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The problem is as follows:

Sketch of the problem

$\begin{array}{ll} 1.&10^\circ\\ 2.&12^\circ\\ 3.&15^\circ\\ 4.&16^\circ\\ \end{array}$

I'm not sure which sort of construction can be used here to solve this problem?.

I've attempted to draw a perpendicular line from $B$ to segment $AC$. But this did not yielded good results.

Then I've attempted tracing a perpendicular segment to $BC$ intersecting $AC$, from looking on this possibility. I got still stuck.

I don't know how to use the given angles, they suspiciously add up to $3\alpha$.

I still don't know how to use $QC=2HC$. Can someone give me some ideas on what to do?. Should congruence be used here?.

I'm assuming that the intended approach is try to spot triangles such as:

$3-4-5$ or $1-\sqrt{3}-2$ or something along those special right triangles.

I do hope someone could help me on how to solve this problem relying on euclidean geometry can this be done?.

Quanto
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Chris Steinbeck Bell
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  • To make things specific, I'd probably take $|QC|=2$ and $|HC|=1$. (It's arbitrary but it eliminates ambiguity.) – Semiclassical Jan 16 '21 at 02:21
  • @Semiclassical I tend to use auxiliary variables such as $a$ but it mostly is as you indicate. But the question remains there. How to do it trig-free?. Is this possible?. – Chris Steinbeck Bell Jan 16 '21 at 05:46

3 Answers3

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enter image description here

Construct isosceles triangle $CQD$ and connect $PD$. Note that $$\angle HPD = \angle HPB + \angle BPD =(90^\circ +\angle HCP) + (90^\circ -\angle BDP)=180^\circ $$ Thus, $H$, $P$ and $D$ are colinear and $$\cos\angle HCD = \cos5\alpha = \frac{HC}{DC} = \frac12$$ which yields $$\alpha = 12^\circ$$

Quanto
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    I was wondering if there was a geometric way to get to the angle of $5\alpha$, and this does so quite well. +1 – Semiclassical Jan 16 '21 at 06:36
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    Or you could say that since, in $\triangle HCD$, $DC=QC=2HC$ and $\angle DHC=90^{\circ}$, it is a $30-60-90$ triangle. Thus, $ 5\alpha=60^{\circ}$ and $\alpha=12^{\circ}$ – Limestone Jan 16 '21 at 07:17
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Hint: Using trigonometry and $|QC|=2|HC|$, express the lengths of $BQ,BP,PC$, and $BC$ in terms of $|HC|$. Now observe that $B,P,C$ are collinear.

Semiclassical
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Since they are complementary to angles that subtend $3\alpha$, $$ \overbrace{\ \angle HPC\ }^{\text{co-}HCP}=\frac\pi2-3\alpha=\overbrace{\ \angle QPB\ }^{\text{co-}PQB}\tag1 $$ Since $\angle BPC=\pi$, we get $$ \angle QPH=6\alpha\tag2 $$

enter image description here

Extend $\overline{QB}$ by an equal length to $R$; that is, $\overline{BR}=\overline{QB}$. Since $\triangle QCR$ is isosceles, $$ \angle BCR=\angle BCQ=2\alpha\tag3 $$ Likewise, $$ \angle BPR=\angle BPQ=\frac\pi2-3\alpha\tag4 $$ Thus, $\angle RPH=\pi$. That is, $R$, $P$, and $H$ are colinear.

Furthermore, $$ \overline{RC}=\overline{QC}\tag5 $$

Extend $\overline{HC}$ by an equal length to $S$; that is, $\overline{HS}=\overline{HC}$. Thus, $$ 2\overline{HC}=\overline{SC}\tag6 $$ and since $\triangle SRC$ is isosceles, we have $$ \overline{RS}=\overline{RC}\tag7 $$ Now we use that $\color{#C00}{\overline{QC}=2\overline{HC}}$. $\color{#090}{(7)}$, $\color{#44F}{(5)}$, and $\color{#CA0}{(6)}$ say $$ \overline{RS}\color{#090}{=}\overline{RC}\color{#44F}{=}\overline{QC}\color{#C00}{=}2\overline{HC}\color{#CA0}{=}\overline{SC}\tag8 $$ Thus, $\triangle RCS$ is equilateral, and therefore, $$ 5\alpha=\angle RCS=\frac\pi3=60^{\large\circ}\tag9 $$ Thus, $$ \alpha=\frac\pi{15}=12^{\large\circ}\tag{10} $$

robjohn
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