Euclidean Geometry
Solutions

1. The area of quadrilateral $ABCD$ is the sum of the areas of $\mathrm{△}ABC$ and $\mathrm{△}ACD$.

   Since $\mathrm{△}ABC$ is right-angled at $B$, its area equals ${\displaystyle \frac{1}{2}}(AB)(BC)={\displaystyle \frac{1}{2}}(3)(4)=6$.

   Since $\mathrm{△}ABC$ is right-angled at $B$, by the Pythagorean Theorem, $$AC=\sqrt{A{B}^2+B{C}^2}=\sqrt{3^2+4^2}=\sqrt{25}=5$$ because $AC>0$. (We could have also observed that $\mathrm{△}ABC$ must be a "3-4-5" triangle.)

   Since $\mathrm{△}ACD$ is right-angled at $A$, by the Pythagorean Theorem, $$AD=\sqrt{C{D}^2-A{C}^2}=\sqrt{{13}^2-5^2}=\sqrt{144}=12$$ because $AD>0$. (We could have also observed that $\mathrm{△}ACD$ must be a "5-12-13" triangle.)

   Thus, the area of $\mathrm{△}ACD$ equals ${\displaystyle \frac{1}{2}}(AC)(AD)={\displaystyle \frac{1}{2}}(5)(12)=30$. Finally, the area of quadrilateral $ABCD$ is thus $6+30=36$.

2. Solution 1

   Suppose that the rectangular prism has dimensions $a\text{ cm}$ by $b\text{ cm}$ by $c\text{ cm}$.

   Suppose further that one of the faces that is $a\text{ cm}$ by $b\text{ cm}$ is the face with area $27{\text{ cm}}^2$ and that one of the faces that is $a$ cm by $c$ cm is the face with area $32{\text{ cm}}^2$. (Since every pair of non-congruent faces shares exactly one side length, there is no loss of generality in picking these particular variables for these faces.) Therefore, $ab=27$ and $ac=32$.

   Further, we are told that the volume of the prism is $144{\text{ cm}}^3$, and so $abc=144$.

   Thus, $bc={\displaystyle \frac{a^2{b}^2{c}^2}{a^{2}bc}}={\displaystyle \frac{(abc{)}^2}{(ab)(ac)}}={\displaystyle \frac{{144}^2}{(27)(32)}}=24$. (We could also note that $abc=144$ means $a^2{b}^2{c}^2={144}^2$ or $(ab)(ac)(bc)={144}^2$ and so $bc={\displaystyle \frac{{144}^2}{(27)(32)}}$.)

   In other words, the third type of face of the prism has area $24{\text{ cm}}^2$.

   Thus, since the prism has two faces of each type, the surface area of the prism is equal to $2(27{\text{ cm}}^2+32{\text{ cm}}^2+24{\text{ cm}}^2)$ or $166{\text{ cm}}^2$.

   Solution 2

   Suppose that the rectangular prism has dimensions $a\text{ cm}$ by $b\text{ cm}$ by $c\text{ cm}$.

   Suppose further that one of the faces that is $a\text{ cm}$ by $b\text{ cm}$ is the face with area $27{\text{ cm}}^2$ and that one of the faces that is $a\text{ cm}$ by $c\text{ cm}$ is the face with area $32{\text{ cm}}^2$. (Since every pair of non-congruent faces shares exactly one side length, there is no loss of generality in picking these particular variables for these faces.) Therefore, $ab=27$ and $ac=32$.

   Further, we are told that the volume of the prism is $144{\text{ cm}}^3$, and so $abc=144$.

   Since $abc=144$ and $ab=27$, we have $c={\displaystyle \frac{144}{27}}={\displaystyle \frac{16}{3}}$.

   Since $abc=144$ and $ac=32$, we have $b={\displaystyle \frac{144}{32}}={\displaystyle \frac{9}{2}}$.

   This means that $bc={\displaystyle \frac{16}{3}}\cdot {\displaystyle \frac{9}{2}}=24$.

   In ${\text{cm}}^2$, the surface area of the prism equals $2ab+2ac+2bc=2(27)+2(32)+2(24)=166$.

   Thus, the surface area of the prism is $166{\text{ cm}}^2$.

3. Let the radius of the smaller circle be $r\text{ cm}$ and let the radius of the larger circle be $R\text{ cm}$.

   Thus, the circumference of the smaller circle is $2\pi r\text{ cm}$, the circumference of the larger circle is $2\pi R\text{ cm}$, the area of the smaller circle is $\pi r^2{\text{ cm}}^2$, and the area of the larger circle is $\pi R^2{\text{ cm}}^2$.

   Since the sum of the radii of the two circles is $10\text{ cm}$, we have $r+R=10$.

   Since the circumference of the larger circle is $3\text{ cm}$ larger than the circumference of the smaller circle, it follows that $2\pi R-2\pi r=3$, or $2\pi (R-r)=3$.

