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Find angles in right triangle if it is known that $\tan \theta = {3\over4}$, where $\theta$ is the angle between catheti medians. I have tried drawing orthogonal projections and making new right triangles with that same angle but it did not seem to lead anywhere, as well as trying to find something that could be useful for sine theorem, but again got nothing good.

Have tried to make new right triangles from centroid, but did not get anything useful.

Any help would be much appreciated. As it was a problem on a competition, calculators were not allowed

enter image description here

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    $\begingroup$ Have you tried using vectors? You should get the angles in the main triangle as $45^o$ $\endgroup$ – David Quinn Dec 30 '16 at 0:00
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I wish if you had named the objects in the picture, but, this is my opinion anyway.

In the picture, there are two edges that make the right angle. Call the horizontal one $B$ and the vertical one $A$. The angle of the main triangle, that faces the edge $A$ is called $\alpha$. The median partitions $\alpha$ into two angles. The lower angle, I call it $\alpha_1$.

$tan(\alpha_1)=\frac{A}{2B}$

Also, we have an angle, with its vertex at the median point on $B$, that faces the edge $A$. The angle is nothing but $\theta+\alpha_1$. Therefore

$tan(\theta+\alpha_1)=\frac{A}{B/2}=\frac{2A}{B}$

Now, we may use a famous formula to write $tan(\theta+\alpha_1)$ in a different way.

$tan(\theta+\alpha_1)=\frac{tan(\theta)+tan(\alpha_1)}{1-tan(\theta)tan(\alpha_1)}$

Plugging in the value of $tan(\theta)$ and what we found for $tan(\alpha_1)$,we finally get

$tan(\theta+\alpha_1)=\frac{2\times (2A+3B)}{8B-3A}$

Put the same quantities equal to each other

$\frac{2A+3B}{8B-3A}=\frac{A}{B}$

A bit of simplification gives

$A^2+B^2-2AB=0$

I Think the rest is easy.

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Have a look at the graphics below.

enter image description here

Using formula $\cos^2\theta=\frac{1}{1+\tan^2\theta}$, the condition on $\tan \theta$ is equivalent to

$$\tag{1}\cos(\theta)=4/5$$

Let $O$ be the right angle vertex, $A$ the rightmost vertex, and $B$ the upper vertex of the right triangle.

Let $A'$ be the midpoint of [OB] and $B'$ the midpoint of [OA].

Let $G$ denotes the intersection of medians (center of mass).

Take OA and OB as the $x$ and $y$ axes of an orthonormal coordinate system where we can assume WLOG $A(1,0)$ and $B(0,b)$. Thus the coordinates of $G$ are $(1/3, b/3)$.

Thus, the coordinates of $\vec{GB'}$ are $\binom{1/6}{-b/3}$ and the coordinates of $\vec{GA}$ are $\binom{2/3}{-b/3}$.

The following dot product can be expressed in two ways:

$$\vec{GB'}.\vec{GA}=\begin{cases}\frac16\frac23+\frac{-b}{3}\frac{-b}{3}=\frac{1}{9}+\frac{b^2}{9}\\ \sqrt{\frac{1}{36}+\frac{b^2}{9}}\sqrt{\frac{4}{9}+\frac{b^2}{9}} \ cos(\theta)\end{cases}$$

Equating the squares of the 2 expressions above, using $(1)$, then multiplying both sides by $9^2$, gives, after some simplifications: $b^4-2 b^2+1=0$ which evidently has the unique positive solution $b=1$.

Conclusion: triangle $AOB$ is an isosceles right triangle.

Remark: the last computation is equivalent to an application of cosine law to triangle GB'A.

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enter image description here

The length of one median is $\sqrt{a^2 + 4b^2}$ and the other is $\sqrt{4a^2 + b^2}$

The intersection of the medians split each other with a ratio of $2:1$

Law of sines

$\frac {\sin x}{a} = \frac {\sin y}{\frac 13 \sqrt{a^2 + 4b^2}}$

$\sin x = \frac 35$

$\sin y = \frac {\sqrt{a^2 + 4b^2}}{5a}$

Aslo $\sin y = \frac b{\sqrt{4a^2 + b^2}}$

$\frac {\sqrt{a^2 + 4b^2}}{5a} = \frac b{\sqrt{4a^2 + b^2}}\\ 5ab = \sqrt {4a^4 + 17a^2b^2 + 4b^4}\\ 25a^2b^2 = 4a^4 + 17a^2b^2 + 4b^4\\ 4a^4 - 8a^2b^2 + b^4) = 0\\ 4(a^2 - b^2)^2 = 0$

$a = b$

you have an isoceles right triangle.

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