🧭 Problem (Right Angle - Equidistance point on hypotenous )

In \(\triangle ABC\), \(\angle ABC = 90^\circ\) and \(BA = BC = \sqrt{2}\). Points \(P_1, P_2, \dots, P_{2024}\) lie on hypotenuse \(\overline{AC}\) so that

What is the length of the vector sum

✏️ Amiya Sir’s Cheat Code

Refer his Note:

First for 1 cevian, then for 3, then for \(n\) equidistant cevians – and he wrote the rule:

In our AMC problem, the triangle is right isosceles with legs \(\sqrt{2}\), so \(|BC| = \sqrt{2}\), but more importantly the hypotenuse length turns out to be 2, and the pairing will give exactly that 2 every time. So if we use this concept anser of out question is \[ \frac{2024}{2}*2 = 2024, \]

General rule (from Amiya Sir’s note)

For an isosceles right triangle with hypotenuse \(AC\) and equally spaced points on \(AC\):

• Number of equal parts = \(n+1\) ⇒ points = \(n\)
• Cevian sum from the right angle = \(\dfrac{n}{2} \times \text{(pair vector)}\) if \(n\) is even
• If \(n\) is odd, there is one middle cevian that is already along the diagonal, so add it once more

That’s exactly what is written in your screenshot: “for \(n\) equidistant cevians” → use half if even, or \((n+1)/2\) if odd.

📝 Quick Quiz

Same triangle, but now suppose we take only 4 equally spaced points on \(AC\) (so \(P_1, P_2, P_3, P_4\)). What is \(\bigl|\vec{BP_1} + \vec{BP_2} + \vec{BP_3} + \vec{BP_4}\bigr|\)?

🧪 Extra Practice

If instead of 2024 points we had 2025 equally spaced points on \(AC\), what would the sum be?