Irish Mathematical Olympiad

For $\triangle ABC$, let the altitude $AD$ standing on the base $BC$ have length $p$, let the altitude $BE$ standing on the base $AC$ have length $q$, and let the altitude $CF$ standing on the base $AB$ have length $r$.

Step 1: Then $\triangle ADC$ is similar to $\triangle BEC$ and so $\frac{p}{q} = \frac{b}{a}$ using the usual notation for the sides of the triangle $ABC$.

Step 2: Let $\triangle A_1B_1C_1$ have side lengths $B_1C_1 = p$, $C_1A_1 = q$ and $A_1B_1 = r$. Let the altitude standing on the base $B_1C_1$ have length $p_1$, let the altitude standing on the base $C_1A_1$ have length $q_1$, and let the altitude standing on the base $A_1B_1$ have length $r_1$. Then, similar reasoning to that of Step 1 yields $\frac{p_1}{q_1} = \frac{q}{p}$. Using the result of Step 1 then yields $\frac{p_1}{q_1} = \frac{a}{b}$.

Step 3: Reasoning as in the previous Step, we obtain $\frac{q_1}{r_1} = \frac{b}{c}$ and $\frac{r_1}{p_1} = \frac{c}{a}$, and so $\triangle A_2B_2C_2$ is similar to $\triangle ABC$.

Step 1: Let $\Delta$ denote the area of $\triangle ABC$. Then the altitudes of $\triangle ABC$ are equal to $p = \frac{2\Delta}{a}$, $q = \frac{2\Delta}{b}$ and $r = \frac{2\Delta}{c}$, using the usual notation for the sides of the triangle $ABC$.

Step 2: Next let $\Delta_1$ denote the area of $\triangle A_1B_1C_1$. Then the altitudes of $\triangle A_1B_1C_1$ are equal to $p_1 = \frac{2\Delta_1}{p} = \frac{\Delta_1}{\Delta}a$, $q_1 = \frac{2\Delta_1}{q} = \frac{\Delta_1}{\Delta}b$ and $r_1 = \frac{2\Delta_1}{r} = \frac{\Delta_1}{\Delta}c$.

Step 3: From the above, $\frac{a}{p_1} = \frac{b}{q_1} = \frac{c}{r_1}$, so that $\triangle A_2B_2C_2$ is similar to $\triangle ABC$.

Step 1: Let $\Delta$ denote the area of $\triangle ABC$. Then the altitudes of $\triangle ABC$ are equal to $p = \frac{2\Delta}{a}$, $q = \frac{2\Delta}{b}$ and $r = \frac{2\Delta}{c}$, using the usual notation for the sides of the triangle $ABC$.

Step 2: Let $\triangle A_1B_1C_1$ have side lengths $B_1C_1 = p$, $C_1A_1 = q$ and $A_1B_1 = r$. Let the altitude standing on the base $B_1C_1$ have length $p_1$, let the altitude standing on the base $C_1A_1$ have length $q_1$, and let the altitude standing on the base $A_1B_1$ have length $r_1$. Then, similar reasoning to that of Step 1 yields $\frac{p_1}{q_1} = \frac{q}{p}$. Using the result of Step 1 then yields $\frac{p_1}{q_1} = \frac{a}{b}$.

Step 3: Reasoning as in the previous Step, we obtain $\frac{q_1}{r_1} = \frac{b}{c}$ and $\frac{r_1}{p_1} = \frac{c}{a}$, and so $\triangle A_2B_2C_2$ is similar to $\triangle ABC$.