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The exercise is to construct a right triangle given $a+b-c$ and $\alpha$.

I know we then have $\beta=90^\circ -\alpha$. I tried to draw the right triangle $\triangle ABC$ and find where I can use $a+b-c$. By adding $a$ to $b$ to get $\overline{DA}, |DA|=a+b$, I get an isosceles triangle $\triangle DCB$ so $|\angle CDB|=45^\circ$. I then subtract the length $c$ from $\overline{DA}$ and have $|DN|=a+b-c$. I thought I could construct the triangle $\triangle DNB$ so I could find $C$ as the intersection of the line $DN$ and the perpendicular bisector of $\overline{DB}$. However, I'm missing an element in order to be able to even construct $\triangle DNB$. Additionally, I tried adding lengths to the hypotenuse, but also without much success.

user562834
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    Huge hint: $a+b-c$ is the diameter of the inscribed circle. – Oscar Lanzi Apr 29 '22 at 23:27
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    Yes, the question has been answered and when I saw that we can use the radius of the inscribed circle, it dawned on me. Unfortunately I hadn't thought of it myself before asking. – user562834 Apr 29 '22 at 23:30

3 Answers3

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I assume that $c$ is the length of the hypotenuse, as usual. If so, we can use that $r=(a+b-c)/2$ is the radius of the inscribed circle to do the following.

Start by constructing a right triangle $\triangle AIQ$ with $\angle QAI=\alpha/2$ and $\overline{IQ}=(a+b-c)/2$. (For instance, trace two parallel lines with distance $(a+b-c)/2$, and then a line with angle $\alpha/2$.) Construct a external square $QIRC$ over the segment $\overline{IQ}$. Now, construct a line passing through $A$ and making an angle $\alpha/2$ with $\overline{AI}$. This line intersects the line $\overline{CR}$ at some point $B$. Clearly $\triangle ABC$ has $I$ as its incenter, which concludes the question.

Situation of the problem

Nicolás Vilches
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  • Thank you. Do you perhaps know a shorter way to show that $\frac{a+b-c}{2}$ is the radius of the inscribed circle for a right triangle other than getting it as $\frac{ab}{a+b+c}$ through the proof itself and then expanding $\frac{ab}{a+b+c}\cdot\frac{a+b-c}{a+b-c}$? – user562834 Apr 29 '22 at 23:22
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    Yes! Let $\triangle ABC$ be any triangle (not necessarily a right triangle), with $D, E, F$ the tangent points of the inscribed circle at $\overline{BC}, \overline{CA}, \overline{AB}$. Then you can compute $\overline{AE}=\overline{AF}=(b+c-a)/2$ and similar for the others (is a $3 \times 3$ system). At last, note that in our situation the right angle gives us a square, so the radius equals one of these segments. – Nicolás Vilches Apr 29 '22 at 23:33
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    @user562834: "Do you perhaps know a shorter way to show [the formula for] the radius of the inscribed circle for a right triangle[?]". See this question, and, perhaps in particular, my answer. – Blue Apr 30 '22 at 00:30
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The sides of the right triangle are $a,b,c$ where $c$ is the hypotenuse, and the angle $\alpha$ is the angle opposite vertex $A$. Therefore,

$ a = c \sin(\alpha) $

$ b = c \cos(\alpha) $

We are given the value of $a + b - c $. Let $X$ be that value, then

$ c( \sin(\alpha) + \cos(\alpha) - 1 ) = X $

Hence,

$ c = \dfrac{X}{\sin(\alpha) + \cos(\alpha) - 1} $

Now the sides are all known.

Hosam Hajeer
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Alternative approach.

It is assumed that $c$ is the hypotenuse, that $a$ is the side opposite the angle $\alpha$, and that the value $(a+b-c)$ equals a given fixed value $T$.

Then, you have the following two equations in the two unknowns, $a,b$.

  • $c^2 = (a + b - T)^2 = a^2 + b^2.$
  • $\displaystyle \frac{a}{b} = \tan(\alpha)$.

    Here, I am assuming that since $\alpha$ is known, that $\tan(\alpha)$ is known.

Once $a,b$ are solved, then $c$ is immediately solved by $a + b - c = T.$ Therefore, the problem reduces to using the above two equations to solve for $a,b$.

Using the first equation above, you have that

$$T^2 + 2ab = 2T(a+b). \tag1 $$

Let $R = \tan(\alpha).$
This implies that

$$a = bR \tag2 .$$

Using (1) and (2) above, you have that

$$T^2 + 2Rb^2 = 2Tb(R+1) \implies$$

$$b^2(2R) - b(2T[R+1]) + T^2 = 0. \tag3 $$

(3) above represents a quadratic in $b$. Therefore,

$$b = \frac{1}{4R}\left[2T(R+1) \pm \sqrt{[2T(R+1)]^2 - 8RT^2}\right]. \tag4 $$

(4) above simplifies to

$$b = \frac{T}{2R}\left[(R+1) \pm \sqrt{R^2 + 1}\right]. \tag5 $$

Temporarily, both roots in (5) will be carried, until $c$ is evaluated. Since $c$ must be $> 0$, one of the roots will then be eliminated.

$$a = \frac{T}{2R}\left[(R+1) \pm \sqrt{R^2 + 1}\right] \times R. \tag6 $$

Using that $c = a + b - T$, this implies that

$$c = \frac{T}{2R}\left[(R^2+1) \pm (R+1)\sqrt{R^2 + 1}\right]. \tag7 $$

At this point, since $R > 0 \implies \sqrt{R^2 + 1} < (R+1)$

you can infer that the smaller roots in (5), (6), and (7) are inappropriate.

You can then manually verify that the resulting values for $a,b,c$ satisfy $a^2 + b^2 = c^2.$

Edit
After reading the answer of Grab a Coffee, I was intrigued enough to sanity-check my answer.

$\displaystyle R = \frac{\sin(\alpha)}{\cos(\alpha)} \implies $

  • $\displaystyle R^2 + 1 = \frac{1}{\cos^2(\alpha)}.$

  • $\displaystyle (R + 1) = \frac{\sin(\alpha) + \cos(\alpha)}{\cos(\alpha)}.$

Therefore,

$$c = \frac{T}{2} \times \frac{\cos(\alpha)}{\sin(\alpha)} \times \left[\frac{1}{\cos^2(\alpha)} + \frac{\sin(\alpha) + \cos(\alpha)}{\cos(\alpha)} \frac{1}{\cos(\alpha)}\right]. $$

This simplifies to

$$\frac{T}{2} \times \frac{1}{\sin(\alpha) \cos(\alpha)} \times [\sin(\alpha) + \cos(\alpha) + 1] \times \frac{\sin(\alpha) + \cos(\alpha) - 1}{\sin(\alpha) + \cos(\alpha) - 1}$$

$$ = \frac{T}{2} \times \frac{\sin^2(\alpha) + \cos^2(\alpha) + 2\sin(\alpha)\cos(\alpha) - 1}{\sin(\alpha)\cos(\alpha) \times [\sin(\alpha) + \cos(\alpha) - 1]}$$

$$ = \frac{T}{2} \times \frac {2}{\sin(\alpha) + \cos(\alpha) - 1}.$$

user2661923
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