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In this below diagram, $\angle ABC=60^\circ, \angle DCB=30^\circ $, $AD$ is parallel to $BC$ and $AP$ is perpendicular to $BC$. Both the area and perimeter of $ABCD$ and $APQD$ are equal . What is the value of /angle $DQC$?

At first, I started with the condition of area of the both the trapezium (mentioned in the question) being equal. As $AD$ is parallel to $BC$, So we can show that,

$\frac{1}{2}(AD+BC)$ ×$AP$ = $\frac{1}{2}(AD+PQ)$ ×$AP$

$AD$+$BC$ = $AD$+$PQ$

$BC$ = $PQ$ .................. (1)

And now, we can show their perimeter are equal with the below equation:

$AB+BC+CD+AD$ = $AP+PQ+DQ+AD$

$AB+BC+CD$ = $AP+PQ+DQ$

$AB+PQ+CD$ = $AP+PQ+DQ$............(from 1)

$CD+AB$ = $AP+DQ$.............(2)

I drew a vertical line $DE$ which is parallel to $AP$. $DE$ is equal to $AP$ as $AD$ and $BC$ are parallel to each other. Here /angle $APB$ and /angle $DEQ$ are respectively right-angle. So, we get two right-angled triangle $APB$ and $DEQ$.

By trigonometry and from the traingle $APB$, we can write that:-

$\frac{AB}{AP}$ = $\mathrm{cosec} 60^\circ$

$AB$ = $\frac{2AP}{\sqrt 3}$..........(3)

Similarly, from $DEC$, we can write that:-

$\frac{DC}{DE}$ = $\mathrm{cosec} 30^\circ$

$DC$ = 2$DE$

$CD$ = 2$DE$..........(4)

Let us denote the $\angle DQE = $$\theta$

So, now from the equation (2), we can get:-

$2AP$ + $\frac{2AP}{\sqrt 3}$ = $AP+ DQ$.........(from 3 and 4)

$\frac{2\sqrt 3AP + 2AP - \sqrt 3AP}{\sqrt 3} $ = $DQ$

$AP × \frac{2+ \sqrt 3}{\sqrt 3} $ = $DQ$

$\frac{2+ \sqrt 3}{\sqrt 3} $ = $\frac{DQ}{AP} $

$\frac{DQ}{DE} $ = $\frac{2+ \sqrt 3}{\sqrt 3} $......($AP = DE$ according to the diagram)

$\frac{DE}{DQ} $ = $\frac{\sqrt 3}{2 +\sqrt 3} $

$\sin \theta$ = $\frac{\sqrt 3}{2} + 1 $.................(5)

We know that if function of x is described as f(x) = $\sin^\text{-1} $x, than function of x will be real and valid if and only if its domain is [-1,1]. But there is a little bit problem in my calculation.

(5) is invalid because the real value of $\sin \theta$ is above 1 which is impossible in this case. For this reason, the value of $\theta$ is unreal. So, there is an extensive mistake either in my calculation or in the question. I have been unable to find my error. So, I need some help to identify the mistake to solve the problem properly.

EDIT: My fault is making equation (5) from the past.

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    $\begingroup$ There is no solution. You cannot have both perimeters and areas the same. $\endgroup$ – Andrei Jan 24 at 7:42
  • $\begingroup$ That's why I got stucked. I often get this kind of faulty question but find no satisfied answer. The source of the problem is also enigmatic. Thank you for telling me that point. I frankly didn't know that point. Should I delete this post? Otherwise anyone will be confused. $\endgroup$ – Anirban Niloy Jan 24 at 7:48
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    $\begingroup$ I did not know it either. But it's easy to prove. I will post that as an answer. $\endgroup$ – Andrei Jan 24 at 7:51
  • $\begingroup$ Okay. I will check that out later. I'll be eagerly waiting. $\endgroup$ – Anirban Niloy Jan 24 at 7:53
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The problem has no solution. Let's assume that there is one. Then we notice that increasing or decreasing the length of $AD$ will not change the fact that areas/ perimeters are the same. Indeed, if we increase the length by $x$, the perimeter of both trapezoids will increase by $2x$, and the areas will both increase by $x\cdot AE$. Then let's do $x=-|AD|$. Then in your picture $A=D$ and $P=E$. I will therefore drop any reference to $D$ and $E$ from now on.

