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Triangle ABC has sides BC = 36, CA = 29 and AB = 25…

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Triangle $ABC$ has sides $BC = 36$ , $CA = 29$ and $AB = 25$ . For a point $P$ inside the triangle or on its sides, let $d_a$ , $d_b$ and $d_c$ be the distances from $P$ to the lines $BC$ , $CA$ and $AB$ , as in the figure. As $P$ moves, the three distances change together, and this problem looks for the extreme values of two sums made from them. A point on a side has distance $0$ to the line of that side. A first guess is that a central point, such as the incentre, gives the extreme values. Check that guess as you work. Each part may use the results of the earlier parts, and every claim about a least or greatest value needs a proof: show that the value cannot be passed, and show a point that reaches it.

Triangle ABC with B at the bottom left, C at the bottom right and A above, nearer to B. The sides are marked BC = 36, CA = 29 and AB = 25. A point P inside the triangle is joined by dashed perpendiculars to the three sides, labelled d_a to BC, d_b to CA and d_c to AB, with right-angle marks at their feet.

Find the area of triangle $ABC$ .
Prove that $36d_a + 29d_b + 25d_c$ has the same value for every such point $P$ , and find that value.
Find the least and the greatest values of $d_a + d_b + d_c$ , and the points $P$ where they occur.
Find the least value of $d_a^2 + d_b^2 + d_c^2$ , and prove that it is reached at a point inside the triangle.
At the point of (d), find $d_a$ .

Give each answer that is not a whole number to two decimal places.

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Splitting the triangle at $P$ gives one linear equation for $d_a$ , $d_b$ , $d_c$ . Comparing coefficients bounds the plain sum, and Cauchy-Schwarz bounds the sum of squares.

(a) $s = 45$ , so Heron's formula gives $\Delta = \sqrt{45 \times 9 \times 16 \times 20} = \sqrt{129\,600} = 360$ .

(b) Joining $P$ to $A$ , $B$ and $C$ cuts the triangle into triangles $PBC$ , $PCA$ and $PAB$ (one of them may be flat when $P$ is on a side). Their areas are $\dfrac12 \cdot 36\,d_a$ , $\dfrac12 \cdot 29\,d_b$ and $\dfrac12 \cdot 25\,d_c$ , and they add up to $360$ . So $36d_a + 29d_b + 25d_c = 720$ for every $P$ .

(c) All three distances are at least $0$ . So $720 = 36d_a + 29d_b + 25d_c \le 36(d_a + d_b + d_c)$ , which gives $d_a + d_b + d_c \ge 20$ . Equality needs $d_b = d_c = 0$ , so $P$ is on both lines $CA$ and $AB$ : $P = A$ , where $d_a$ is the altitude $\dfrac{720}{36} = 20$ . In the same way $720 \ge 25(d_a + d_b + d_c)$ , so the sum is at most $28.8$ , with equality only when $d_a = d_b = 0$ , that is, $P = C$ . The least value $20$ is at the corner $A$ , opposite the longest side, and the greatest value $28.8$ is at the corner $C$ , opposite the shortest side. The incentre gives $3r = 3 \times \dfrac{360}{45} = 24$ , which is neither.

(d) By Cauchy-Schwarz, $(36^2 + 29^2 + 25^2)(d_a^2 + d_b^2 + d_c^2) \ge (36d_a + 29d_b + 25d_c)^2 = 720^2$ . Here $36^2 + 29^2 + 25^2 = 2762$ , so $d_a^2 + d_b^2 + d_c^2 \ge \dfrac{518\,400}{2762} \approx 187.69$ . Equality needs $(d_a, d_b, d_c) = \lambda(36, 29, 25)$ , and (b) then gives $\lambda = \dfrac{720}{2762}$ . Such a point exists inside the triangle. Put $u = \dfrac{36d_a}{720}$ , $v = \dfrac{29d_b}{720}$ and $w = \dfrac{25d_c}{720}$ , three positive numbers that add up to $1$ , and let $P$ be the weighted average $uA + vB + wC$ . Inside the triangle, the distance to the line $BC$ changes at a steady rate along any straight segment, so the distance from a weighted average of points is the same weighted average of their distances. This distance is $20$ at $A$ and $0$ at $B$ and $C$ , so the distance from $P$ to $BC$ is $20u = d_a$ , and in the same way the other two distances are $d_b$ and $d_c$ . The point $P$ is inside the triangle because $u$ , $v$ , $w$ are all positive. (It is the symmedian point of the triangle.)

(e) $d_a = 36\lambda = \dfrac{36 \times 720}{2762} \approx 9.38$ .

Check. With $B = (0, 0)$ , $C = (36, 0)$ and $A = (15, 20)$ , the distances were computed at twenty thousand points of the triangle: (b) held at every point, the sum ran from $20$ to $28.8$ , and the sum of squares never went below the value in (d).

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