   Then the difference, in ${\text{cm}}^2$, between the area of the larger circle and the area of the smaller circle is $$\pi R^2-\pi r^2=\pi (R-r)(R+r)={\displaystyle \frac{1}{2}}[2\pi (R-r)](R+r)={\displaystyle \frac{1}{2}}(3)(10)=15$$ Therefore, the difference between the areas is $15{\text{ cm}}^2$.

4. Since $ABC$ is a quarter of a circular pizza with centre $A$ and radius $20\text{ cm}$, we have $AC=AB=20\text{ cm}$. We are also told that $\mathrm{\angle }CAB=90\mathrm{°}$ (one-quarter of $360\mathrm{°}$).

   Since $\mathrm{\angle }CAB=90\mathrm{°}$ and $A$, $B$ and $C$ are all on the circumference of the circle, it follows that $CB$ is a diameter of the pan. (This is a property of circles: if $X$, $Y$ and $Z$ are three points on a circle with $\mathrm{\angle }ZXY=90\mathrm{°}$, then $YZ$ must be a diameter of the circle.)

   Since $\mathrm{△}CAB$ is a right-angled isosceles triangle, we have $CB=\sqrt{2}AC=20\sqrt{2}\text{ cm}$. Therefore, the radius of the circular plate is ${\displaystyle \frac{1}{2}}CB$ or $10\sqrt{2}\text{ cm}$. Thus, the area of the circular pan is $\pi (10\sqrt{2}\text{ cm}{)}^2=200\pi {\text{ cm}}^2$.

   The area of the slice of pizza is one-quarter of the area of a circle with radius $20\text{ cm}$, or ${\displaystyle \frac{1}{4}}\pi (20\text{ cm}{)}^2=100\pi {\text{ cm}}^2$.

   Finally, the fraction of the pan that is covered is the area of the slice of pizza divided by the area of the pan, or ${\displaystyle \frac{100\pi {\text{ cm}}^2}{200\pi {\text{ cm}}^2}}={\displaystyle \frac{1}{2}}$.

5. Solution 1

   Since $\mathrm{△}AFD$ is right-angled at $F$, by the Pythagorean Theorem, $$AD=\sqrt{A{F}^2+F{D}^2}=\sqrt{4^2+2^2}=\sqrt{20}=2\sqrt{5}$$ since $AD>0$.

   Let $\mathrm{\angle }FAD=\beta$. Since $ABCD$ is a rectangle, we have $\mathrm{\angle }BAF=90\mathrm{°}-\beta$.

   Since $\mathrm{△}AFD$ is right-angled at $F$, we have $\mathrm{\angle }ADF=90\mathrm{°}-\beta$.

   Since $ABCD$ is a rectangle, we have $\mathrm{\angle }BDC=90\mathrm{°}-(90\mathrm{°}-\beta )=\beta$.

   

   Therefore, $\mathrm{△}BFA$, $\mathrm{△}AFD$, and $\mathrm{△}DFE$ are all similar as each is right-angled and has either an angle of $\beta$ or an angle of $90\mathrm{°}-\beta$ (and hence both of these angles).
   Therefore, ${\displaystyle \frac{AB}{AF}}={\displaystyle \frac{DA}{DF}}$ and so $AB={\displaystyle \frac{4(2\sqrt{5})}{2}}=4\sqrt{5}$.

   Also, ${\displaystyle \frac{FE}{FD}}={\displaystyle \frac{FD}{FA}}$ and so $FE={\displaystyle \frac{2(2)}{4}}=1$.

   Since $ABCD$ is a rectangle, we have $BC=AD=2\sqrt{5}$, and $DC=AB=4\sqrt{5}$.

   Finally, the area of quadrilateral $BCEF$ equals the area of $\mathrm{△}DCB$ minus the area $\mathrm{△}DFE$. Thus, the required area is $${\displaystyle \frac{1}{2}}(DC)(CB)-{\displaystyle \frac{1}{2}}(DF)(FE)={\displaystyle \frac{1}{2}}(4\sqrt{5})(2\sqrt{5})-{\displaystyle \frac{1}{2}}(2)(1)=20-1=19$$

   Solution 2

   Since $\mathrm{△}AFD$ is right-angled at $F$, by the Pythagorean Theorem, $$AD=\sqrt{A{F}^2+F{D}^2}=\sqrt{4^2+2^2}=\sqrt{20}=2\sqrt{5}$$ since $AD>0$.

   Let $\mathrm{\angle }FAD=\beta$. Since $ABCD$ is a rectangle, we have $\mathrm{\angle }BAF=90\mathrm{°}-\beta$. Since $\mathrm{△}BAF$ is right-angled at $F$, we have $\mathrm{\angle }ABF=\beta$.

   Since $\mathrm{△}AFD$ is right-angled at $F$, we have $\mathrm{\angle }ADF=90\mathrm{°}-\beta$.

   Since $ABCD$ is a rectangle, we have $\mathrm{\angle }BDC=90\mathrm{°}-(90\mathrm{°}-\beta )=\beta$.

