A Golden Path from Right to Isosceles

The inradius $r$ of a triangle with sides $a$ , $b$ , $c$ and area $A$ is $r=\dfrac{2A}{a+b+c}$ , thus the inradius of the $3$ - $4$ - $5$ right triangle is $r=\dfrac{2\times \frac{1}{2}\times 3\times 4}{3+4+5}=1$ .
The hypotenuse of a right triangle is a diameter of its circumcircle, thus the circumradius of the $3$ - $4$ - $5$ right triangle is $R=\dfrac{5}{2}$ .

Let $O$ be the circumcenter and $I$ the incenter of our isosceles $\triangle TPQ$ , ( $TP=TQ$ ) and $D$ , $E$ the intersection points of the incircle with its sides $PQ$ , $TP$ , as seen in the figures. By Euler's theorem the distance between the circumcenter and the incenter is $d=\sqrt{R\left( R-2r \right)}$ , hence $OI=\sqrt{\dfrac{5}{2}\left( \dfrac{5}{2}-2\times 1 \right)}=\dfrac{\sqrt{5}}{2}$ This means that there are two ways point $I$ can be placed on the axis of symmetry of the isosceles $\triangle TPQ$ : $I$ may lie either inside segment $OT$ , or outside it. The first case occurs when $\triangle TPQ$ is an obtuse triangle, the second when it is an acute one. This can be verified by showing that $\angle ITE$ is greater than, or less than $45{}^\circ$ respectively.
Indeed, on $\triangle ITE$ , in the first case (figure 1) we have $\sin \dfrac{T}{2}=\dfrac{IE}{IT}=\dfrac{r}{R-d}=\dfrac{1}{\dfrac{5}{2}-\dfrac{\sqrt{5}}{2}}=\dfrac{5+\sqrt{5}}{10}>\dfrac{\sqrt{2}}{2}=\sin 45{}^\circ$ Since $0{}^\circ <\dfrac{T}{2}<90{}^\circ$ , we get that $\frac{T}{2}>45{}^\circ$ , thus $\triangle TPQ$ is an obtuse triangle.

In the second case (figure 2), $\sin \dfrac{T}{2}=\dfrac{IE}{IT}=\dfrac{r}{R+d}=\dfrac{1}{\dfrac{5}{2}+\dfrac{\sqrt{5}}{2}}<\dfrac{\sqrt{2}}{2}$ thus $\triangle TPQ$ is an acute triangle.

In the problem we are interested in the latter case. But then, on right $\triangle TPD$ , $\sec P=\csc \frac{T}{2}=\frac{5}{2}+\frac{\sqrt{5}}{2}=\frac{1+\sqrt{5}}{2}+2$ For the answer, $n=\boxed{2}$ .

Intuition.

Without loss of generality, since the right triangle $\bigtriangleup ABC$ (of side lengths $3$ , $4$ and $5$ ) is not isosceles, the circumcenter $O_1$ does not lie within the incircle of center $O_2$ , which automatically produces one isosceles acute triangle and one isosceles obtuse triangle both having the congruent inradii and circumradii. Thus, for this problem, we look at $\bigtriangleup PNR$ as the circumcenter $O_1$ does not lie inside the obtuse isosceles triangle.

By Euler's theorem , since the circumradius and the inradius of $\bigtriangleup ABC$ are respectively $R = \dfrac{5}{2}$ and $r = 1$ , the distance between the circumcenter and the incenter is

$\begin{array}{rl} |\overline{O_1O_2}|^2 &= R(R - 2r)\\ &= \dfrac{5}{2}\left(\dfrac{5}{2} - 2\right)\\ |\overline{O_1O_2}| &= \dfrac{\sqrt{5}}{2} \end{array}$

which shows that the height of the isosceles triangle is

$|\overline{RM}| = |\overline{RO_1}| + |\overline{O_1O_2}| + |\overline{O_2M}| = \dfrac{5}{2} + \dfrac{\sqrt{5}}{2} + 1 = \dfrac{1}{2}(7 + \sqrt{5})$

So since

$\text{circumradius of isosceles triangle} = \dfrac{1}{8}\left(\dfrac{|\overline{PN}|^2}{|\overline{RM}|} + 4|\overline{RM}|\right)$

With some algebra, we found that $|\overline{PN}| = 2\sqrt{4 - \sqrt{5}}$ . Thus, as $0^{\circ} < \angle P < 90^{\circ}$ ,

$\begin{array}{rl} \sec^2 \angle P &= 1 + \tan^2 \angle P\\ \sec \angle P &= \sqrt{1 + \tan^2 \angle P}\\ &= \sqrt{1 + \tan^2 \left(\arctan\left(\dfrac{|\overline{RM}|}{\frac{1}{2}|\overline{PN}|} \right) \right)}\\ &= \dfrac{1}{2}(5 + \sqrt{5})\\ &= \dfrac{1}{2}(1 + \sqrt{5}) + 2 \end{array}$

where the answer to this problem is $\boxed{n = 2}$ .

