Consider all the triangles $ABC$ which have a fixed base $AB$ and whose altitude from $C$ is a constant $h$. For which of these triangles is the product of its altitudes a maximum?
Problem
Source: APMO 1990
Tags: calculus, trigonometry, geometry, geometry unsolved
Davron
12.03.2006 07:42
This problem can be tackled using calculus but the following approach is more direct. Let $h_a$ and $h_b$ be the altitudes from $A$ and $B$, respectively. Then $AB\cdot h\cdot AC\cdot h_b\cdot BC\cdot h_a=8[ABC]^2=(AB\cdot h)^3$ which is a constant. So the product $h_a\cdot h_b\cdot h$ attains its maximum when the product $AC\cdot BC $ attains minimum. Since $AC\cdot BC\cdot \sinh = BC\cdot h_a=2[ABC]$, which is constant, $AC\cdot BC$ attains its minimum when $\sin C$ reaches its maximum. There are two cases: 1) $h\le\frac{AB}{2}$. Then there exists a triangle $ABC$ which has a right angle at $C$, and for precisely such a triangle $\sin C$ attains its maximum namely $1$. 2) $h>\frac{AB}{2}$. In this case the angle at $C$ is acute and assume its maximum when $AC=BC$.
BOGTRO
17.03.2013 19:01
Let the altitude from C meet AB at D, and let AD=a, BD=b, so that $a+b$ is constant. Then,
$AC=\sqrt{a^2+h^2}, BC=\sqrt{b^2+h^2}$
and so the product of the altitudes (noting that the area is constant) is
$h(\frac{(a+b)h}{\sqrt{a^2+h^2}})(\frac{(a+b)h}{\sqrt{b^2+h^2}})=\frac{(a+b)^2h^3}{\sqrt{(a^2+h^2)(b^2+h^2)}}$.
As $(a+b)^2h^3$ is constant, maximizing this is equivalent to minimizing
$(a^2+h^2)(b^2+h^2)=a^2b^2+h^2(a^2+b^2)+h^4=a^2b^2-2h^2ab+h^2(a+b)^2+h^4$.
As $h^2(a+b)^2+h^4$ is fixed, this is equivalent to minimizing
$a^2b^2-2h^2ab=(ab-h^2)^2-h^4$, which is equivalent to minimizing $(ab-h^2)^2$.
Clearly $(ab-h^2)^2$ attains its minimum at 0, at which point $ab=h^2$. Note that this is possible iff $\frac{a+b}{2} \geq \sqrt{ab} \implies a+b \geq 2h$.
Furthermore, this implies that D lies in a circle on diameter AB going through C by Power of a Point. In particular, this means that ABC is right.
If $a+b<2h$, then we must maximize $ab$ given $a+b$, or $a=b$ implying that ABC is isosceles. We can write this more concisely as
$h \leq \frac{AB}{2} \implies \triangle ABC~\text{right}, h>\frac{AB}{2} \implies \triangle ABC~\text{isosceles with AC=BC}$.
TheDarkPrince
14.10.2018 09:28
Product of altitudes $= c^2\sin A\sin B$, thus we need to maximise $\sin A\sin B$. We have $\frac{abc}{4R} = \frac{1}{2}\cdot c\cdot h = $ constant say $k$. Thus $\sin A\sin B = \frac{k}{c\cdot R}$ $\implies$ we need to maximise $\frac{1}{R}$ or minimise $R$. [asy][asy] import olympiad; import geometry; unitsize(3cm); pair C = dir(130), B = dir(210), A = -1/B; triangle tri = triangle((point) A, (point) B, (point) C); label("A", A, A); label("B", B, B*dir(10)); label("$C'$",C,C); draw(tri); pair M = midpoint(A--B); dot(M); pair D = A+C-B; pair E = foot(C,A,B); pair F = A+C-E; pair G = 2F-D; pair C_1 = (G+C)/2; dot(C_1); pair D_1 = 2C-C_1; draw(C_1--M,red);draw(D--D_1,blue); label("$C$",C_1,C_1); label("M",M,M); draw(A--C_1);draw(B--C_1); pair O = circumcenter(C_1,A,B); dot(O);label("O",O,O*dir(90)); pair O_1 = circumcenter(C,A,B); dot(O);label("$O'$",O_1,O_1*dir(-90)); dot(O_1); [/asy][/asy] Let $C$ be such that $AC = BC$ and $C'$ be a point on line parallel to $AB$ through $C$ and $C'\not\equiv C$. Let $O$ be the circumcenter of $ABC$ and $O'$ be the circumcenter of $ABC'$. If $CO<CO'$ then $CO<C'O'$ and $AO>AO'$. Thus $CO<CO'=AO'<AO=CO$ which is not possible. Thus we have that $AO<AO'$ for all $C'$ which gives that smallest circumradius is when $AC = BC$.
joshualiu315
30.04.2024 22:19
Sketch: Do some manipulations to get that we wish to minimize $R$. Then, we can prove that $\triangle ABC$ is isosceles by varying $C$ to another point $C'$ and proving that the latter has a smaller circumradius.