Find the sum of the shaded areas in the figure below

Question

Calculate : $A1 + A2 + A3$, if $PQ=2$.

(S:$\frac{\pi}{2}-1)$

I confess that I was not able to develop this exercise... I found it very difficult..

Apparently $ PAIQ $ would be a parallelogram and this could be useful in the resolution but would need to be demonstrated

Answer 1

Observe that the letters for points $I$ and $D$ are interchanged to be in line with common notation for the incenter.

Sketch of the proof:

Let point $Q'$ be the intersection of $MN$ and $AI$. From the lemma mentioned in the comment and proved elsewhere we know $\displaystyle\angle AQ'C=\frac\pi2$. This means the point lies on the circle $ABC$ and hence coincides with $Q$. By the same argument the point $P$ lies on the line $CI$.

It follows $\displaystyle\angle POQ=\pi-\angle A-\angle C=\angle B=\frac\pi2$. This implies (a) $PQ=\sqrt2 r$, where $r$ is the radius of the circle $ABC$, and (b) $\displaystyle\angle PCQ=\frac\pi4$.

Now somewhat trigonometry (possibly there is a smarter way).

With $\angle COQ=\angle A=\alpha$ we have: $$\begin{align} A1&=\frac12r^2\Big(\frac\pi2-\alpha-\cos\alpha\Big),\\ A2&=\frac12r^2\Big(\alpha-\sin\alpha\Big),\\ A3&=\frac12 AC\cdot CD\cdot\sin(\angle ACD)\\ &=\frac12\Big[2r\Big]\left[2r\sin\frac\alpha2\cdot\cos\frac\pi4\right]\sin\left(\frac\pi4-\frac\alpha2\right)\\ &=r^2\sin\frac\alpha2\Big[\cos\frac\alpha2-\sin\frac\alpha2\Big]\\ &=\frac12r^2(\sin\alpha+\cos\alpha-1), \end{align} $$ and the claim $$A1+A2+A3=\frac\pi2-1$$ easily follows.

Answer 2

In line with the accepted answer, but without trigonometry, since $IFBG$ is a square, segment $PBQ$ of circle $ABC$ equals segment $PIQ$ of the equal circle through $P$, $I$, $Q$. Join $IP$, $IQ$.

Granting that $POQ$, $IQC$ and $API$ are isosceles right triangles, then segment $PI=A_1$, and segment $QI=A_2$. Further, in triangles $PQI$, $ACD$, we have, subtended by arc $AP$,$$\angle PQI=\angle DCA$$and$$\frac{DC}{QI}=\frac{QP}{CA}=\frac{1}{\sqrt 2}$$Therefore, $\triangle PQI=\triangle ACD$ (Euclid, Elements, VI, 15), and$$A_1+A_2+A_3=PIQ=PBQ$$And since segment $PBQ$ is sector $PBQO$ minus triangle $PQO$, or $\frac {\pi}{2}-1$, then$$A_1+A_2+A_3=\frac {\pi}{2}-1$$

Answer 3

We're told nothing about the precise location of point $B$, so we will move it.

Slide $B$ along the semicircle towards $A$. The diagram below shows $B$ close to $A$ and is given to clarify how the various points move as $B$ moves towards $A$.

In the limit when $B=A$ we find that $P, M, N$ and $D$ are also at $A$ and that $I$ is at $O$. It's given that $\angle DIQ$ is a right angle so $\angle AOQ$ is also a right angle placing $Q$ at the midpoint of the semicircular arc.

Then area $A_1=0$, area $A_3=0$, and $A_2$ is a segment across a quarter circle. Since $PQ=QC=2$ it's easily shown that area $A_1+A_2+A_3=A_2=\frac{\pi}{2}-1$.