   

   Looking at $\mathrm{△}AFD$, we see that $\sin \beta ={\displaystyle \frac{FD}{AD}}={\displaystyle \frac{2}{2\sqrt{5}}}={\displaystyle \frac{1}{\sqrt{5}}}$, $\cos \beta ={\displaystyle \frac{AF}{AD}}={\displaystyle \frac{4}{2\sqrt{5}}}={\displaystyle \frac{2}{\sqrt{5}}}$, and $\tan \beta ={\displaystyle \frac{FD}{AF}}={\displaystyle \frac{2}{4}}={\displaystyle \frac{1}{2}}$.

   Since $AF=4$ and $\mathrm{\angle }ABF=\beta$, we have $AB={\displaystyle \frac{AF}{\sin \beta }}={\displaystyle \frac{4}{\frac{1}{\sqrt{5}}}}=4\sqrt{5}$.

   Since $FD=2$ and $\mathrm{\angle }FDE=\beta$, we have $FE=FD\tan \beta =2\cdot {\displaystyle \frac{1}{2}}=1$.

   Since $ABCD$ is a rectangle, we have $BC=AD=2\sqrt{5}$, and $DC=AB=4\sqrt{5}$.

   Finally, the area of quadrilateral $EFBC$ equals the area of $\mathrm{△}DCB$ minus the area $\mathrm{△}DFE$. Thus, the required area is $${\displaystyle \frac{1}{2}}(DC)(CB)-{\displaystyle \frac{1}{2}}(DF)(FE)={\displaystyle \frac{1}{2}}(4\sqrt{5})(2\sqrt{5})-{\displaystyle \frac{1}{2}}(2)(1)=20-1=19$$

6. Join $BE$.

   

   Since $\mathrm{△}FBD$ is congruent to $\mathrm{△}AEC$, we have $FB=AE$.

   Since $\mathrm{△}FAB$ and $\mathrm{△}AFE$ are each right-angled, share a common side $AF$ and have equal hypotenuses ($FB=AE$), it follows that these triangles are congruent, and so $AB=FE$.

   Now $BAFE$ has two right angles at $A$ and $F$ (so $AB$ and $FE$ are parallel) and has equal sides $AB=FE$ so must be a rectangle. This means that $BCDE$ is also a rectangle.

   Now the diagonals of a rectangle partition it into four triangles of equal area. (Diagonal $AE$ of the rectangle splits the rectangle into two congruent triangles, which have equal area. The diagonals bisect each other, so the four smaller triangles all have equal area.)

   Since one quarter of rectangle $ABEF$ is shaded and one quarter of rectangle $BCDE$ is shaded, it follows that one quarter of the total area is shaded. (If the area of $ABEF$ is $x$ and the area of $BCDE$ is $y$, then the total shaded area is ${\displaystyle \frac{1}{4}}x+{\displaystyle \frac{1}{4}}y$, which is one quarter of the total area $x+y$.)

   Since $AC=200$ and $CD=50$, the area of rectangle $ACDF$ is $200(50)=10000$, so the total shaded area is ${\displaystyle \frac{1}{4}}(10000)=2500$.

7. Suppose that $M$ is the midpoint of $YZ$. Suppose that the centre of the smaller circle is $O$ and the centre of the larger circle is $P$. Suppose that the smaller circle touches $XY$ at $C$ and $XZ$ at $D$, and that the larger circle touches $XY$ at $E$ and $XZ$ at $F$. Join $OC$, $OD$ and $PE$.

   Since $OC$ and $PE$ are radii that join the centres of circles to points of tangency, it follows that $OC$ and $PE$ are perpendicular to $XY$.

   Construct $XM$. Since $\mathrm{△}XYZ$ is isosceles, $XM$ (which is a median by construction) is an altitude (that is, $XM$ is perpendicular to $YZ$) and an angle bisector (that is, $\mathrm{\angle }MXY=\mathrm{\angle }MXZ$).

   Now $XM$ passes through $O$ and $P$. (Since $XC$ and $XD$ are tangents from $X$ to the same circle, we have $XC=XD$. This means that $\mathrm{△}XCO$ is congruent to $\mathrm{△}XDO$ by side-side-side. This means that $\mathrm{\angle }OXC=\mathrm{\angle }OXD$ and so $O$ lies on the angle bisector of $\mathrm{\angle }CXD$, and so $O$ lies on $XM$. Using a similar argument, $P$ lies on $XM$.)
   Draw a perpendicular from $O$ to $T$ on $PE$. Note that $OT$ is parallel to $XY$ (since each is perpendicular to $PE$) and that $OCET$ is a rectangle (since it has three right angles).

   

   Consider $\mathrm{△}XMY$ and $\mathrm{△}OTP$. Each triangle is right-angled (at $M$ and at $T$, respectively). Also, $\mathrm{\angle }YXM=\mathrm{\angle }POT$. (This is because $OT$ is parallel to $XY$, since both are perpendicular to $PE$.) Therefore, $\mathrm{△}XMY$ is similar to $\mathrm{△}OTP$. Thus, ${\displaystyle \frac{XY}{YM}}={\displaystyle \frac{OP}{PT}}$.

