Given two angle and a segment, can you find $h$?

Question

While designing another problem I came up with the following question:

Considering the picture below, can you determine $h$ based only on the information of $\alpha$, $\beta$ and $x$ given the fact that the segment $x$ is the continuation of the height?

My guess is that the problem is not solvable (in the sense that one can construct multiple $h$ with the same $\alpha,\beta$ and $x$), but cannot see how to show this.

So far I gave the problem a try by splitting the angle and combining some trigonometric identities but couldn't conclude.

So far I only obtained the following \begin{align*} \tan(\alpha)=\frac{(h_1+h_2)(x+y)}{(x+y)^2-h_1h_2} \end{align*} and $$\tan(\beta)=\frac{(h_1+h_2)\cdot y}{y^2-h_1h_2}.$$

Many thanks in advance.

Answer 1

It is not possible to determine $h$ using only $\alpha, \beta, x$.

For example, if we take $\alpha = \alpha_1, \beta= \beta_1$ (i.e. making a right triangle), we get $$h = \frac{x \tan \alpha \tan \beta}{\tan \beta - \tan \alpha}$$

On the other hand, if we take $\frac{\alpha}{2} = \alpha_1 = \alpha_2, \frac{\beta}{2} = \beta_1 = \beta_2$ (i.e. making an isosceles triangle), then we can cut it in two to get right triangles, so we get $$h = 2\frac{x \tan(\frac{\alpha}{2})\tan(\frac{\beta}{2})}{\tan(\frac{\beta}{2}) - \tan(\frac{\alpha}{2})}$$

Trying a few simple values for $x, \alpha, \beta$ (I used $1, 30^\circ, 45^\circ$) shows that $h$ takes different values.

Answer 2

I like using the law of sines for problems like this. Consider the upper leftmost triangle in your diagram, the one whose angles are $\alpha_1$, $\pi - \beta_1$, and $\beta_1 - \alpha_1$. Let's write $r$ for the length of the side opposite the angle $\pi - \beta_1$. By the law of sines, $$ \frac{\sin(\beta_1 - \alpha_1)}{x} = \frac{\sin(\pi-\beta_1)}{r} $$ Notice that $r$ is determined by the values of $x$, $\alpha_1$ and $\beta_1$. But $h_1$ is the product $r\, \sin(\alpha_1)$, so $h_1$ too is determined by the values of $x$, $\alpha_1$ and $\beta_1$.

Answer 3

This may not be the easiest solution but it is a different one which I thought would be interesting to share as well. Looking at those points we know that $\angle ACB=\angle ALB=\beta,$ $\angle AMB=\angle AKB=\alpha$ and by construction $MC$ and $KL$ are prependicular to $AB$. Moreover, we have $h=\overline{AB}$ and as we can see in the image $\overline{MC}< \overline{KL}.$ So, we have two triangles with the same $h,\alpha,\beta$ but different $x$.

Let us now think of the homothety centered at $B$ with ratio $\frac{\overline{KL}}{\overline{MC}}$ and let $C',M',A'$ be the images of $C,M,A$ respectively. By the homothty's properties we have \begin{align} \angle BC'A'&=\angle BCA=\beta\\ \angle BM'A'&=\angle BMA=\alpha\\ \overline{M'C'}&=\frac{\overline{KL}}{\overline{MC}}\overline{MC}=\overline{KL}\\ \overline{A'B}&=\overline{AB}\frac{\overline{KL}}{\overline{MC}}>\overline{AB}. \end{align} Furthermore, $M'C'$ will be prependicular to $A'B.$

Hence, we have $\angle A'C'B=\angle ALB=\beta,$ $\angle A'M'B=\angle AKB=\alpha$ and $x'=\overline{M'C'}=\overline{KL}=x$ but $h'=\overline{A'B}>\overline{AB}=h,$ therefore having $x$ and both angles is not enough.

Answer 4

This is what I would practically do upfront before attempting a solution.

Take $ 30 ^{\circ}$ and $ 45 ^{\circ}$ set square triangles from a geometry box as a particular example.

Place two vertices at distance $x$ along $x-$ axis.

No matter how you rotate the triangles about the fulcrum vertices,line connecting side intersections do not occur with perpendicularity in the manner you suggest. So drop further general solution attempts.

Answer 5

Draw the circumscribed circles of $\triangle ABC$ and $\triangle EBC$, select the point $A_1$ and $E_1$ on these circles such that $A_1E_1\parallel AE$ and $|A_1E_1|\ne|AE|$. We know that then $\angle CA_1B=\alpha$. $\angle CE_1B=\beta$.

Construct $E_2\in A_1E_1:\ |A_1E_2|=|AE|$. Draw the line $L_1$ through $E_2$ parallel to $E_1B$ and the line $L_2$ through $E_2$ parallel to $E_1C$. Point $B_1=L_1 \cap A_1B$, point $C_1=L_2 \cap A_1C$. point $D_1=BC \cap A_1E_2$, point $D_2=B_1C_1 \cap A_1E_2$.

This new construction that includes $\triangle A_1B_1C_1$ and$\triangle E_2B_1C_1$ has the same properties as the original one, namely $\angle C_1A_1B_1=\alpha$, $\angle C_1E_1B_1=\beta$, $A_1E_1\perp B_1C_1$, $|A_1E_1|=x$, but $|B_1C_1|\ne|BC|$.

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Edit

As @pm2595 noted, this construction and conclusion is valid only if the point $E_1:\ |A_1E_1|\ne x$ exists. And such a point can always be found excluding a special case, when the circumradius $R_{ABC}$ of $\triangle ABC$ and the circumradius $R_{EBC}$ of $\triangle EBC$ are the same:

\begin{align} R_{ABC}&=R_{EBC} ,\\ \frac{h}{2\sin\alpha} &= \frac{h}{2\sin\beta} , \end{align}

in other words, unless we are given that

\begin{align} \beta&=180^\circ-\alpha . \end{align}

In this special case for any point $A_1\in\operatorname{arc}(CAB)$ the distance to the corresponding point $E_1\in\operatorname{arc}(CEB)$ will always be the same, $|A_1E_1|=x$ and $E_2\equiv E_1$. Surprisingly, in such a case we indeed can find a unique value for $h$, since it would be the same for all possible constructions, and we can easily find it from the configuration, in which both $\triangle ABC$ and $\triangle EBC$ are isosceles:

In this case we have

\begin{align} \triangle ADC:\quad \tfrac12h&=|AD|\tan\tfrac\alpha2 \tag{1}\label{1} ,\\ \triangle EDC:\quad \tfrac12h&=|ED|\tan\tfrac\beta2 =(|AD|-x)\tan\left(\tfrac12(180^\circ-\alpha)\right) \tag{2}\label{2} , \end{align}

so we can solve the system \eqref{1}-\eqref{2} for $|AD|$ and $h$.

Thus in case when $\beta=180^\circ-\alpha$, we have the unique answer

\begin{align} h&=x\tan\alpha . \end{align}.

Note that even in this case the value of $|AD|$ is still uncertain: that one we've found from \eqref{1}-\eqref{2} is valid for isosceles configuration only and it would be different otherwise.