7

The side of the square measures $1\ \mathrm{cm}$ , and $AC = 1\ \mathrm{cm}$, find the value of $AB$

picture

Michael Hardy
  • 1
Felipe
  • 201
  • 1
  • 3
  • At first glance I'd suggest using similar triangles and Pythagoras and then bashing it out with algebra but that would be no fun (nor am I sure whether it'd work). Hoping someone posts some simplistically marvelous proof – Sp3000 Mar 01 '12 at 12:37

3 Answers3

2

First try:

There are 3 rectangular triangles here, where we get 3 Pythagoras from: $$ \begin{eqnarray*} 1^2+(1-d)^2&=&c^2\\ a^2+d^2&=&1^2\\ (1+a)^2+1^2&=&(1+c)^2, \end{eqnarray*} $$ with $a=\overline{AB}$,$d=\overline{BC}$ and $c=\overline{0C}$, where $0$ is the upper left point. With this, we ask Wolfram to get $$ d=\overline{AB}\approx 0.883204 = \frac{1}{2}\left( \sqrt{2}-1+\sqrt{2\sqrt{2}-1} \right) $$

Second try:

We have $\sin \alpha= \frac{a}{1}= \frac{1+a}{1+c}$ and $1+(1-d)^2=c^2$. Further, since $d=\cos( \arcsin (a) )=\sqrt{1-a^2}$, we get $$ a=\frac{1+a}{1+\sqrt{1+(1-d)^2}}=\frac{1+a}{1+\sqrt{1+(1-\sqrt{1-a^2})^2}}, \tag{*} $$ with the same solution as above. $(*)$ can be reformulated to $$ a^8-2a^6+7a^4-6a^2+1=A^4-2A^3+7A^2-6A+1=0, $$ where the roots of polynomial of $4$th order can be found here. But because it works out nice, let's go one step further and substitute $a=u+1/2$, so we get: $$ u^4+\frac{11}{2}u^2-\frac{7}{16}=0. $$ From here on, the rest is left to you. Nice Question!

draks ...
  • 18,449
2

Let D and E be the endpoints of the upper side of the square. We know that $\triangle DEC \cong \triangle ABC. $ Therefore, we have the following: $$EC= \frac{1}{AB+1} ; 1=(1-EC)^2 + AB^2 $$ Solving for $AB$, we have $$ AB^4 + 2AB^3+AB^2-2AB-1 =0$$ Since we know that $0 \le AB \le 1,$ we apply the newton's method by letting $1$ be our initial value in the first iteration. Then we have: $$AB= 0.8832035059$$

Keneth Adrian
  • 1,118
  • 1
    The triangles should be similar which implies I make use of its idea that corresponding ratios are equal to simplify the messy algebraic manipulation – Keneth Adrian Mar 02 '12 at 02:42
1

OK, maybe not a perfect solution, but I managed to transform it into a degree-4 equation.

Let's call the top-left vertex D. Let's call DC x and AB y. Using pythagoras theorem, one can see easily that $(1 + x)^2 = (1 + y)^2 + 1^2$. If we look at the angles CAB and CDE (where E is the top right vertex). They are the same. We can compute their cosine from both angles, once as $1/x$ and the other as $(1+y)/(1+x)$.

If we do the substitution and simplify we get $x^4 + 2x^3 - x^2 -2x - 1 = 0$. Remembering we are trying to compute y.We substitute $y = 1/x$ and get the same answer as draks'.

aelguindy
  • 2,698