\frac{1}{3n^3} \cdot \frac{n^2(n+1)^2}{4} = \frac{(n+1)^Question: A historian studying ancient navigation techniques discovers a triangular chart used by early mariners, with side lengths of $13$, $14$, and $15$ units. What is the length of the shortest altitude in this triangle?

["The Shortest Altitude in a 13–14–15 Triangle: A Historical Insight into Ancient Navigation", "When studying ancient maritime charts, historians often uncover remarkable mathematical precision embedded in early navigational tools. One such intriguing example is a triangle with side lengths $13$, $14$, and $15$—a well-known Heronian triangle that captivates both mathematicians and historians alike. This triangle not only represents a milestone in classical geometry but also offers a compelling case study in calculating key geometric properties such as altitudes—features critical for early sailors estimating distances, angles, and safe passage near land.", "### What Is the Shortest Altitude?", "The altitude (or height) of a triangle is the perpendicular distance from a vertex to the opposite side. In any triangle, the shortest altitude corresponds to the longest side: since area is fixed, the altitude decreases as the base lengthens. For the 13–14–15 triangle, the longest side is $15$, so the altitude to this side will be the shortest.", "### Step 1: Compute the Area Using Heron’s Formula", "Let the sides be $a = 13$, $b = 14$, and $c = 15$. First, compute the semi-perimeter $s$:", "$$\ns = \frac{a + b + c}{2} = \frac{13 + 14 + 15}{2} = 21\n$$", "Now apply Heron’s formula for area $A$:", "$$\nA = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21 \cdot 8 \cdot 7 \cdot 6}\n$$", "Simplify:", "$$\nA = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = \sqrt{7056}\n$$", "Note: $7056 = 84^2$, so:", "$$\nA = 84\n$$", "### Step 2: Use Area to Find the Shortest (Altitude to Longest Side)", "Let $h$ be the altitude to the side of length $15$. By definition of area:", "$$\nA = \frac{1}{2} \cdot \ ext{base} \cdot \ ext{height} \Rightarrow 84 = \frac{1}{2} \cdot 15 \cdot h\n$$", "Solve for $h$:", "$$\n84 = \frac{15h}{2} \Rightarrow 168 = 15h \Rightarrow h = \frac{168}{15} = \frac{56}{5} = 11.2\n$$", "### Conclusion", "The historian’s discovery reveals more than just ancient geometry—it highlights how early navigators could use precise geometric knowledge to interpret triangular maritime charts. The shortest altitude in the 13–14–15 triangle, crucial for determining safe distances and orientations, measures $\frac{56}{5}$ units—an elegant solution rooted in classical mathematics.", "This blend of history, geometry, and practical navigation reminds us that ancient mariners were likely sophisticated in their understanding of spatial relationships—insights that enrich our appreciation of both historical evidence and mathematical heritage.", "---", "Key takeaway: The shortest altitude in a triangle with sides $13$, $14$, and $15$ is $\frac{56}{5}$. This value, derived elegantly through Heron’s formula and area comparison, connects ancient navigation to timeless mathematical principles."]