Do You Truly Know Right Triangles?

By the Thales theorem a right triangle is defined by a circle with diameter as the hypotenuse of the right triangle. If you draw a point through one of the arcs and make a triangle, you will get a right angle. Look the following image:

If the diameter is 12, the highest height that this triangle can have by this definition is the radius of the circle forming an isosceles triangle, which is 6.

Thus, a right triangle with hypotenuse measuring 12 and an altitude to the hypotenuse measuring 8 doesn't exist.

We know that: $m\cdot n={ h }^{ 2 }$ , where $m+n$ is the hypotenuse and $h$ is the altitude to the hypotenuse. Also, we have $m+n=12$ and $h=8$ . So, $m(12-m)={ h }^{ 2 }$

Solving for $m$ : $m(12-m)=8^{ 2 }$

${ m }^{ 2 }-12m+64=0\\ { (m-6) }^{ 2 }=-28$

Thus, we have an imaginary distance.

The relation that I applied is also known as geometric mean theorem or right triangle altitude theorem. I will explain the proof: Let $\angle C$ be the right angle of $\triangle ACB$ . Let $h$ be the altitude to the hypotenuse. Note that, in the image, $\measuredangle ACH+\measuredangle BCH=90°$ and $\measuredangle ABC+\measuredangle CAB=90°$ . As $\measuredangle CHB=90°=\measuredangle CHA$ , we have that $\measuredangle BCH+\measuredangle ABC=90°$ and, thus, $\measuredangle CAB=\measuredangle BCH$ . Similarly, we infer that $\measuredangle ACH=\measuredangle CBA$ . Hence, $\triangle ACH\sim \triangle CBH$ . Therefore, we have the following equation: $\frac { AH }{ CH } =\frac { CH }{ HB }$ which is the same that: $\frac { m }{ h } =\frac { h }{ n }$ , so ${ h }^{ 2 }=mn$ . The converse is also true.

https://brilliant.org/problems/easy-and-i-know-it/?group=eeQPEItCpxwU

Same plot as that of my question.