Triangles

The hint gives away the problem. We solve the problem by using similarity of triangles.

(i) \( \triangle ADC \sim \triangle EFC \)

Equality AE = EC \(\implies\) AC=2EC \(\implies\) AD=2EF.

(ii) \( \triangle BMD \sim \triangle BEF \)

Equality BM = ME \(\implies\) MD=EF/2.

From (i) and (ii), we have 4MD=AD. So, the ratio AM:MD=3.

Claim. \(\angle E P D=\angle D P Q=\angle E D P\)

Proof. The first equality comes from the problem statement.

Consider the second equality. Alternate angles made by line \(P D\) intersecting parallel lines \(D E\) and \(P Q\) are equal. Thus \(\triangle E P D\) is isosceles with \(E P=E D \). Similarly \(E D=D Q\) using the fact that \(Q E\) bisects \(\angle D Q P\) also intersects parallel lines \(D E\) and \(P Q\). Therefore \(E P=\) \(E D=D Q .\) Now by the basic proportionality theorem, \(\frac{R E}{E P}=\frac{R D}{D Q} .\) As the denominators \(E P\) and \(D Q\) are equal, the numerators must be equal as well, i.e., \(R E=R D .\) Finally, \(R P=R E+E P=R D+D Q=R Q,\) so \(\triangle P Q R\) is isosceles.

Let the length of AB be one unit and BC be two units, respectively. We draw a perpendicular from vertex P to X. Since APX is a right angled triangle: \begin{align} AX^2 &= AP^2 - PX^2 \\\\ AX^2 &= 2^2 - 1^2 \\\\ AX &= \sqrt{3} \\\\ \end{align} Hence, the length of CP is \( 2-\sqrt{3} \). The angle CPD is \( \arctan \frac{1}{CP} = \arctan \frac{1}{2 - \sqrt{3}} \) . We can verify that the answer is \( 75^{\circ} \).

It is clear that \(1< a \leq b \leq c<10\). Now, \(c< a+b\) and \(c=20-a-b\) implies \(10< a+b ;\) this also means that \(b \geq a\) and \(b \geq 11-a\). Moreover, we also have \(b \leq 20-a-b .\) One can further conclude that \(a \leq 6,\) otherwise \(7 \leq b \leq 6 .\) So as \(a\) ranges from 2 to 6 we have that \(b\) takes the following values \(a=2, b=9 ; a=3, b=\) \(8 ; a=4, b \in\{7,8\} ; a=5, b \in\{6,7\} ; a=6, b \in\{6,7\} .\) The total number of possible triangles is 8.

1. None. The triangle inequality fails since \(AB+AC< BC\).
2. \(\infty\). The angles sum to \(180^{\circ}\) and triangle of any scale is possible.
3. 1.
4. 2. Draw a line segment \(AB\) with the given length. Consider the line passing through \(A\) and making \(30^\circ\) with \(AB\). If we draw a circle with center as \(B\), it touches the line at two points (see Fig. below). Hence, there are two choices for the vertex \(C\).

   

Note that triangles \(A B A^{\prime}, C A C^{\prime}\) and \(B C B^{\prime}\) are congruent by the SAS test. Triangles \(B A^{\prime} Q, C B^{\prime} R\) and \(A C^{\prime} P\) are a lso congruent. By using the property of opposite angles we get that all the three angles of the triangle \(P Q R\) are the same. Hence it is an equilateral triangle. Dropping the perpendicular bisector \(A O\) on the side \(B C\) we get the following : \[ \begin{aligned} A A^{\prime 2} &=A O^{2}+A^{\prime} A^{2} \\ &=(1-k)^{2}+(\sqrt{3})^{2} \\ &=k^{2}-2 k+4 \end{aligned} \] Observe that triangles \(A B A^{\prime}\) and \(B Q A^{\prime}\) are similar by the AAA test: \(A^{\prime} Q B\) and \(A^{\prime} B A\) are 60 degrees and \(A^{\prime} B Q\) and \(A^{\prime} A B\) are corresponding angles. Therefore: \[ \begin{aligned} \frac{A B}{B Q} &=\frac{B A^{\prime}}{Q A^{\prime}}=\frac{A^{\prime} A}{A^{\prime} B} \\ \frac{2}{B Q} &=\frac{k}{Q A^{\prime}}=\frac{\sqrt{k^{2}-2 k+4}}{k} \\ B Q &=\frac{2 k}{\sqrt{k^{2}-2 k+4}} \\ Q A^{\prime} &=\frac{k^{2}}{\sqrt{k^{2}-2 k+4}} \end{aligned} \] Now using \(A A^{\prime}=A P+P Q+Q A^{\prime}\) we get \[ P Q=\frac{4(1-k)}{\sqrt{k^{2}-2 k+4}} \]

We want to find the side length of the rhombus \(BDEF\). We will find the length of \(AB\) first. Let \(M\) be the mid-point of \(BC\). So \(BM=1/2\,\mbox{cm}\). We know that \( AM = BC = 1\,\mbox{cm}\). By applying Pythagoras theorem to \(\Delta ABM\), we get \(AB = \sqrt{BM^2 + AM^2} = \frac{\sqrt{5}}{2}\).