   Now $XY=a$ and $YM={\displaystyle \frac{1}{2}}b$. Also, $OP$ is the line segment joining the centres of two tangent circles, so $OP=r+R$.

   Lastly, $PT=PE-ET=R-r$, since $PE=R$, $ET=OC=r$, and $OCET$ is a rectangle. Therefore, $$\begin{array}{rl}{\displaystyle \frac{a}{b/2}}& ={\displaystyle \frac{R+r}{R-r}}\\ {\displaystyle \frac{2a}{b}}& ={\displaystyle \frac{R+r}{R-r}}\\ 2a(R-r)& =b(R+r)\\ 2aR-bR& =2ar+br\\ R(2a-b)& =r(2a+b)\\ {\displaystyle \frac{R}{r}}& ={\displaystyle \frac{2a+b}{2a-b}}& \text{(since }2a>b\text{ we have }2a-b\ne 0\text{, and }r>0\text{)}\end{array}$$ Therefore, ${\displaystyle \frac{R}{r}}={\displaystyle \frac{2a+b}{2a-b}}$.

8. Let $\mathrm{\angle }PEQ=\theta$. Join $P$ to $B$.

   We use the fact that the angle between a tangent to a circle and a chord in that circle that passes through the point of tangency equals the angle inscribed by that chord. We prove this fact below.

   More concretely, $\mathrm{\angle }DEP=\mathrm{\angle }PBE$ (using the chord $EP$ and the tangent through $E$) and $\mathrm{\angle }ABP=\mathrm{\angle }PEQ=\theta$ (using the chord $BP$ and the tangent through $B$).

   Now $\mathrm{\angle }DEP$ is exterior to $\mathrm{△}FEP$ and so $\mathrm{\angle }DEP=\mathrm{\angle }FPE+\mathrm{\angle }EFP=25\mathrm{°}+30\mathrm{°}$, and so $\mathrm{\angle }PBE=\mathrm{\angle }DEP=55\mathrm{°}$. Furthermore, $\mathrm{\angle }AQB$ is an exterior angle of $\mathrm{△}PQE$. Thus, $\mathrm{\angle }AQB=\mathrm{\angle }QPE+\mathrm{\angle }PEQ=25\mathrm{°}+\theta$.

   

   In $\mathrm{△}ABQ$, we have $\mathrm{\angle }BAQ=35\mathrm{°}$, $\mathrm{\angle }ABQ=\theta +55\mathrm{°}$, and $\mathrm{\angle }AQB=25\mathrm{°}+\theta$.

   Thus, $35\mathrm{°}+(\theta +55\mathrm{°})+(25\mathrm{°}+\theta )=180\mathrm{°}$ or $115\mathrm{°}+2\theta =180\mathrm{°}$, and so $2\theta =65\mathrm{°}$. Therefore $\mathrm{\angle }PEQ=\theta ={\displaystyle \frac{1}{2}}(65\mathrm{°})=32.5\mathrm{°}$.

   As an addendum, we prove that the angle between a tangent to a circle and a chord in that circle that passes through the point of tangency equals the angle inscribed by that chord.

   Consider a circle with centre $O$ and a chord $XY$, with tangent $ZX$ meeting the circle at $X$. We prove that if $ZX$ is tangent to the circle, then $\mathrm{\angle }ZXY$ equals $\mathrm{\angle }XWY$ whenever $W$ is a point on the circle on the opposite side of $XY$ as $XZ$ (that is, the angle subtended by $XY$ on the opposite side of the circle).
   We prove this in the case that $\mathrm{\angle }ZXY$ is acute. The cases where $\mathrm{\angle }ZXY$ is a right angle or an obtuse angle are similar.
   Draw diameter $XOV$ and join $VY$.

   

   Since $\mathrm{\angle }ZXY$ is acute, points $V$ and $W$ are on the same arc of chord $XY$. This means that $\mathrm{\angle }XVY=\mathrm{\angle }XWY$, since they are angles subtended by the same chord.

   Since $OX$ is a radius and $XZ$ is a tangent, it follows that $\mathrm{\angle }OXZ=90\mathrm{°}$. Therefore, we have $\mathrm{\angle }OXY+\mathrm{\angle }ZXY=90\mathrm{°}$.

   Since $XV$ is a diameter, we have $\mathrm{\angle }XYV=90\mathrm{°}$.

   From $\mathrm{△}XYV$, we see that $\mathrm{\angle }XVY+\mathrm{\angle }VXY=90\mathrm{°}$.

   But $\mathrm{\angle }OXY+\mathrm{\angle }ZXY=90\mathrm{°}$ and $\mathrm{\angle }XVY+\mathrm{\angle }VXY=90\mathrm{°}$ and $\mathrm{\angle }OXY=\mathrm{\angle }VXY$ tells us that $\mathrm{\angle }ZXY=\mathrm{\angle }XVY$. This gives us that $\mathrm{\angle }ZXY=\mathrm{\angle }XWY$, as required.

