# The triangle must be obtuse if two of its medians are perpendicular?

In problem #577679, the question says if two of the medians of triangle ABC are perpendicular, then .... ($5a^2 = b^2 + c^2$, the result).

In the course of solving it, I drew the following picture (slightly different to that question).

I found that angle ABC 'must be obtuse' before I can get a satisfactory (hand) drawing. (I did not use any tool from Wolfram.)

My statement is "Prove or disprove the triangle must be obtuse under the mentioned given".

I tried to use the cosine law to show that $cos ABC$ (or $a^2 + c^2 - b^2$) is negative, but was not successful.

Any idea?

• Consider the case where $\triangle ABC$ is isosceles with base $\overline{BC}$.
– Blue
Feb 2, 2014 at 20:44
• A bit expanding on the idea by Blue $\angle BMC$ is also a right angle. Feb 2, 2014 at 21:29

## 3 Answers

(This old question came up again through an edit by @JonMarkPerry.)

Let's find all acute triangles with two perpendicular medians. We may assume the following situation (see the figure): $$A=(0,0), \quad B=(c,0),\quad C=(u,v), \quad M_c=\left({c\over2},0\right), \quad M_b=\left({u\over2},{v\over2}\right)$$ with $v>0$. We want that $$BM_b\perp CM_c\ ,\tag{1}$$ which means that $$\overrightarrow{BM_b}\cdot\overrightarrow{CM_c}=\left({u\over2}-c,{v\over2}\right)\cdot\left({c\over2}-u,-v\right)=0$$ or $$\left(u-{5\over4}c\right)^2+ v^2={9\over16}c^2\ .$$ It follows that the points $C$ satisfying $(1)$ are lying on the circle $K$ with center $\bigl({5\over4}c,0\bigr)$ and radius ${3\over4}c$. The triangle $ABC$ is acute iff $C$ lies outside of the Thales circle over $AB$ and to the left of the line $u=c$. This part of $K$ is drawn in bold red in the figure.

• Maybe because this is an old post, my memory about that is not that active. Hope that you can say a few more on the conclusion you are trying to make.
– Mick
Aug 8, 2015 at 15:09
• Could there be typos? I don’t understand (1) why do I need “The triangle $AM_cM_b$ is acute”; and (2) if $M_c$ is an endpoint of the Thales circle over $AM_c$, how can “$M_c$ lies outside of the Thales circle over $AM_c$. I think you are suggesting that the locus of $M_b$ is on the bolded arc and $C$ should be moved to further right such that the triangle thus formed meets the said requirement.
– Mick
Aug 8, 2015 at 18:15

A counterexample: a triangle with a right angle between a and c, and c=a*sqrt(2), b=a*sqrt(3)

I found this by expressing the medians as vector sums of a and c, and requiring that their dot product be zero. This leads to the condition Cos(<(a,c)) = (c^2 - 2a^2)/ac. This is negative only if c

Hope this answers your question.

• I should have the scope broaden by asking "one of the angle must be greater than or equal to 90 degrees" such that your counter-example(s) can be excluded. Anyway, as @blue pointed out, a suitable isosceles triangle will serve as a counter-example. Sorry for posting an in-valid finding.
– Mick
Feb 3, 2014 at 5:08

Going down the route of using cosine rule, if all the angles are less than $90$ then $c^2+a^2>b^2, c^2+b^2>a^2, b^2+a^2>c^2$. Additionally, $c^2+b^2=5a^2$. Hence one gets the inequalities $(6b^2)/5>(4c^2)/5/>(8b^2)/15$ and a similar one where $b$ and $c$ are interchanged. As an example, with $b=1, c=0.9, a=\frac{\sqrt{905}}{50}$, all the inequalities, including the triangle inequality are satisfied, providing an example of medians perpendicular to each other without being obtuse.

• Why the downvote? Mar 4, 2018 at 13:58
• I did not downvote your answer. I verified that your counter-example is reasonable. I just don't understand how the inequalities are setup.
– Mick
Mar 4, 2018 at 16:25
• Sorry I was asking the person who downvoted. I've edited it so it becomes a little clearer. If all the angles are acute, then the cosines are positive, and by the cosine rule the only term that can give a negative part is the numerator. The three inequalities are stating that each of the cosines are positive, so the angles are acute Mar 4, 2018 at 20:44
• Like to elaborate a bit more on "$(6b^2)/5>(4c^2)/5/>(8b^2)/15$"?
– Mick
Mar 5, 2018 at 10:40
• Substitue for a^2=(c^2+b^2)/5 in the inequalities of the sides Mar 5, 2018 at 18:14