Let ABC be a triangle with rational sides and rational area. Let \( B\) be the largest angle.

TFFT
(i) BP and CP are angle bisectors meeting at P, so AP bisects \(\angle\)A since the angle bisectors are concurrent.
(iv) The angles marked with symbol o at point P are all \(60^{\circ}\) because \(\angle\)EPD is twice this common value. It follows that half the sum of \(\angle\)B and \(\angle\)C is \(60^{\circ}\). So \(\angle\)A is \(60^{\circ} \).
The others are false, in fact check that any triangle with \(\angle A=60^{\circ},\) angle bisectors BD and CE, their point of intersection P and PF bisecting \(\angle\)BPC will satisfy the given data.

For any configuration, there are only finitely many pairings. Choose one with least possible total length of segments. Here no two of the \(n\) segments can interest, because if \(R B\) and \(R^{\prime} B^{\prime}\) intersect in point \(X\) then we get a contradiction as follows. Using triangle inequality in triangles \(R X B^{\prime}\) and \(R^{\prime} X B,\) we get \(R B^{\prime}+R^{\prime} B<R B+R^{\prime} B^{\prime}\) (draw a picture). So replacing \(R B\) and \(R^{\prime} B^{\prime}\) with \(R^{\prime} B\) and \(R B^{\prime}\) would give a pairing with smaller total length.

Given \(n\) red points, find a triangle \(A B C\) such that \(A\) is a red point and all other red points are inside triangle \(ABC\). This is always possible since \(B\) and \(C\) can be placed anywhere, as long as \(ABC\) is a triangle. Place one blue point at \(B\) and \(n-1\) blue points at \(C\). This will ensure that there will be a non-intersecting pair of lines involving vertex \(A\).

(a) Suppose the segment \(XY\) meets opposite sides \(QR\) and \(TU\) of the hexagon. Let \(O\) be the midpoint of \(XY\). In the figure below, the midpoint of the chord must pass through the center. We show that if \(\frac{QX}{XR}=\frac{TY}{YU}\), then \(O\) is the center of the hexagon.

Consider triangles \(OXQ\) and \(OTY\). By the \(ASA\) postulate they are congruent. So, \(QO=OT\). Therefore, \(O\) is the midpoint of \(QT\), which makes it the center of the hexagon. The other direction can be proved with a similar argument.

(b) Suppose we have inscribed a square \(A B C D\) in a hexagon \(P Q R S T U\), as shown in the figure below. In the previous problem, we proved that a chord that cuts the opposite sides of the hexagon in equal ratios must pass through the center of the hexagon. Here we show that the diagonals of the square are chords that satisfy that property. This will imply that both diagonals pass through the center of the hexagon.

The diagonals of the square pass through the center of the hexagon. We have used the result of the previous problem.
Lemma. \(\triangle A Q B \cong \triangle C T D \)
Proof. We know that \(PQ \| S T\) and \(A C\) is a transversal. So \(\angle Q A C=\angle T C A\), also \(\angle B A C=\angle D C A=45^{\circ} \). So \(\angle Q A B=\angle D C T\). Similarly, \(\angle Q B A=\angle C D T\). Also, \(\angle A Q B=\angle C T D,\) since they are angles in a regular hexagon. Moreover, \(A B=C D\). As a result we get that \(\triangle Q B A \cong\) \(\triangle T D C.\;\;\;\;\) \(\square\)
So we have \(Q B=T D\) and \(Q A=T C\). This in turn implies that \(B R=D U\) and \(P A=C S\). Thus, \[ \frac{Q B}{B R}=\frac{T D}{D U} \text { and } \frac{P A}{A Q}=\frac{S C}{C T} \]

(c) In the figure above, let \(PA=x\) and \(AQ=1-x\). The vertical distance between \(A\) and \(P\) is \(x\sin 30^\circ\). So, \[ AD = 2x\sin 30^{\circ} + 1 = x+1 \] Also, we have \(AB=2AQ\sin 30^\circ\). \[ AB = \sqrt{3}(1-x) \] Since \(AB=AD\), we have \(x = (\sqrt{3}-1)/(\sqrt{3}+1) \). Hence, the side of the square is \(AD=x+1=\sqrt{3}(\sqrt{3}-1)\) units.