9. Solution 1

   Draw a line segment through $M$ in the plane of $\mathrm{△}PMN$ parallel to $PN$ and extend this line until it reaches the plane through $P$, $A$ and $D$ at $Q$ on one side and the plane through $N$, $B$ and $C$ at $R$ on the other side.

   Join $Q$ to $P$ and $A$. Join $R$ to $N$ and $B$.

   

   So the volume of solid $ABCDPMN$ equals the volume of solid $ABCDPQRN$ minus the volumes of solids $PMQA$ and $NMRB$.

   Solid $ABCDPQRN$ is a trapezoidal prism. This is because $NR$ and $BC$ are parallel (since they lie in parallel planes), which makes $NRBC$ a trapezoid. Similarly, $PQAD$ is a trapezoid. Also, $PN$, $QR$, $DC$, and $AB$ are all perpendicular to the planes of these trapezoids and equal in length, since they equal the side lengths of the squares.

   Solids $PMQA$ and $NMRB$ are triangular-based pyramids. We can think of their bases as being $\mathrm{△}PMQ$ and $\mathrm{△}NMR$. Their heights are each equal to $2$, the height of the original solid. (The volume of a triangular-based pyramid equals $\frac{1}{3}$ times the area of its base times its height.)

   The volume of $ABCDPQRN$ equals the area of trapezoid $NRBC$ times the width of the prism, which is 2.

   That is, this volume equals ${\displaystyle \frac{1}{2}}(NR+BC)(NC)(NP)={\displaystyle \frac{1}{2}}(NR+2)(2)(2)=2\cdot NR+4$.

   So we need to find the length of $NR$.

   Consider quadrilateral $PNRQ$. This quadrilateral is a rectangle since $PN$ and $QR$ are perpendicular to the two side planes of the original solid. Thus, $NR$ equals the height of $\mathrm{△}PMN$.

   Join $M$ to the midpoint $T$ of $PN$. Since $\mathrm{△}PMN$ is isosceles, $MT$ is perpendicular to $PN$.

   

   Since $NT={\displaystyle \frac{1}{2}}PN=1$ and $\mathrm{\angle }PMN=90\mathrm{°}$ and $\mathrm{\angle }TNM=45\mathrm{°}$, it follows that $\mathrm{△}MTN$ is also right-angled and isosceles with $MT=TN=1$.

   Therefore, $NR=MT=1$ and so the volume of $ABCDPQRN$ is $2\cdot 1+4=6$.

   The volumes of solids $PMQA$ and $NMRB$ are equal. Each has height 2 and their bases $\mathrm{△}PMQ$ and $\mathrm{△}NMR$ are congruent, because each is right-angled (at $Q$ and at $R$) with $PQ=NR=1$ and $QM=MR=1$.

   Thus, using the formula above, the volume of each is ${\displaystyle \frac{1}{3}}({\displaystyle \frac{1}{2}}(1)(1))2={\displaystyle \frac{1}{3}}$.

   Finally, the volume of the original solid equals $6-2\cdot {\displaystyle \frac{1}{3}}={\displaystyle \frac{16}{3}}$.

   Solution 2

   We determine the volume of $ABCDPMN$ by splitting it into two solids: $ABCDPN$ and $ABNPM$ by slicing along the plane of $ABNP$.

   Solid $ABCDPN$ is a triangular prism, since $\mathrm{△}BCN$ and $\mathrm{△}ADP$ are each right-angled (at $C$ and $D$), $BC=CN=AD=DP=2$, and segments $PN$, $DC$ and $AB$ are perpendicular to each of the triangular faces and equal in length.

   Thus, the volume of $ABCDPN$ equals the area of $\mathrm{△}BCN$ times the length of $DC$. Therefore, ${\displaystyle \frac{1}{2}}(BC)(CN)(DC)={\displaystyle \frac{1}{2}}(2)(2)(2)=4$. (This solid can also be viewed as "half" of a cube.)

   Solid $ABNPM$ is a pyramid with rectangular base $ABNP$. (Note that $PN$ and $AB$ are perpendicular to the planes of both of the side triangular faces of the original solid, that $PN=AB=2$ and $BN=AP=\sqrt{2^2+2^2}=2\sqrt{2}$, by the Pythagorean Theorem.)

   Therefore, the volume of $ABNPM$ equals ${\displaystyle \frac{1}{3}}(AB)(BN)h={\displaystyle \frac{4\sqrt{2}}{3}}h$, where $h$ is the height of the pyramid (that is, the distance that $M$ is above plane $ABNP$).

   So we need to calculate $h$.

   Join $M$ to the midpoint, $T$, of $PN$ and to the midpoint, $S$, of $AB$. Join $S$ and $T$.

