Infinite Altitudes

Geometry Level 2

In the diagram shown below, isosceles right $\triangle{ABC}$ with right $\angle{B}$ and hypotenuse $\overline{AC}$ has a leg length of $1$ . Furthermore, Point $D$ is on $\overline{AC}$ , Point $E$ is on $\overline{AB}$ , and Point $F$ is on $\overline{BD}$ such that $\overline{BD}$ is an altitude to $\overline{AC}$ , $\overline{DE}$ is an altitude to $\overline{AB}$ , and $\overline{EF}$ is an altitude to $\overline{BD}$ . If the process of altitudes being drawn to neighboring hypotenuses in a coiling manner is repeated infinitely many times within $\triangle{ABC}$ , the sum of the infinite number of altitudes can be written in the form $p+\sqrt{q}$ , where $p$ and $q$ are coprime positive integers.

What is $p+q$ ?

The answer is 3.

1 solution

Aidan Poor
Sep 25, 2018

Let the sum of the infinite number of altitudes be $S$ . Using geometric properties of $45°-45°-90°$ triangles, the first few altitudes can be listed:

$S=(\frac{1}{\sqrt{2}}+\frac{1}{2}+\frac{1}{2\sqrt{2}}+\frac{1}{4}+\frac{1}{4\sqrt{2}}+\frac{1}{8}+\cdots)$

Notice that each following term is being multiplied by a factor of $\frac{1}{\sqrt{2}}$ from the preceding term. Because of this, solving for $S$ can be done as follows:

$S=(\color{#D61F06}\frac{1}{\sqrt{2}}+\frac{1}{2}+\frac{1}{2\sqrt{2}}+\frac{1}{4}+\frac{1}{4\sqrt{2}}+\frac{1}{8}+\cdots\color{#333333})$

$S=\frac{1}{\sqrt{2}}(1+\color{#D61F06}\frac{1}{\sqrt{2}}+\frac{1}{2}+\frac{1}{2\sqrt{2}}+\frac{1}{4}+\frac{1}{4\sqrt{2}}+\frac{1}{8}+\cdots\color{#333333})$

$S=\frac{1}{\sqrt{2}}(1+\color{#D61F06}S\color{#333333})$

$S\sqrt{2}=1+S$

$S\sqrt{2}-S=1$

$S(\sqrt{2}-1)=1$

$S=\frac{1}{\sqrt{2}-1}$

$S=1+\sqrt{2}$

$\therefore p+q=\boxed{3}$

Infinite Altitudes

The answer is 3.

1 solution

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