A= \angle ELA$

From here, $\angle LBB'= 90^{\circ}$.

$LB'= 2LB$, $\implies \angle BLB'=60^{\circ}$ and thus $\angle CLA= \angle BLB'= 60^{\circ}$.

Answer 2

Draw a circle with center $L$ and radius $LB$, intersecting $AL$ at $E$, and extend $LC$ to $F$ on the circle.

Draw a second circle with center $E$ and radius $EL=LB=EA$. Since the circles have equal radii and intersect at $F$, then triangle $ELF$ is equilateral (Euclid, Elements I, 1), and$$\angle CLA=\angle FLE=60^o$$

Answer 3

Let $A,B,C,D$ be the vertices of the given square. We may and do assume that its side is normed to the unit $$ AB=BC=CD=DA=1\ . $$ There is a theorem of Appolonius, that shows that the geometric locus of the points $X$ with the property $$ XA=k\cdot XB $$ is a circle. In our case, the constant $k$ is two, and the corresponding circle passes through the points $S,T$ marked in the following picture, and it is tangent to $AB$. The point $S$ is on the segment $AB$ and satisfies the proportion $AS:SB=2:1$. The point $T$ is on the ray $[AB$, it is beyond $B$ and from $TA:TB=2$ we infer $AB=BT$, so $T$ is the reflection of $A$ in $B$. The red circle has as center the point $\Omega$, taken so that $$ AS=S\Omega=\Omega T=\frac 23\ . $$ Let us draw the equilateral triangle $AUV$ with height $AB$. Then $AU=UV=2UB$, so $U$ is also on the red circle. Similarly, $V$ is also on the red circle.

The point $\color{red}L$ is by definition the intersection of the red circle $\odot(\Omega)=\odot(USVT)$ with the red half-circle of diameter $BC$.

We observe also that the green circle $\odot (AUC)$ has the chord $AC$ opposed to the angle $\widehat{AUV}=60^\circ$. We may want to also construct an equilateral triangle with height $CB$ and vertex in $C$, let $W$ be the vertex between $B$ and $T$. So $BU=BV=BW=\frac 1{\sqrt 3}$. Using the reflected argument $W$, we have $\widehat{AWC}=60^\circ$, so $AUWC$ cyclic.

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So it is enough to show:

The three circles $\color{red}{\odot(\Omega)=\odot(SUTV)}$ and $\color{blue}{\odot(BLC)}$ and $\color{darkgreen}{\odot AUC}$ intersect in a point.

Bonus: $VBWL$ is cyclic with right angles in $B,L$, and $\widehat{VLB}=45^\circ$. So $L$ is on a further circle.

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Proof: We use an inversion denoted by a star, $\color{red}*$, with center in $C$ and with power $$ \Pi =CU\cdot CV=\left(1-\frac 1{\sqrt3}\right)\left(1+\frac 1{\sqrt3}\right) =\frac 23\ , $$ and show that the transformed objects intersect in a point $L^*$, which arguably may feel simpler to handle.

The images of the circles are obtained by figuring out images of points on them. The following picture may look complicated, but this is only because a lot of points are on the list of examination. We introduce also $U',V'\in CD$, the reflections of $U,V\in CB$ w.r.t. the square diagonal $AC$, so that $\Delta AU'V'$ is equilateral with height $AD$.

• Because of $CU\cdot CV=\Pi$, the points $U,V$ are transformed in each other, $U^*=V$, $V^*=U$.
• Similarly $U',V'$ transform in each other, and $V'$ is on the green circle.
• The distance $CA$ is $\sqrt2$, so $A$ transforms into the point $A^*$ which is on the diagonal $CA$ of the square $ABCD$ at one third of it distance from $C$. In particular, $AS:AB=2:3=AA^*:AC$, so $SA^*\|BC$.
• The reflected argument w.r.t. $BC$ shows that the transformed point $T^*$ is on the reflected diagonal $AT$ (of the reflected square) at the same one third distance from $C$, and $\Omega T^*\|BC$, too. Observe that $\Omega T^*=\frac 23 BC=\frac23=\Omega S=\Omega T$, so $T^*$ is also on the red circle.
• Where is the point $B^*$? It is on $BC$ at distance $CB^*=\frac 23$ from $C$, for it satisfies $CB\cdot CB^*=\frac 23$. So $BS=BB^*=B\Omega=\frac 13$.
• The distance $CW$ satisfies $CW^2= CB^2+BW^2=1+\frac 13=\frac 43$, so $CW=\frac 2{\sqrt 3}$, and so $W^*$ is the mid point of $CW$.

