Lines and Angles Practice problems

Solution 1 for question 6:

$$ \phi=\angle EAB,\ BE = \tan\phi\,,\ AE = \frac{1}{\cos\phi}, $$ $$ $\psi=\angle DAF,\ DF = \tan\psi\,,\ AF = \frac{1}{\cos\psi}, $$ $$Because\ AB\perp AD,\ we\ have\ \angle EAF \;=\;\angle EAB+\angle BAD+\angle DAF \;=\;\phi+40^\circ+\psi \;=\;45^\circ $$ $$ \quad\Longrightarrow\quad \phi+\psi=45^\circ. $$ The tangent‐addition formula gives $$ \tan(\phi+\psi) \;=\;\tan45^\circ\;=\;1 \;=\;\frac{\tan\phi+\tan\psi}{1-\tan\phi\,\tan\psi} \quad\Longrightarrow\quad \tan\phi+\tan\psi=1-\tan\phi\,\tan\psi. $$ 3. Show EF=BE+DF. Pythagoras in the little right–triangle CEF then delivers exactly $$ EF \;=\;\sqrt{(1-\tan\phi)^2+(1-\tan\psi)^2} \;=\;\tan\phi+\tan\psi \;=\;BE+DF. $$

Solution 2 for Question 6:

Label the sides of ΔAEF as follows: $$ b=AE, c=AF, a=EF, $$ so that $$ \cos\angle EAF=\frac{b^2+c^2-a^2}{2bc} =\cos45^\circ=\frac{\sqrt2}{2}. $$ Set up coordinates. Take the square ABCD of side 1 with $$ A=(0,0),\quad B=(1,0),\quad C=(1,1),\quad D=(0,1), $$ and let $$ E=(1,x),\quad F=(y,1), $$ so that BE=x and DF=y. 2. Compute the squared lengths. $$ \begin{aligned} b^2&=AE^2=(1-0)^2+(x-0)^2=1+x^2,\\ c^2&=AF^2=(y-0)^2+(1-0)^2=y^2+1,\\ a^2&=EF^2=(1-y)^2+(x-1)^2=(1-y)^2+(1-x)^2. \end{aligned} $$ 3. Plug into the cosine‐rule formula. $$ \frac{b^2+c^2-a^2}{2bc} =\frac{(1+x^2)+(1+y^2)-\bigl[(1-y)^2+(1-x)^2\bigr]} {2\sqrt{1+x^2}\,\sqrt{1+y^2}} =\frac{2(x+y)}{2\sqrt{(1+x^2)(1+y^2)}} \;\;=\;\frac{\sqrt2}{2}. $$ Hence $$ \frac{x+y}{\sqrt{(1+x^2)(1+y^2)}}=\frac{\sqrt2}{2}. $$ Squaring and rearranging gives the unique solution (with 0 <= x,y <= 1) $$ y=\frac{1-x}{1+x}. $$ Compute the required ratio. $$ BE+DF=x+y =x+\frac{1-x}{1+x} =\frac{1+x^2}{1+x}, $$ and $$ EF =\sqrt{(1-y)^2+(1-x)^2} =\sqrt{\Bigl(\frac{2x}{1+x}\Bigr)^2+\Bigl(\frac{1-x}{1+x}\Bigr)^2} =\frac{1+x^2}{1+x}. $$ Therefore $$ \frac{EF}{BE+DF} =\frac{\tfrac{1+x^2}{1+x}}{\tfrac{1+x^2}{1+x}} =1. $$ So the answer is 1.
Solution for question 18:

1. Think in terms of “turns” at each vertex. As you draw the star {7/2}, at each corner you pivot your pencil by some exterior e_i so that your line goes off to the next vertex. 2. Total turning = number of times you wind around the center. Because you’re skipping one vertex each time (step size is 2) and 7 is prime, you don’t close up until you’ve gone around the center twice. That means $$ \sum_{i=1}^{7} e_i \;=\; 2\times360^\circ \;=\;720^\circ. $$ 3. Interior + exterior = straight angle. At each vertex the interior angle A_i and the exterior turn e_i lie on a straight line: $$ A_i + e_i = 180^\circ \quad\Longrightarrow\quad A_i = 180^\circ - e_i. $$ 4.Sum up the interiors. $$ \sum_{i=1}^{7} A_i = \sum_{i=1}^{7}(180^\circ - e_i) = 7\cdot180^\circ \;-\;\sum_{i=1}^{7}e_i = 1260^\circ - 720^\circ = \boxed{540^\circ}. $$ So the seven interior (vertex) angles of the {7/2} heptagram add up to 540°.