   

   By symmetry, $M$ lies directly above $ST$. Since $ABNP$ is a rectangle and $S$ and $T$ are the midpoints of opposite sides, we have $ST=AP=2\sqrt{2}$.
   Since $\mathrm{△}PMN$ is right-angled and isosceles, $MT$ is perpendicular to $PN$. Since $NT={\displaystyle \frac{1}{2}}PN=1$ and $\mathrm{\angle }TNM=45\mathrm{°}$, it follows that $\mathrm{△}MTN$ is also right-angled and isosceles with $MT=TN=1$.

   Also, $MS$ is the hypotenuse of the triangle formed by dropping a perpendicular from $M$ to $U$ in the plane of $ABCD$ (a distance of 2) and joining $U$ to $S$.

   

   Since $M$ is 1 unit horizontally from $PN$, we have $US=1$.
   Thus, $MS=\sqrt{2^2+1^2}=\sqrt{5}$ by the Pythagorean Theorem.

   We can now consider $\mathrm{△}SMT$. $h$ is the height of this triangle, from $M$ to base $ST$.

   

   Now $h=MT\sin (\mathrm{\angle }MTS)=\sin (\mathrm{\angle }MTS)$.

   By the cosine law in $\mathrm{△}SMT$, we have $$M{S}^2=S{T}^2+M{T}^2-2(ST)(MT)\cos (\mathrm{\angle }MTS)$$ Therefore, $5=8+1-4\sqrt{2}\cos (\mathrm{\angle }MTS)$ or $4\sqrt{2}\cos (\mathrm{\angle }MTS)=4$.

   Thus, $\cos (\mathrm{\angle }MTS)={\displaystyle \frac{1}{\sqrt{2}}}$ and so $\mathrm{\angle }MTS=45\mathrm{°}$ which gives $h=\sin (\mathrm{\angle }MTS)={\displaystyle \frac{1}{\sqrt{2}}}$.

   (Alternatively, we note that the plane of $ABCD$ is parallel to the plane of $PMN$, and so since the angle between plane $ABCD$ and plane $PNBA$ is $45\mathrm{°}$, it follows that the angle between plane $PNBA$ and plane $PMN$ is also $45\mathrm{°}$, and so $\mathrm{\angle }MTS=45\mathrm{°}$.)

   Finally, this means that the volume of $ABNPM$ is ${\displaystyle \frac{4\sqrt{2}}{3}}\cdot {\displaystyle \frac{1}{\sqrt{2}}}={\displaystyle \frac{4}{3}}$, and so the volume of solid $ABCDPMN$ is $4+{\displaystyle \frac{4}{3}}={\displaystyle \frac{16}{3}}$.

10. Let $\mathrm{\angle }EAF=\theta$. Since $ABCD$ is a parallelogram, $AB$ and $DC$ are parallel with $AB=DC$, and $DA$ and $CB$ are parallel with $DA=CB$.

    Since $AE$ is perpendicular to $DC$ and $AB$ and $DC$ are parallel, it follows that $AE$ is perpendicular to $AB$. In other words, $\mathrm{\angle }EAB=90\mathrm{°}$, and so $\mathrm{\angle }FAB=90\mathrm{°}-\theta$.

    Since $\mathrm{△}AFB$ is right-angled at $F$ and $\mathrm{\angle }FAB=90\mathrm{°}-\theta$, we have $\mathrm{\angle }ABF=\theta$.

    Using similar arguments, we obtain that $\mathrm{\angle }DAE=90\mathrm{°}-\theta$ and $\mathrm{\angle }ADE=\theta$.

    

    Since $\cos (\mathrm{\angle }EAF)=\cos \theta ={\displaystyle \frac{1}{3}}$ and ${\cos }^2\theta +{\sin }^2\theta =1$, we have $$\sin \theta =\sqrt{1-{\cos }^2\theta }=\sqrt{1-{\displaystyle \frac{1}{9}}}=\sqrt{{\displaystyle \frac{8}{9}}}={\displaystyle \frac{2\sqrt{2}}{3}}$$ (Note that $\sin \theta >0$ since $\theta$ is an angle in a triangle.)

    In $\mathrm{△}AFB$, $\sin \theta ={\displaystyle \frac{AF}{AB}}$ and $\cos \theta ={\displaystyle \frac{FB}{AB}}$.

    Since $AF=32$ and $\sin \theta ={\displaystyle \frac{2\sqrt{2}}{3}}$, we have $AB={\displaystyle \frac{AF}{\sin \theta }}={\displaystyle \frac{32}{2\sqrt{2}/3}}={\displaystyle \frac{48}{\sqrt{2}}}=24\sqrt{2}$.

    Since $AB=24\sqrt{2}$ and $\cos \theta ={\displaystyle \frac{1}{3}}$, we have $FB=AB\cos \theta =24\sqrt{2}({\displaystyle \frac{1}{3}})=8\sqrt{2}$.

    In $\mathrm{△}AED$, $\sin \theta ={\displaystyle \frac{AE}{AD}}$ and $\cos \theta ={\displaystyle \frac{DE}{AD}}$.

