Let $ABC$ be an acute-angled triangle and variable $D \in [BC]$ . Let's denote by $E, F$ the feet of the perpendiculars from $D$ to $AB$, $AC$ respectively . a) Show that $$\frac{4S^2}{b^2+c^2}\le DE^2 + DF^2\le max \{h_B^2 + h_C^2 \}.$$b) Proved that, if $D_0 \in [BC]$ is the point where the minimum of the sum $DE^2 + DF^2$ is achieved, then $D_0$ is the leg of the symmetrical median of $A$ facing the bisector of angle $A$. c) Specify the position, of $D \in [BC]$ for which the maximum of the sum $DE^2 + DF^2$ is achieved. (The area of the triangle $ABC$ was denoted by $S$ and $h_b, h_c$ are the lengths of the altitudes from $B$ and $C$ respectively)
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Tags: geometry, geometric inequality
rms
22.08.2024 13:06
parmenides51 wrote: Let $ABC$ be an acute-angled triangle and variable $D \in [BC]$ . Let's denote by $E, F$ the feet of the perpendiculars from $D$ to $AB$, $AC$ respectively . a) Show that $$\frac{4S^2}{b^2+c^2}\le DE^2 + DF^2\le max \{h_B^2 + h_C^2 \}.$$b) Proved that, if $D_0 \in [BC]$ is the point where the minimum of the sum $DE^2 + DF^2$ is achieved, then $D_0$ is the leg of the symmetrical median of $A$ facing the bisector of angle $A$. c) Specify the position, of $D \in [BC]$ for which the maximum of the sum $DE^2 + DF^2$ is achieved. (The area of the triangle $ABC$ was denoted by $S$ and $h_b, h_c$ are the lengths of the altitudes from $B$ and $C$ respectively)
**Problem 3.**
a). Let \( S_1 \) and \( S_2 \) be the areas of triangles \( ADB \) and \( ADC \). Since \( S = S_1 + S_2 \), it follows that \( 2S = cDE + bDF \), so \( 4S^2 = (cDE + bDF)^2 \leq (b^2 + c^2)(DE^2 + DF^2) \), from which we obtain the first inequality stated in the problem.
Let \( x = BD \), \( x \in [0, a] \). Since \( \frac{DE}{c} = \frac{x}{a} \) and \( \frac{DF}{b} = \frac{a-x}{a} \), it follows that
\[
DE^2 + DF^2 = \frac{1}{a^2}\left[ h_c^2x^2 + h_b^2(a-x)^2 \right] \leq \max\{h_b^2, h_c^2\} \cdot \frac{2x^2 - 2ax + a^2}{a^2} \leq \max\{h_b^2, h_c^2\}
\]for \( \frac{2x^2 - 2ax + a^2}{a^2} \leq 1 \).
b). Let \( M \) be the midpoint of \( [BC] \) and \( T \) the foot of the altitude from \( A \). From Steiner's theorem, it follows that
\[
\frac{BT \cdot BM}{CT \cdot CM} = \frac{c^2}{b^2}, \text{ so } \frac{BT}{CT} = \frac{c}{b}.
\]If \( E \) and \( F \) are the feet of the perpendiculars from \( T \) to \( AB \) and \( AC \) respectively, then from the relation obtained in 1, we deduce:
\[
TE^2 + TF^2 = \frac{1}{a^2}\left[ h_b^2 \cdot CT^2 + h_c^2 \cdot BT^2 \right] = \frac{1}{a^2}\left[ \frac{h_b^2 \cdot a^2b^4}{(b^2 + c^2)^2} + \frac{h_c^2 \cdot a^2c^4}{(b^2 + c^2)^2} \right] = \frac{b^2(h_b^2) + c^2(h_c^2)}{(b^2 + c^2)^2} = \frac{4S^2}{b^2 + c^2}.
\]From 1, it follows that \( \frac{4S^2}{b^2 + c^2} \) is the minimum expression for \( DE^2 + DF^2 \) when \( D \in [BC] \) and it is attained when \( D = T \). It remains to show that \( T \) is the only point where the minimum is attained. Let \( P \in [BC] \) such that
\[
PE^2 + PF^2 = \frac{4S^2}{b^2 + c^2}.
\]Since \( 4S^2 = (cPE + bPF)^2 = (b^2 + c^2)(PE^2 + PF^2) \), we get
\[
\frac{PE}{PF} = \frac{c}{b}, \text{ so } \frac{BP}{CP} = \frac{BP}{CP} = \frac{a}{c} \cdot \frac{PE}{PF} = \frac{h_b}{h_c} = \frac{b \cdot h_b}{c \cdot h_c} = \frac{c^2}{b^2}.
\]Since \( T \) and \( P \in [BC] \), it follows that \( T = P \).
c). Let's assume \( h_B \leq h_C \). Then, if \( E \) is the foot of the perpendicular from \( h_C \), we have:
\[ CE^2 + CC^2 = CE^2 = h_C^2 = \max \{h_B^2, h_C^2\} \]Thus, \( h_C^2 \) is the maximum of the expression \( DE^2 + DF^2 \) when \( D \) belongs to \( [BC] \) and is reached when \( D = C \). Let \( P \) be a point where the maximum is reached. Then, from 1, it follows that:
\[
\frac{2x^2 - 2ax + a^2}{a^2} = 1
\]where \( x = BP \). It follows that \( x(x - a) = 0 \), so \( P = B \) or \( P = C \). Since for \( P = B \) we have:
\[ BE^2 + BF^2 = h_B^2 \leq h_C^2 \]we obtain:
i) If \( h_B < h_C \), the only point where the maximum is reached is \( C \);
ii) If \( h_B = h_C \), the maximum is reached for both \( B \) and \( C \);
iii) If \( h_B > h_C \), the maximum is reached only when \( D = B \).