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Computation.

If suppose we are interested in a more algebraic approach (assuming we don't have some tool to bypass by geometry), we can simply apply the following formulas:

$\begin{array}{rl} \text{circumradius of isosceles triangle} &= \dfrac{1}{8}\left(\dfrac{a^2}{h} + 4h\right)\\ \text{inradius of isosceles triangle} &= \dfrac{a}{4h}\left(\sqrt{a^2 + 4h^2} - a\right) \end{array}$

where $a$ is the side length opposite from the apex vertex, and $h$ is the height. In terms of $(a,h)$ , the solutions are $S_1 = \left(2\sqrt{4 - \sqrt{5}}, \dfrac{1}{2}(7 + \sqrt{5})\right)$ and $S_2 = \left(2\sqrt{4 + \sqrt{5}}, \dfrac{1}{2}(7 - \sqrt{5})\right)$ . But since $\dfrac{1}{2}(7 - \sqrt{5}) < \dfrac{5}{2}$ , $S_2$ gives the obtuse triangle. So the rest of the steps involve using $S_1$ to find $\angle P$ . This result agrees with my solution that inspects all possible triangles with less algebra.

We'll use the identity cos(A)+cos(B)+cos(C)=1+(r/R) where A,B,C are angles of triangle and r and R are the inradius and circumradius respectively. We can easily calculate that r=1,R=5/2 and takeA=B=P. Thus we have 2cos(P)-cos(2P)=7/5 => 2cos²(P)-2cos(P)+2/5=0 => 5cos²(P)-5cos(P)+1=0 =>t²-5t+5=0....taking sec(P)=t =>x²-x-1=0.....taking t-2=x => sec(P)=x+2. Now we the above quadratic has root of the golden ratio...thus sec(P)=x+2, hence n=2 is our answer.

The circumradius of a triangle is $R = \frac{abc}{4T}$ , where $a$ , $b$ , and $c$ are the sides of a triangle and $T$ is its area. For the right triangle with sides $3$ , $4$ , and $5$ , $a = 3$ , $b = 4$ , $c = 5$ , and $T = \frac{1}{2}\cdot 3 \cdot 4 = 6$ , so $R = \frac{abc}{4T} = \frac{3 \cdot 4 \cdot 5}{4 \cdot 6} = \frac{5}{2}$ .

The inradius of a triangle is $r = \frac{2T}{a + b + c}$ , where $a$ , $b$ , and $c$ are the sides of a triangle and $T$ is its area. For the right triangle with sides $3$ , $4$ , and $5$ , $a = 3$ , $b = 4$ , $c = 5$ , and $T = \frac{1}{2}\cdot 3 \cdot 4 = 6$ , so $r = \frac{2T}{a + b + c} = \frac{2 \cdot 6}{3 + 4 + 5} = 1$ .

Let the base of the acute isosceles triangle be $2x$ , its legs be $y$ , and its height be $h$ .

Then the sides of the acute isosceles triangle are $a = 2x$ , $b = y$ , $c = y$ , and $T = \frac{1}{2} \cdot 2x \cdot h = xh$ , so $R = \frac{abc}{4T} = \frac{2x \cdot y \cdot y}{4xh} = \frac{y^2}{2h} = \frac{5}{2}$ , which means $y^2 = 5h$ .

and $r = \frac{2T}{a + b + c} = \frac{2xh}{2x + y + y} = \frac{xh}{x + y} = 1$ , which means $xh = x + y$ .

By the Pythagorean Theorem, $x^2 + h^2 = y^2$ . These three equations solve to $(h, x, y) = (\frac{1}{2}(7 + \sqrt{5}), \sqrt{4 - \sqrt{5}}, \frac{1}{2}\sqrt{4 - \sqrt{5}}(5 + \sqrt{5}))$ or $(h, x, y) = (\frac{1}{2}(7 - \sqrt{5}), \sqrt{4 + \sqrt{5}}, \frac{1}{2}\sqrt{4 + \sqrt{5}}(5 - \sqrt{5}))$ , but only $(h, x, y) = (\frac{1}{2}(7 + \sqrt{5}), \sqrt{4 - \sqrt{5}}, \frac{1}{2}\sqrt{4 - \sqrt{5}}(5 + \sqrt{5}))$ leads to an acute isosceles triangle.

For this acute isosceles triangle, $\sec P = \frac{y}{x} = \frac{\frac{1}{2}\sqrt{4 - \sqrt{5}}(5 + \sqrt{5})}{\sqrt{4 - \sqrt{5}}} = \frac{1}{2}(5 + \sqrt{5}) = \phi + 2$ for $\phi = \frac{1}{2}(1 + \sqrt{5})$ . Therefore, $n = \boxed{2}$ .