    Since $AE=20$ and $\sin \theta ={\displaystyle \frac{2\sqrt{2}}{3}}$, we have $AD={\displaystyle \frac{AE}{\sin \theta }}={\displaystyle \frac{20}{2\sqrt{2}/3}}={\displaystyle \frac{30}{\sqrt{2}}}=15\sqrt{2}$.

    Since $AD=15\sqrt{2}$ and $\cos \theta ={\displaystyle \frac{1}{3}}$, we have $DE=AD\cos \theta =15\sqrt{2}\left({\displaystyle \frac{1}{3}}\right)=5\sqrt{2}$. (To calculate $AD$ and $DE$, we could also have used the fact that $\mathrm{△}ADE$ is similar to $\mathrm{△}ABF$.)

    Finally, the area of quadrilateral $AECF$ equals the area of parallelogram $ABCD$ minus the combined areas of $\mathrm{△}AFB$ and $\mathrm{△}ADE$.

    The area of parallelogram $ABCD$ equals $AB\cdot AE=24\sqrt{2}\cdot 20=480\sqrt{2}$.

    The area of $\mathrm{△}AFB$ equals ${\displaystyle \frac{1}{2}}(AF)(FB)={\displaystyle \frac{1}{2}}(32)(8\sqrt{2})=128\sqrt{2}$.

    The area of $\mathrm{△}AED$ equals ${\displaystyle \frac{1}{2}}(AE)(DE)={\displaystyle \frac{1}{2}}(20)(5\sqrt{2})=50\sqrt{2}$.

    Thus, the area of quadrilateral $AECF$ is $480\sqrt{2}-128\sqrt{2}-50\sqrt{2}=302\sqrt{2}$.

11. Solution 1

    Consider $\mathrm{△}BCE$ and $\mathrm{△}ACD$.

    

    Since $\mathrm{△}ABC$ is equilateral, we have $BC=AC$. Since $\mathrm{△}ECD$ is equilateral, we have that $CE=CD$.

    Since $BCD$ is a straight line and $\mathrm{\angle }ECD=60\mathrm{°}$, we have $\mathrm{\angle }BCE=180\mathrm{°}-\mathrm{\angle }ECD=120\mathrm{°}$.

    Since $BCD$ is a straight line and $\mathrm{\angle }BCA=60\mathrm{°}$, we have $\mathrm{\angle }ACD=180\mathrm{°}-\mathrm{\angle }BCA=120\mathrm{°}$.

    Therefore, $\mathrm{△}BCE$ is congruent to $\mathrm{△}ACD$ ("side-angle-side").

    Since $\mathrm{△}BCE$ and $\mathrm{△}ACD$ are congruent and $CM$ and $CN$ are line segments drawn from the corresponding vertex ($C$ in both triangles) to the midpoint of the opposite side, we have $CM=CN$.

    Since $\mathrm{\angle }ECD=60\mathrm{°}$, $\mathrm{△}ACD$ can be obtained by rotating $\mathrm{△}BCE$ through an angle of $60\mathrm{°}$ clockwise about $C$. This means that after this $60\mathrm{°}$ rotation, $CM$ coincides with $CN$. In other words, $\mathrm{\angle }MCN=60\mathrm{°}$.

    But since $CM=CN$ and $\mathrm{\angle }MCN=60\mathrm{°}$, we have $$\mathrm{\angle }CMN=\mathrm{\angle }CNM={\displaystyle \frac{1}{2}}(180\mathrm{°}-\mathrm{\angle }MCN)=60\mathrm{°}$$ Therefore, $\mathrm{△}MNC$ is equilateral, as required.

    Solution 2

    We prove that $\mathrm{△}MNC$ is equilateral by introducing a coordinate system.

    Suppose that $C$ is at the origin $(0,0)$ with $BCD$ along the $x$-axis, with $B$ having coordinates $(-4b,0)$ and $D$ having coordinates $(4d,0)$ for some real numbers $b,d>0$.

    Drop a perpendicular from $E$ to $P$ on $CD$.

    

    Since $\mathrm{△}ECD$ is equilateral, $P$ is the midpoint of $CD$.

    Since $C$ has coordinates $(0,0)$ and $D$ has coordinates $(4d,0)$, it follows that the coordinates of $P$ are $(2d,0)$.

    Since $\mathrm{△}ECD$ is equilateral, we have $\mathrm{\angle }ECD=60\mathrm{°}$ and so $\mathrm{△}EPC$ is a $30\mathrm{°}$-$60\mathrm{°}$-$90\mathrm{°}$ triangle and so $EP=\sqrt{3}CP=2\sqrt{3}d$.

    Therefore, the coordinates of $E$ are $(2d,2\sqrt{3}d)$.

    In a similar way, we can show that the coordinates of $A$ are $(-2b,2\sqrt{3}b)$.