(d) Suppose the square \(ABCD\) is inscribed in a regular hexagon \(PQRSTU\). The positions of \(A\) and \(B\) on \(PQ\) and \(QR\), respectively, determine the square. If there was a square other than the one shown in the figure above, we would have \(|QA|\neq |QB|\). Without loss of generality, assume that \(|QA|>|QB|\). Let \(O\) be the center of the hexagon and the square. We have proved that the centers coincide in part (b). Since \(O\) is the center of \(ABCD\), we have \(|OA|=|OB|\). Let \(A^\prime\) be a point on \(QR\) such that \(|QA|=|QA^\prime|\). Due to symmetry, \(|OA|=|OA^\prime|\) which implies that \(|OA^\prime|=|OB|\). Thus, \(\triangle B O A^{\prime}\) is isosceles. Let \(\angle QAB=\alpha\) and \(\angle QBA=\beta\). Since \(\angle AQB = 120^\circ\), we have \(\alpha+\beta=60^\circ\). \begin{align*} 180^{\circ}&=\angle O B A^{\prime}+\angle O B Q \\ &=\angle O B Q+\angle O A^{\prime} B=\angle O B Q+\angle O A Q \\ &=45^{\circ}+\beta+45^{\circ}+\alpha \end{align*} so \(\alpha+\beta=90^{\circ},\) a contradiction since \(\alpha+\beta=60^{\circ}\).

Lemma. \(EFGH\) is a parallelogram.

Proof. Consider the diagonal \(AC\). By the basic proportionality theorem:

• \({EF}\) and \({AC}\) are parallel.
• \({AC}=2 {EF} \)
• \(\Delta ABC \sim \Delta EBF\).

A similar argument to diagonal \(BD\) implies the lemma. \(\quad\square\)

Let \((X)\) denote the area of shape \(X\).

\begin{align} (ABCD) = (ABC) + (ACD) \\ 1/2 \cdot (ABCD) = (EFB) + (HGD) \end{align}

By applying the above argument to the diagonal \(BD\) and triangles \(CFG\) and \(AEH\), we get the following:

\begin{align} (ABCD) = (ABD) + (BCD) \\ 1/2\cdot (ABCD) = (CFG) + (AEH) \end{align}

Together we get:

\begin{align} 1/2\cdot (ABCD) &= (CFG) + (AEH) + (EFB) + (HGD) \\ (EFGH) &= \frac{1}{2} (ABCD) \end{align}

Let us draw the square \(EFGH\) on the coordinate plane (see the figure below). From the previous problem we know that the diagonals of \(ABCD\) must be parallel to the axes. Suppose the two diagonals lie on the lines \(x=p\) and \(y=-q\), respectively. Since the diagonals pass through \(EFGH\), we must have \( p\in (0,1) \) and \( q\in(0,1) \).

Notation. Let \(\mbox{proj}_x PQ\) (resp. \(\mbox{proj}_y PQ\)) denote the length of the projection of segment \(PQ\) on the \(x\)-axis (resp. \(y\)-axis).

Lemma 1. The coordinates of vertices \(A\) and \(B\) are \( (p,q) \) and \( (-p,-q) \), respectively.

Since \(E\) is the mid-point of \(AB\), we have \(|AE|=|BE|\).

\begin{align*} \mbox{proj}_y BE &= q = \mbox{proj}_y AE \implies A=(p,q) \\ \mbox{proj}_x AE &= p = \mbox{proj}_x BE \implies B=(-p,-q) \quad\square \end{align*}

Lemma 2. The coordinates of vertices \(C\) and \(D\) are \( (p,2-q) \) and \( (2-p,-q) \), respectively.

Let us calculate the \(y\)-coordinate of \(C\) first. Since \(F\) is the mid-point of \(BC\) we have \(|BF|=|CF|\).

\begin{align*} \mbox{proj}_y BF &= 1-q = \mbox{proj}_y CF \implies C=(p,2-q) \\ \end{align*}

Now we calculate the \(x\)-coordinate of \(D\). Since \(G\) is the mid-point of \(CD\) we have \(|CG|=|GD|\).

\begin{align*} \mbox{proj}_x CG &= 1-p = \mbox{proj}_x GD \implies D=(2-p,-q) \quad\square \end{align*}

From the above lemmas we infer that once the diagonals are fixed, the quadrilateral \(ABCD\) also gets fixed. There is a one-to-one correspondence between the pair of numbers \( (p,-q) \) and the quadrilateral \(ABCD\). The vertex \(A\) can take a value \((p,q)\) with \(p\in(0,1) \) and \(q \in (0,1) \). The set of all possible points for vertex \(A\) is shaded in the above figure.