Solve irregular quadrilateral given 3 sides, 2 angles and a constraint

Question

In this quadrilateral, $|AB|, |CD|$ and $|DA|$ are given and $\measuredangle ABC=90^\circ$. There is also a rectangle $BPQR$ with given dimensions. The point $Q$lies on a vector parallel to $DC$ at a height of $2$. This vector cannot pass through the rectangle, so we know that angle $C\leqslant 90^\circ$ and $RC\geqslant 2$.

My question is in two parts:

• With this information, is it possible to solve this quadrilateral; and if so, is there a single solution or a range of possible solutions?
• If there is no solution, can we find a minimum equal value of CD and DA that will yield a solution?

Answer 1

For a solution to exist, it is necessary that

(1) $AD = DC = 100$

(2) that the line $DC$ be tangent at a point $Q'$ to the circumference of center $Q$ and radius $2$.

In sum, the position of three points $D, Q'$ and $C$ subject to constraints (1) and (2) must be determined.

The determination of point $D$ requires that the center $C$ be known at the vertical $x = 180$ of a circumference of radius $100$, but there is no way of predicting that then constraint (2) will be satisfied. In the same way, if from a point $C$ on the vertical $x = 180$ the tangent to the circumference described in the restriction (2) is drawn, it cannot be expected that it passes through the intersection $D$ of the two circumferences involved of radius $100$.

In the attached figure it is shown three cases where we can see that the $DC$ side crosses the vertical $x = 180$ at a point whose second coordinate is located higher than the second coordinate of the center $C$ of the respective circumference.

Thus the problema has no solution.

Answer 2

To approach the problem let us first derive the equation of the tangents from the point $(x_0,y_0)$ to a circle of radius $R$ centered at the point $(X,Y)$.

Let us look for the tangent line equation in the form: $$ \sin\alpha(x-x_0)-\cos\alpha(y-y_0)=0.\tag1 $$ To find the angle $\alpha$ (which is assumed to be in the range $(-\frac\pi2,\frac\pi2]$) we should solve the equation: $$ \sin\alpha(X-x_0)-\cos\alpha(Y-y_0)=\pm R\tag2, $$ where the choice of the sign in the RHS determines if the center of the circle is above ($-$) or under ($+$) the line. In the considered problem only the upper tangent ($+$) is of interest.

The equation (2) can be easily solved using the following result valid for any $(a,b,c):\; a^2+b^2\ge c^2$: $$ A\sin\alpha-B\cos\alpha=C\\ \implies \sin\alpha=\frac{AC\pm B\sqrt{A^2+B^2-C^2}}{A^2+B^2}; \cos\alpha=\frac{-BC\pm A\sqrt{A^2+B^2-C^2}}{A^2+B^2},\tag3 $$ which can be straightforwardly verified.

Plugging into expression (3) the values of parameters from (2) one obtains: $$ \cos\alpha=\frac{-R(Y-y_0)+(X-x_0)\sqrt{(Y-y_0)^2+(X-x_0)^2-R^2}}{(Y-y_0)^2+(X-x_0)^2}, $$ where we choose positive sign for the root to ensure that $\cos\alpha>0$.

Let $\measuredangle BAD=\phi$, $a=180$, $b=100$, $R=2$, $Q=(X,Y)$. Then the coordinates of the point $D$ are $(b\cos\phi,b\sin\phi)$. Let $C$ be intersection of the line $CD$ with the perpendicular to $x$-axis erected from the point $B$. Then according to (4): $$ CD=\frac{a-b\cos\phi}{\cos\alpha}\\ =\frac{(a-b\cos\phi)[(X-b\cos\phi)^2+(Y-b\sin\phi)^2]} {-R(Y-b\sin\phi)+(X-b\cos\phi)\sqrt{(X-b\cos\phi)^2+(Y-b\sin\phi)^2-R^2}}.\tag5 $$ The plot of the function is given below:

The further strategy would be to minimize (5) wrt. $\phi$ and confirm that $$\min_\phi(CD(a,b,R,X,Y,\phi))>b$$ For the second part one would need to solve the equation: $$ \min_\phi(CD(a,b,R,X,Y,\phi))=b. $$ However I see here no possibility to proceed further without computer assistance.

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The numerical results are: $$ \min_\phi(CD(180,b,2,116,66,\phi))\approx 102.459\\ \min_\phi(CD(180,b,2,116,66,\phi)=b\implies b\approx101.177 $$