    Now $M$ is the midpoint of $B(-4b,0)$ and $E(2d,2\sqrt{3}d)$, and therefore, the coordinates of $M$ are $({\displaystyle \frac{1}{2}}(-4b+2d),{\displaystyle \frac{1}{2}}(0+2\sqrt{3}d))$ or $(-2b+d,\sqrt{3}d)$.

    Also, $N$ is the midpoint of $A(-2b,2\sqrt{3}b)$ and $D(4d,0)$, and therefore, the coordinates of $N$ are $({\displaystyle \frac{1}{2}}(-2b+4d),{\displaystyle \frac{1}{2}}(2\sqrt{3}b+0))$ or $(-b+2d,\sqrt{3}b)$.

    To show that $\mathrm{△}MNC$ is equilateral, we show that $CM=CN=MN$ or equivalently that $C{M}^2=C{N}^2=M{N}^2$: $$\begin{array}{rl}C{M}^2& =(-2b+d-0{)}^2+(\sqrt{3}d-0{)}^2\\ & =(-2b+d{)}^2+(\sqrt{3}d{)}^2\\ & =4b^2-4bd+d^2+3d^2\\ & =4b^2-4bd+4d^2\\ \\ C{N}^2& =(-b+2d-0{)}^2+(\sqrt{3}b-0{)}^2\\ & =(-b+2d{)}^2+(\sqrt{3}b{)}^2\\ & =b^2-4bd+4d^2+3b^2\\ & =4b^2-4bd+4d^2\\ \\ M{N}^2& =((-2b+d)-(-b+2d){)}^2+(\sqrt{3}d-\sqrt{3}b{)}^2\\ & =(-b-d{)}^2+3(d-b{)}^2\\ & =b^2+2bd+d^2+3d^2-6bd+3b^2\\ & =4b^2-4bd+4d^2\end{array}$$ Therefore, $C{M}^2=C{N}^2=M{N}^2$ and so $\mathrm{△}MNC$ is equilateral, as required.

12. Since $PQ$ is parallel to $AB$, it is parallel to $DC$ and is perpendicular to $BC$.

    Drop perpendiculars from $A$ to $E$ on $PQ$ and from $P$ to $F$ on $DC$.

    

    Then $ABQE$ and $PQCF$ are rectangles. Thus, $EQ=x$, which means that $PE=r-x$ and $FC=r$, which means that $DF=y-r$.

    Let $BQ=b$ and $QC=c$. Thus, $AE=b$ and $PF=c$.

    The area of trapezoid $ABQP$ is ${\displaystyle \frac{1}{2}}(x+r)b$.

    The area of trapezoid $PQCD$ is ${\displaystyle \frac{1}{2}}(r+y)c$.

    Since these areas are equal, we have ${\displaystyle \frac{1}{2}}(x+r)b={\displaystyle \frac{1}{2}}(r+y)c$, which gives ${\displaystyle \frac{x+r}{r+y}}={\displaystyle \frac{c}{b}}$.

    Since $AE$ is parallel to $PF$, we have $\mathrm{\angle }PAE=\mathrm{\angle }DPF$ and $\mathrm{△}AEP$ is similar to $\mathrm{△}PFD$.

    Thus, ${\displaystyle \frac{AE}{PE}}={\displaystyle \frac{PF}{DF}}$ which gives ${\displaystyle \frac{b}{r-x}}={\displaystyle \frac{c}{y-r}}$ or ${\displaystyle \frac{c}{b}}={\displaystyle \frac{y-r}{r-x}}$.

    Combining ${\displaystyle \frac{x+r}{r+y}}={\displaystyle \frac{c}{b}}$ and ${\displaystyle \frac{c}{b}}={\displaystyle \frac{y-r}{r-x}}$ gives ${\displaystyle \frac{x+r}{r+y}}={\displaystyle \frac{y-r}{r-x}}$ or $(x+r)(r-x)=(r+y)(y-r)$.

    From this, we get $r^2-x^2=y^2-r^2$ or $2r^2=x^2+y^2$, as required.

13. Suppose that the parallel line segments $EF$ and $WX$ are a distance of $x$ apart.

    This means that the height of trapezoid $EFXW$ is $x$.

    Since the side length of square $EFGH$ is $10$ and the side length of square $WXYZ$ is $6$, it follows that the distance between parallel line segments $ZY$ and $HG$ is $10-6-x$ or $4-x$.

    Recall that the area of a trapezoid equals one-half times its height times the sum of the lengths of the parallel sides.

    Thus, the area of trapezoid $EFXW$ is ${\displaystyle \frac{1}{2}}x(EF+WX)={\displaystyle \frac{1}{2}}x(10+6)=8x$.

    Also, the area of trapezoid $GHZY$ is ${\displaystyle \frac{1}{2}}(4-x)(HG+ZY)={\displaystyle \frac{1}{2}}(4-x)(10+6)=32-8x$.

    Therefore, the sum of the areas of trapezoids $EFXW$ and $GHZY$ is $8x+(32-8x)=32$.

    This sum is a constant and does not depend on the position of the inner square within the outer square, as required.