An Isosceles right triangle $ABC$ , $AB=BC= 4 cm$
Point $p$ is a midpoint of $BC$ , points $q$, $s$ lies on $AC$,$AB$ respectively , such that the triangle $pqs$ is an equilateral triangle ;
what is the area of triangle $pqs$
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@Arnaldo i have tried using analytic geometry to find equations and intersection points but i stuck, angle ∠B=90 – user373141 Apr 12 '17 at 13:07
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Rotate $\Delta Bps$ clockwise $60^\circ$ about $p$ as shown, to form $\Delta B'pq$.
Using $\Delta B'Cp$, find that $B'C=2\sqrt3~\mathrm{cm}$.
Using Sine Law on $\Delta B'Cq$, find that $\dfrac{B'q}{\sin 15^\circ}=\dfrac{B'C}{\sin 105^\circ}$, so $B'q=(4\sqrt3-6)~\mathrm{cm}$.
Use Pythagoras' theorem on $\Delta B'pq$ to find that $pq=2\sqrt{22-12\sqrt3}~\mathrm{cm}$.
Therefore the area is $\dfrac12a^2\sin60^\circ=(22\sqrt3-36)~\mathrm{cm}^2$.
DHMO
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Brilliant solution , i have question; how did you get angle $qpB'$=$15$ – user373141 Apr 12 '17 at 13:44
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Using Sine Law on $\Delta B'Cq$: $\dfrac{B'q}{\sin\angle B'CQ} = \dfrac{B'C}{\sin\angle B'qC}$ – DHMO Apr 12 '17 at 13:54
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How do you know that rotating the triangle $pBs$ by $60^\circ$ about $p$ will cause $s$ to land on the hypotenuse? – Hoc Ngo Apr 12 '17 at 14:18
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1@HocNgo $\Delta pqs$ is equilateral and the angle of rotation to cause $s$ to land on the hypotenuse is $\angle spq = 60^\circ$. – DHMO Apr 12 '17 at 14:19
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Thanks. I can see now that you can compute $B'C$ from the triangle $B'pC$, from which you can compute $B'q$. That's very elegant. – Hoc Ngo Apr 12 '17 at 19:49
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@DHMO I was initially not convinced because if point $s$ were arbitrary on edge $AB$, rotating the triangle $pBs$ by $60^\circ$ about $p$ would not land $s$ on $AC$. It's now easier to understand if we show how to construct the equilateral triangle first and prove it later. First, rotate the triangle $pBA$ by $60^\circ$ about $p$. The new edge $B'A'$ will intersect the edge $AC$ at $q$. Next, draw an arc of radius 2 centered at $p$; let the intersection between the arc and edge $AB$ be $s$. Finally, proceed to prove that $pqs$ is the equilateral triangle as you did. – Hoc Ngo Apr 12 '17 at 20:26
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@HocNgo you're over-complicating everything. I simply glued two triangles together, which is guaranteed by the fact that pqs is equilateral. Then, I found the angles by letting one of the components to be $\theta$ then finding that the other components are $120^\circ-\theta$ (which combines with $\theta$ to form the $120^\circ$ I marked), $90^\circ-\theta$ and $15^\circ+\theta$ (which form $115^\circ$). I only said to rotate "by $60^\circ$" for extra rigor. – DHMO Apr 12 '17 at 23:55
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Based on your description, let's define:
$$ \begin{align} \mbox{Hypotenuse AC}:\; y &= -x +4 \\ p &= (2,0) \\ q &= (x, y=-x+4) \\ s & = (0,h) \end{align}$$
The three sides are equal, hence:
$$ \begin{align} \mbox{sp = pq} \Rightarrow 4+h^2 &= (x-2)^2 + (x-4)^2 \quad (1)\\ \mbox{sp = sq} \Rightarrow 4+h^2 &= x^2 + (h+x-4)^2, \quad(2) \\ \mbox{pq = sq} \Rightarrow (x-2)^2 + (x-4)^2 &= x^2 + (h+x-4)^2. \quad(3)\\ \end{align}$$
There are 3 equations with only 2 variables $x$ and $h$ but only two equations are needed to solve for the variables. Let's say we solve $(2)$ and $(3)$ for $x$ and $h$. Then we plug the solution in to $(1)$ and if $(1)$ is not satisfied, then the problem has no solution.
From $(3)$:
$$ x = \frac{4 + 8h -h^2}{4 + 2h}. \quad(4) $$
Plugging $(4)$ into $(2)$ will result in an equation in $h$ alone. This equation can be solved numerically. Once $h$ is obtained, $x$ will be found by $(4)$.
The next step is to plug the results for $x$ and $h$ into $(1)$. If the equality fails, the problem has no solution.
If the problem has a solution, then the area of the triangle $spq$ is given by
$$ A = \frac{\sqrt3}{4}(sp)^2 = \frac{\sqrt3}{4}(4+h^2). $$
Hoc Ngo
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