$ABC$ is a right angle triangle, with the acute angles $30^\circ$ and $60^\circ$. Then $$AC=\frac{\sqrt 3}2 BC$$ and $$AB=\frac 12 BC$$ In the right angle triangle $ABC$ I can write the area in two ways to get $$BC\cdot AP=AB\cdot AC=BC^2\frac{\sqrt 3}{4}$$ Therefore $AP=BC \frac{\sqrt 3}{4}$. You already showed that $PQ=BC$. So in the right angle triangle $APQ$ you can write $$\tan\theta=\frac{AP}{PQ}=\frac{\sqrt 3}4$$

Now lets see that this does not verify equal perimeter requirement: $$AB+AC+BC=BC\left(\frac 12+\frac{\sqrt 3}2+1\right)$$ $$AP+PQ+QA=BC\left(\frac{\sqrt 3}{4}+1+\sqrt{1^2+\left(\frac{\sqrt 3}{4}\right)^2}\right)$$ You can see that there is something with $\sqrt{19}$ in the second equation, that it's not there in the first.

Note: you could get to the same conclusion even if you explicitly carry around $AD\ne 0$.

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  • $\begingroup$ I didn't understand the fact of triangle ABC being a right angled triangle. Would you please explain that? $\endgroup$ – Anirban Niloy Jan 24 at 8:41
  • $\begingroup$ $AP\perp BC$. For the other one, if you have in a triangle an angle of $60^\circ$ and an angle of $30^\circ$, the last one has to be $90^\circ$ $\endgroup$ – Andrei Jan 24 at 8:44
  • $\begingroup$ But here we can see in that diagram that $\angle DCE = 30^\circ$. Steven gregory solved the value of $\angle DQC but how can we justify that? $\endgroup$ – Anirban Niloy Jan 24 at 8:46
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    $\begingroup$ Make $A=D$. In the picture in the other answer use $y=0$ $\endgroup$ – Andrei Jan 24 at 8:48
  • $\begingroup$ Oops, sorry. You told me in your answer. It's a great honour that you gave your precious time to me. Tnx for your help and support. $\endgroup$ – Anirban Niloy Jan 24 at 8:50
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enter image description here

\begin{align} \operatorname{perimeter}(APQD) &= \operatorname{perimeter}(ABCD) \\ (3+\sqrt 3)x + 2y + CQ+DQ &= (6+2\sqrt 3)x + 2y \\ CQ +DQ &= (3+\sqrt 3)x \end{align}

\begin{align} \operatorname{area}(APQD) &= \operatorname{area}(ABCD) \\ \frac 12(3x+2y+CQ)(\sqrt 3x) &= \frac 12(4x+2y)((\sqrt 3x)) \\ CQ &= x \\ \hline DQ &= (2+\sqrt 3)x \\ (DQ)^2 &= (7+4\sqrt 3)x^2 \end{align}

But, then, $(7+4\sqrt 3)x^2=(DQ)^2 = (\sqrt 3x)^2+(4x)^2=19x^2$

So, by contradiction, there is no solution.

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    $\begingroup$ In triangle $DEQ$ you have $EQ=EC+CQ=4x$. Then $DQ^2=DE^2+EQ^2=3x^2+16x^2=19x^2$ but $x\sqrt{19}\ne(2+\sqrt 3)x$ $\endgroup$ – Andrei Jan 24 at 8:52
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    $\begingroup$ @stevengregory If you use the tangent function, $\tan\angle DQC=\frac{\sqrt 3 x}{4x}$ you get a different value for the angle, only about $23.4^\circ$ $\endgroup$ – Andrei Jan 24 at 9:15
  • $\begingroup$ Yeah, you're right. If we consider the condition as a right one (including area and perimeter are equal of both the field mentioned in the question), then $27.65^\circ$ $\approx$ $23.4^\circ$. That makes no sense. $\endgroup$ – Anirban Niloy Jan 24 at 10:20

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