An altitude $AH$ of triangle $ABC$ bisects a median $BM$. Prove that the medians of triangle $ABM$ are sidelengths of a right-angled triangle. by Yu.Blinkov
Problem
Source: Sharygin Geometry Olympiad 2016 Final Round problem 1 grade 8
Tags: geometry, Medians, right triangle, altitude
RagvaloD
22.07.2018 21:17
Let medians of $ABM$ are $AD,ME,BF$ and $G$ is centroid. $ME\parallel BC$ so $\angle MGD=90$. As $MD=BD$ then $\triangle MGD = \triangle BHD$. So $GH=2GD=AG$ and so $\triangle AGM= \triangle GHB \to GB=AM$ $FB^2=\frac{9}{4} GB^2=\frac{9}{4} AM^2=\frac{9}{4}(AG^2+MG^2)=\frac{9}{4}( \frac{4}{9}AD^2+\frac{4}{9}ME^2)=AD^2+ME^2$
pinkpig
16.07.2022 00:39
Lemma: In triangle $ABC$, the medians from vertices $A$ and $B$ are perpendicular if and only if $a^2 + b^2 = 5c^2$ ($AB=c,BC=a,AC=b$).
Proof: If $a^2+b^2=5c^2,$ then $$(\overrightarrow{B}-\overrightarrow{C})\cdot(\overrightarrow{B}-\overrightarrow{C})+(\overrightarrow{A}-\overrightarrow{C})\cdot(\overrightarrow{A}-\overrightarrow{C})=5(\overrightarrow{A}-\overrightarrow{B})\cdot(\overrightarrow{A}-\overrightarrow{B})\Leftrightarrow$$$$|\overrightarrow{B}|^2+|\overrightarrow{C}|^2-2\overrightarrow{B}\cdot\overrightarrow{C}+|\overrightarrow{A}|^2+|\overrightarrow{C}|^2-2\overrightarrow{A}\cdot\overrightarrow{C}=5|\overrightarrow{A}|^2+5|\overrightarrow{B}|^2-10\overrightarrow{A}\cdot\overrightarrow{B}\Leftrightarrow$$$$2|\overrightarrow{C}|^2-2\overrightarrow{B}\cdot\overrightarrow{C}-2\overrightarrow{A}\cdot\overrightarrow{C}+10\overrightarrow{A}\cdot\overrightarrow{B}=4|\overrightarrow{A}|^2+4|\overrightarrow{B}|^2\Leftrightarrow$$$$|\overrightarrow{C}|^2-\overrightarrow{B}\cdot\overrightarrow{C}-\overrightarrow{A}\cdot\overrightarrow{C}+5\overrightarrow{A}\cdot\overrightarrow{B}-2|\overrightarrow{A}|^2-2|\overrightarrow{B}|^2=0\Leftrightarrow$$$$(2\overrightarrow{A}-\overrightarrow{B}-\overrightarrow{C})\cdot(2\overrightarrow{B}-\overrightarrow{A}-\overrightarrow{C})=0\Leftrightarrow$$$$\left(\overrightarrow{A}-\frac{\overrightarrow{B}+\overrightarrow{C}}{2}\right)\cdot\left(\overrightarrow{B}-\frac{\overrightarrow{A}+\overrightarrow{C}}{2}\right)=0.$$The last equality is dot product of the $A$ and $B$ medians. Since the condition $a^2+b^2=5c^2$ is the same as the $A$ and $B$ medians being perpendicular, we conclude that the $A$ and $B$ medians are perpendicular in triangle $ABC$ if and only if $a^2+b^2=5c^2.$
Now, we will use complex numbers to proceed. Note that the midpoint of $BM$ is denoted by $\frac{2b+a+c}{4}.$ This point lies on the $A$-altitiude, which means that $$\frac{a-\frac{2b+a+c}{4}}{b-c}$$is pure imaginary, or equivalently, $$\frac{3a-2b-c}{b-c}$$is imaginary. Now, we claim that the $A$-median and $M$-median (of triangle $ABM$) are perpendicular. This is equivalent to showing that $$\frac{\frac{a+c}{2}-\frac{a+b}{2}}{a-\frac{\frac{a+c}{2}+b}{2}}=2\cdot\frac{b-c}{3a-2b-c}$$is pure imaginary. However, this follows directly from the fact that $$\frac{3a-2b-c}{2}$$is pure imaginary. We may conclude that the $A$-median and $M$- median of triangle $ABM$ are perpendicular. Now, from the Lemma, this implies that $$BM^2+BA^2=5AM^2.$$In order to prove that the medians form the sides of a right triangle, it suffices to prove that $$2(AM^2+BM^2)-(AB)^2+2(AM^2+AB^2)-(BM)^2=(BM^2+BA^2)-(AM)^2,$$which is equivalent from the result from the Lemma. We conclude that the medians of triangle $ABM$ form a right triangle, as desired.
Entrepreneur
03.09.2023 23:06
Let $P$, $Q$ & $N$ be the midpoints of $BM$, $AM$ & $AB$ respectively. Use ratio Lemma to obtain, $$\frac{BP}{PM}=\frac{\sin \angle BAP}{\sin \angle PAM}\cdot \frac{AB}{AM}$$$$\implies 1=\frac{\cos B}{\cos C}\cdot \frac{2c}{b}$$Use Law of Cosines and simplify further to obtain $$a^2+3c^2=3b^2 $$ Also, by Appolonius Theorem, we have $$BM^2 =\frac{2c^2+2a^2-b^2}{4}$$$$=\frac{3c^2+5a^2}{12}$$ Again, by Appolonius Theorem, $$AP^2=\frac{2AB^2+2AM^2-BM^2}{4}$$$$=\frac{9c^2-a^2}{16}$$Similarly, we obtain, $$BQ^2=\frac{9c^2+3a^2}{16}$$$$MN^2=\frac{a^2}{4}$$Now, it is easy to notice that $$AP^2+MN^2=BQ^2$$Hence, $AP, MN$ & $BQ$ are the sides of a right-angled triangle. $QED.$
matinyousefi
10.04.2024 17:24
Just reflect $G$ the centroid of $ABM$ wrt midpoint of $BM$, let it be called $G'$. Note that $MGG'$ is a triangle similar to the triangle with medians of $ABM$ (medians cut each other with $1:2$ ratio.) and $GG'M=90$ ($MG\|BC$). Hence, $MGG'$ is a right triangle and the conclusion follows.