Find C's Coordinates for Right Triangle ABC

[Wilmer]

Right triangle ABC.
BC = a = sqrt(17)
AC = b = sqrt(68)
AB = c = sqrt(85)
A's coordinates: 0,12
B's coordinates: 6,5

What's EASIEST way to get C's coordinates?

 

[Olinguito]

AB is clearly the hypotenuse of the right triangle, so C will lie on a circle with AB as diameter. There will be two possible answers for C, one on either side of AB. You just have to find the two points on the circle that yield the given side lengths for the triangle.

 

[Wilmer]

Agree.
And the 2 solutions for C are (2,4) and (38/5,44/5).

BUT I was trying to solve without involving a circle.
(guess I should have said so!)
Can you tell me it's definitely impossible?

 

[I like Serena]

Right triangle ABC.
BC = a = sqrt(17)
AC = b = sqrt(68)
AB = c = sqrt(85)
A's coordinates: 0,12
B's coordinates: 6,5

What's EASIEST way to get C's coordinates?

Let's make a drawing:
\begin{tikzpicture}[>=stealth']
%preamble \usetikzlibrary{arrows}
\def\a{sqrt(17)}
\def\b{sqrt(68)}
\def\c{sqrt(85)}
\coordinate (A) at (0,12);
\coordinate (B) at (6,5);
\coordinate (C1) at (2,4);
\coordinate (C2) at ({38/5},{44/5});
\coordinate (D) at ({\b*\b/\c/\c * 6}, {12+\b*\b/\c/\c * (5-12)});
\coordinate (E) at ({\b*\b/\c/\c * 6 + \a*\b/\c/\c * (12-5)}, {12+\b*\b/\c/\c * (5-12) + \a*\b/\c/\c * 6});
\draw (A) -- (B) -- (C1) -- cycle;
\draw (A) -- (B) -- (C2) -- cycle;
\draw[->, thick] (A) -- node[right, xshift=6] {$b\cos\alpha\cdot\frac{\overrightarrow{AB}}{c}$} (D);
\draw[->, thick] (D) -- node[above, xshift=-18] {$b\sin\alpha\cdot\frac{\overrightarrow{AB}^\perp}{c}$} (E);
\path (A) node[above left] {A} -- (B) node[below right] {B} -- (C2) node[above right] {C} -- (C1) node[below left] {C'};
\path (A) -- node[below left] {c} (B) -- node

{a} (C2) -- node[above] {b} (A) -- (D) node[below left] {D};
\path (A) node[below right, xshift=10, yshift=-6] {$\alpha$};
\end{tikzpicture}


It follows that:
$$\overrightarrow C = \overrightarrow A + b\cos\alpha\cdot\frac{\overrightarrow{AB}}{c} \pm b\sin\alpha\cdot \frac{\overrightarrow{AB^\perp}}{c}
= \overrightarrow A + b\cdot \frac bc \cdot\frac{\overrightarrow{AB}}{c} \pm b\cdot \frac ac\cdot \frac{\overrightarrow{AB^\perp}}{c} \\
= \binom{0}{12} + \frac {68}{85} \cdot\binom{6-0}{5-12} \pm \frac {\sqrt{17\cdot 68}}{85}\cdot\binom{12-5}{6-0}
$$​

 

[Wilmer]

Nice. How about "without using trigonometry"? :)

 

[I like Serena]

Nice. How about "without using trigonometry"? :)

Trigonometry is only used in the first step.
We can replace it with non-trigonometry by observing that triangles ABC and ACD are similar, meaning that they scale to each other.
(Just added point D to my previous drawing for reference. ;))

 

[Wilmer]

Ah so! Thanks loads.

 

[I like Serena]

Ah so! Thanks loads.

To be honest, I only used trigonometry because it means I have to use fewer words to explain those ratios! (Rofl)

 

[Wilmer]

I'm also lazy!

 

[tkhunny]

While you're at it, how about trying it without words, expressions, or symbols?

 

[I like Serena]

While you're at it, how about trying it without words, expressions, or symbols?

You mean just drawings? (Wondering)

Something like this:
\begin{tikzpicture}[>=stealth']
%preamble \usetikzlibrary{arrows}
\def\a{sqrt(17)}
\def\b{sqrt(68)}
\def\c{sqrt(85)}
\coordinate (A) at (0,12);
\coordinate (B) at (6,5);
\coordinate (C1) at (2,4);
\coordinate (C2) at ({38/5},{44/5});
\coordinate (D) at ({\b*\b/\c/\c * 6}, {12+\b*\b/\c/\c * (5-12)});
\coordinate (E) at ({\b*\b/\c/\c * 6 + \a*\b/\c/\c * (12-5)}, {12+\b*\b/\c/\c * (5-12) + \a*\b/\c/\c * 6});
\draw (A) -- (B) -- (C1) -- cycle;
\draw (A) -- (B) -- (C2) -- cycle;
\draw[->, thick] (A) -- (D);
\draw[->, thick] (D) -- (E);
\path (A) node[above left] {A} -- (B) node[below right] {B} -- (C2) node[above right] {C} -- (C1) node[below left] {C'};
\path (A) -- node[below left] {c} (B) -- node

{a} (C2) -- node[above] {b} (A);
\end{tikzpicture}

Admittedly there are still some symbols in there, but only the ones introduced in the OP!
The rest is left as an exercise to the reader. (Devil)​

 

[I like Serena]

For the record, something like that is called a sangaku. A drawing that exhibits a proof, without any words, expressions, or symbols. Admittedly my drawing is not really a sangaku, since it does contain symbols. (Nerd)

 

[Wilmer]

GOOD idea...never thought of that!

 

[tkhunny]

No good. That's just without narrative description. "...except those introduced..." seems an arbitrary addition to the requirement.

How are lines, arrows, and letters NOT symbols? They all mean things.

I just figured if we were eliminating things, we may as well eliminate it all. It's never been a favorite practice in my opinion - arbitrary elimination of reasonable methods. Anyway, exploration always has value.

 

[Monoxdifly]

For the record, something like that is called a sangaku.

Doesn't sangaku simply literally means triangle, though?

 

[Wilmer]

Nope Mr.Fly:
Sangaku translation: calculation tablet) are Japanese geometrical problems or theorems on wooden tablets which were placed as offerings at Shinto shrines or Buddhist temples during the Edo period by members of all social classes

 

[I like Serena]

Doesn't sangaku simply literally means triangle, though?

Google translates triangle as sankakkei.
And searching for sankakkei shows that this word is also used for a three dimensional puzzle of triangles.
View attachment 8074

However, that is different from a sangaku.
View attachment 8075

 

[Wilmer]

YIKES! I feel guilty for starting this thread :)

 

[Monoxdifly]

I also feel guilty for posting an off-hand comment...

 

[Wilmer]

(x-6)^2 + (y-5)^2 = 17
x^2 + (y-12)^2 = 68

Expand:

x^2 - 12x + y^2 - 10y = -44
x^2 - 00x + y^2 - 24y = -76

Ahhh...ain't that cute!
2 solutions:

x=2, y=4 or x=38/5, y=44/5