Now we can describe the images of the three circles. $$ \begin{aligned} \color{red}{\odot(\Omega)^*} & \color{red}{=\odot(UTT^*V)^*=\odot( U^*T^*T^{**}V^*)=\odot( VT^*TU)=\odot(\Omega)}\ ,\text{ invariated by $*$,} \\ \color{blue}{{\overset\frown{BC}}^*} & \color{blue}{=\text{ray from $B^*$ perpendicular in $B^*$ on $BC$,}} \\ \color{darkgreen}{\odot(CV'AUW)^*} & \color{darkgreen}{=\text{line }{V'}^*A^*U^*W^*=\text{line }U'A*VW^*}\ . \end{aligned} $$ Let now $L^*$ be the intersection of the two lines, the green line, and the blue line. We compute $B^*L^*$ from the similar triangles $\Delta VCU'\sim\Delta VB^*L^*$: $$ B^*L^* =CU'\cdot\frac{VB^*}{VC} =\left(1+\frac 1{\sqrt 3}\right)\cdot\frac{\frac 1{\sqrt 3}-\frac 13}{1-\frac 1{\sqrt 3}}=\left(1+\frac 1{\sqrt 3}\right)\cdot \frac 1{\sqrt 3} =\frac 1{\sqrt 3}+\frac 13\ . $$ We can now compute the hypotenuse $\Omega L^*$ in the right triangle with the other sides on the directions $B^*L^*$ and $\Omega T^*$, with lengths $B^*L^*-B\Omega=\frac 1{\sqrt 3}$ and $BB^*=\frac 13$: $$ {B^*L^*}^2 =\frac 13+\frac 19=\frac 49=\left(\frac 23\right)^2=\Omega S^2\ , $$ so $L^*$ is also on the red circle.

$\square$

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The job is done, but let us make also some remarks.

• The angle between the green line $U'V$ and the direction $AB\|CD$ is $15^\circ$ because of $$ \tan\widehat{VU'C}=\frac{VC}{U'C}=\frac{1-\frac 1{\sqrt 3}}{1+\frac 1{\sqrt 3}}=\frac{\sqrt 3-1}{\sqrt 3+1}=\frac 12(\sqrt 3-1)^2=2-\sqrt 3=\tan 15^\circ\ . $$
• Let $\Psi$ be the point of intersection of the green line $U'V$ with the "diagonal" $CT$. So $\Delta CU'\Psi$ has known angles, $15^\circ$ in $U'$, and $30^\circ$ in $\Psi$, so we can compute $C\Psi$ from the sine theorem: $$ C\Psi =CU'\cdot\frac {\sin 15^\circ}{\sin 30^\circ} =\frac {CU'}{2\cos 30^\circ} =\frac{1+\frac 1{\sqrt 3}}{\frac 1{\sqrt 2}(\sqrt 3+1)}=\sqrt{\frac 23}\ . $$ In particular $C\Psi^2=\frac 23$, so $\Psi^*=\Psi$. In other words, $\Psi$ is also on the green circle, since its transform is on the green line.
• We already know that $L$ is on the green circle, so $\widehat {CLA}=60^\circ$.
• Let us show that $VBWL$ is cyclic. Equivalently, we show the transformed object $(VBWL)^*=UB^*W^*L^*$ is cyclic. The angle $\widehat{B^*CW}$ is $30^\circ$, from the equilateral triangle with height $CB$. In it, $B^*$ is the center of symmetry, being on the height $CB$ at the right place. So we know the angle $\widehat{W^*B^*U}=120^\circ$, $W^*$ being the mid point of $CW$. The opposite angle in our quadrilateral to be shown cyclic is $$ \widehat{W^*L^*U}= \widehat{VL^*U}= \widehat{VTU}= \widehat{VAU}= 60^ \circ\ . $$ So it is the right supplement. This shows $VLWB$ cyclic.
• In particular, $\widehat{VLW}=180^\circ-\widehat{VBW}=90^\circ$, and $\widehat{VLB}=\widehat{VWB}=45^\circ$.

Answer 4

Build an equilateral triangle $ABE$ as in the picture. Let $F$ be the midpoint of $BC$. Let $G$ be the reflection of $B$ in $EF$. Note that $FG = FB = FC$, hence $G$ lies on the circle with diameter $BC$.

Note that $A,B,G$ lie on a circle with center $E$ and radius $AE=BE=GE$. Therefore $\angle AGB = \frac 12 \angle AEB = 30^\circ$. Furthermore, by symmetry, $\angle EGF = \angle FBE = 90^\circ - 60^\circ = 30^\circ$.

Moreover, $\angle FEG = \frac 12 \angle BEG = \angle BAG$. This shows that $\triangle EFG \sim \triangle ABG$ by AAA. Since $EG = EB = AB = BC = 2BF = 2FG$, it follows that $AG=2BG$.

Since $G$ lies on the circle with diameter $BC$ and satisfies $AG=2BG$, it follows that $G$ coincides with $L$.

To finish, simply note that $\angle CLA = \angle CGA = 90^\circ - 30^\circ = 60^\circ$.