Answer to Question #157409 in Geometry for Prince

Question

3. OABC is a tetrahedron and OA = a, OB = b and OC = c. The points P and Q

are such that OA = AP and 2OB = BQ. The point M is the midpoint of P Q. Find

(i) AB, (ii) PQ, (iii) CQ, (iv) QM, (v) MB and (vi) OM in terms of a, b and c.

4. ABC is a triangle and P, Q are the midpoints of AB, AC respectively. If AB = 2x

and AC = 2y, express the vectors (i) BC, (ii) PQ, (iii) PC, (iv) BQ in terms of x

and y. What can you deduce about the directed line-segments BC and PQ?

5. ABC is a triangle. If D is the midpoint of AC, show that BA + BC = 2BD.

6. ABCD is a quadrilateral with AB equal and parallel to DC. Prove that AD is equal

and parallel to BC.

7. ABCD is a square and P, Q are the midpoints of BC, CD respectively. If AP = a

and AQ = b, find in terms of a and b, the directed line segments (i) AB, (ii) AD,

(iii) BD, (iv) AC.

8. ABC is a triangle and P is any point in BC. If PQ is the resultant of AP, PB, PC,

show that ABQC is a parallelogram, and Q is therefore a fixed point.

Answer 1

Solution

Find

3.) (i) AB, (ii) PQ, (iii) CQ, (iv) QM, (v) MB and (vi) OM in terms of a, b and c.

Let $\vec{OA}=a,\ \vec{OB}=b,\ \vec{OC}=c$

Then

$\vec{OP}=2a,\ \vec{OQ}=3b\\ \vec{OA}+\vec{AB}=\vec{OB} \implies \vec{AB} =b-a\\ \vec{OP}+\vec{PQ}=\vec{OQ} \implies \vec{PQ} =3b-2a\\ \vec{OC}+\vec{CQ}=\vec{OQ} \implies \vec{CQ} =3b-c\\ \vec{QM}=-\frac{1}{2}\vec{PQ}=a-\frac{3}{2}b\\ \vec{OM}=\vec{OQ}+\vec{QM}=3b+a-\frac{3}{2}b=a+\frac{3}{2}b\\ \vec{OM}+\vec{MB}=\vec{OB} \implies \vec{MB}=b-(a+\frac{3}{2})=-a-\frac{1}{2}b$

4.) Express the vectors (i) BC, (ii) PQ, (iii) PC, (iv) BQ in terms of x

and y. What can you deduce about the directed line-segments BC and PQ?

So, as $2x + \vec{BC} = 2y$, we get $\vec{BC} = 2y - 2x$

As $x + \vec{PQ} = y$, we get $\vec{PQ} = y - x$

As $x + \vec{PC} = 2y$, we get $\vec{PC} = 2y - x$

As $2x + \vec{BQ} = y$, we get $\vec{BQ} = y - 2x$

Something we can deduce about the directed line-segments BC and PQ is that

$\vec{BC} = \vec{2PQ}$ , so they are parallel.

5.) ABC is a triangle. If D is the midpoint of AC, show that BA + BC = 2BD.

Let $\vec{AB}=-x$ and $\vec{BC}=y$, Then $AC=AB+BC$ which is equal to $-x+y$. Then $BD=BA+AD =x+\frac{1}{2(-x+y)}=\frac{1}{2(x+y)}$ which implies that $BA+BC=2BD= x+y= 2(\frac{1}{2x+y})$ hence proved

6.) Prove that AD is equal and parallel to BC.

Join A and C (diagonal). In triangles ABC and ACD ∠BAC=∠ACD (alternate angles between parallels AB and DC); AC is common; AB=DC (given). So the triangles are congruent (SAS, two sides and included angle). Therefore ∠DAC=∠ACB, and DA=CB in length. But these are alternate angles (opposite sides of the diagonal), so DA is parallel to CB.

7.) Find in terms of a and b, the directed line segments (i) AB, (ii) AD, (iii) BD, (iv) AC.

Let $ABCD$ be a square. Let equal sides of square = x.

Given that $P$ and $Q$ are middle points of $BC$ and $CD$ respectively

$BC=CD=x$

$BP=\frac{x}{2}\:and\:QD\:=\frac{x}{2}$

Given that $AP=a$ and $AQ=b$

Now in right angled triangle $ABP$

By Pythagorous Theorem

$\left(AP\right)^2=\left(AB\right)^2+\left(BP\right)^2$

$a^2=x^2+\left(\frac{x}{2}\right)^2\\a^2=\frac{5x^2}{4}$

Now in right angled triangle ADQ

By Pythagorous Theorem

$\left(AQ\right)^2=\left(AD\right)^2+\left(DQ\right)^2\\ b^2=x^2+\frac{x^2}{4}$

$b^2=\frac{5x^2}{4}$

$a^2=b^2=\frac{5x^2}{4}\:or\:a=b=\frac{\sqrt{5x}}{2}$

$\:x=\frac{2a}{\sqrt{5}}\:or\:x=\frac{2b}{\sqrt{5}}$

$i.)\:AB=x=\frac{2a}{\sqrt{5}}\:or\:\frac{2b}{\sqrt{5}}$$I.\:AB=x=\frac{2a}{\sqrt{5}}\:or\:\frac{2b}{\sqrt{5}}$$I.\:AB=x=\frac{2a}{\sqrt{5}}\:or\:\frac{2b}{\sqrt{5}}$

$ii.)\:AD=x=\frac{2a}{\sqrt{5}}\:or\:\frac{2b}{\sqrt{5}}$

$iii.)\:BD$

$\left(BD\right)^2=\left(BC\right)^2+\left(CD\right)^2$

$=x^2+x^2$

$=2x^2$

$=2\times\:\frac{4a^2}{5}$

$BD=\sqrt{2}\times\frac{2a}{\sqrt{5}}\:or\:\sqrt{2}\times\frac{2b}{\sqrt{5}}$

$iv.)\:AC$

$\left(AC\right)^2=x^2+x^2=2x^2$

$\:AC=\frac{2\sqrt{2a}}{\sqrt{5}}\:or\:\frac{2\sqrt{2b}}{\sqrt{5}}$

8.) If $PQ$ is the resultant of $AP, PB, PC$, show that $ABQC$ is a parallelogram, and $Q$ is therefore a fixed point.

1) By adding all of the vectors we get the point $C'$ and the vector $AC,$ which, when translated, results in the vector $PQ$ at the last drawing.

2) $AP = C'Q$ (the rule of translation), $BC' = PC$ (when added vectors $BP$ was translated into $CC'$ vector). $C'Q$ is parallel to $AP$ so that the $C'$ angle equals to the $P$ angle. Thus, the C`BQ triangle equals the $PCA$ triangle. Consequently, $BQ = AC,$ similarly, $AC = BQ, \ ABQC$ is a parallelogram.