In recent years, television has undergone a true revolution in terms of image quality, sound and interactivity with viewers. This transformation is due to the conversion of the analog signal to the digital signal. However, many cities still do not have this new technology. Seeking to bring these benefits to three cities, a television station intends to build a new transmission tower that sends signal to antennas A, B and C, already existing in these cities. The locations of the antennas are represented in the Cartesian plane.
The tower must be located in a place equidistant from the three antennas.
The appropriate location for the construction of this tower corresponds to the point with coordinates
(A) $(65; 35)$. (B) $(53; 30)$. (C) $(45; 35)$. (D) $(50; 20)$. (E) $(50; 30)$.

QUESTION 150
In a right triangle, one leg measures 6 cm and the hypotenuse measures 10 cm. The other leg measures
(A) 4 cm
(B) 6 cm
(C) 7 cm
(D) 8 cm
(E) 9 cm

In a rectangle ABCD , the length BC is twice the width AB . Pick a point P on side BC such that the lengths of AP and BC are equal. The measure of angle CPD is
(A) $75 ^ { \circ }$
(B) $60 ^ { \circ }$
(C) $45 ^ { \circ }$
(D) none of the above

Let $S$ be a circle with center $O$. Suppose $A , B$ are points on the circumference of $S$ with $\angle A O B = 120 ^ { \circ }$. For triangle $A O B$, let $C$ be its circumcenter and $D$ its orthocenter (i.e., the point of intersection of the three lines containing the altitudes). For each statement below, write whether it is TRUE or FALSE. a) The triangle $A O C$ is equilateral.
Answer: $\_\_\_\_$ b) The triangle $A B D$ is equilateral.
Answer: $\_\_\_\_$ c) The point $C$ lies on the circle $S$.
Answer: $\_\_\_\_$ d) The point $D$ lies on the circle $S$.
Answer: $\_\_\_\_$

26. In the figure below, $\hat{ABF} = C\hat{A}E = B\hat{C}D$, $DF = 2.5$, and $EF = 3$. What is the length of $AB$?
[Figure: Triangle ABC with point D on side AB, point E on side BC, and point F inside the triangle, with lines drawn from vertices through F]

• [(1)] $8.6$
• [(2)] $7.5$
• [(3)] $10.5$
• [(4)] $9.6$

9. A variable plane $x / a + y / b + z / c = 1$ at a unit distance from origin cuts the coordinate axes at $A , B$ and $C$. Centroid $( x , y , z )$ satisfies the equation $1 / x ^ { 2 } + 1 / y ^ { 2 } + 1 / z ^ { 2 } = K$. The value of $K$ is :
(a) 9
(b) 3
(c) $1 / 9$
(d) $1 / 3$

Let $O(0,0)$, $P(3,4)$, $Q(6,0)$ be the vertices of the triangle $OPQ$. The point $R$ inside the triangle $OPQ$ is such that the triangles $OPR$, $PQR$, $OQR$ are of equal area. The coordinates of $R$ are
(A) $\left(\frac{4}{3}, 3\right)$
(B) $(3, \frac{2}{3})$
(C) $(3, \frac{4}{3})$
(D) $\left(\frac{4}{3}, \frac{2}{3}\right)$

45. Let $O ( 0,0 ) , P ( 3,4 ) , Q ( 6,0 )$ be the vertices of the triangle $O P Q$. The point $R$ inside the triangle $O P Q$ is such that the triangles $O P R , P Q R , O Q R$ are of equal area. The coordinates of $R$ are
(A) $\left( \frac { 4 } { 3 } , 3 \right)$
(B) $\left( 3 , \frac { 2 } { 3 } \right)$
(C) $\left( 3 , \frac { 4 } { 3 } \right)$
(D) $\left( \frac { 4 } { 3 } , \frac { 2 } { 3 } \right)$
Answer ◯ [Figure] [Figure] [Figure]
(A)
(B)
(C)
(D)

A circle $C$ of radius 1 is inscribed in an equilateral triangle $P Q R$. The points of contact of $C$ with the sides $P Q , Q R , R P$ are $D , E , F$, respectively. The line $P Q$ is given by the equation $\sqrt { 3 } x + y - 6 = 0$ and the point $D$ is $\left( \frac { 3 \sqrt { 3 } } { 2 } , \frac { 3 } { 2 } \right)$. Further, it is given that the origin and the centre of $C$ are on the same side of the line $P Q$.
Points $E$ and $F$ are given by
(A) $\left( \frac { \sqrt { 3 } } { 2 } , \frac { 3 } { 2 } \right) , ( \sqrt { 3 } , 0 )$
(B) $\left( \frac { \sqrt { 3 } } { 2 } , \frac { 1 } { 2 } \right) , ( \sqrt { 3 } , 0 )$
(C) $\left( \frac { \sqrt { 3 } } { 2 } , \frac { 3 } { 2 } \right) , \left( \frac { \sqrt { 3 } } { 2 } , \frac { 1 } { 2 } \right)$
(D) $\left( \frac { 3 } { 2 } , \frac { \sqrt { 3 } } { 2 } \right) , \left( \frac { \sqrt { 3 } } { 2 } , \frac { 1 } { 2 } \right)$

Tangents are drawn from the point $P ( 3,4 )$ to the ellipse $\frac { x ^ { 2 } } { 9 } + \frac { y ^ { 2 } } { 4 } = 1$ touching the ellipse at points A and B.
The orthocenter of the triangle $P A B$ is
A) $\left( 5 , \frac { 8 } { 7 } \right)$
B) $\left( \frac { 7 } { 5 } , \frac { 25 } { 8 } \right)$
C) $\left( \frac { 11 } { 5 } , \frac { 8 } { 5 } \right)$
D) $\left( \frac { 8 } { 25 } , \frac { 7 } { 5 } \right)$

Let $A ( h , k ) , B ( 1,1 )$ and $C ( 2,1 )$ be the vertices of a right angled triangle with $A C$ as its hypotenuse. If the area of the triangle is 1 , then the set of values which ' k ' can take is given by
(1) $\{ 1,3 \}$
(2) $\{ 0,2 \}$
(3) $\{ - 1,3 \}$
(4) $\{ - 3 , - 2 \}$

If the straight lines $x + 3 y = 4,3 x + y = 4$ and $x + y = 0$ form a triangle, then the triangle is
(1) scalene
(2) equilateral triangle
(3) isosceles
(4) right angled isosceles

If two vertices of a triangle are $(5, -1)$ and $(-2, 3)$ and its orthocentre is at $(0, 0)$, then the third vertex is
(1) $(4, -7)$
(2) $(-4, -7)$
(3) $(-4, 7)$
(4) $(4, 7)$

If the three lines $x - 3 y = p , a x + 2 y = q$ and $a x + y = r$ form a right-angled triangle then :
(1) $a ^ { 2 } - 9 a + 18 = 0$
(2) $a ^ { 2 } - 6 a - 12 = 0$
(3) $a ^ { 2 } - 6 a - 18 = 0$
(4) $a ^ { 2 } - 9 a + 12 = 0$

The $x$-coordinate of the incentre of the triangle that has the coordinates of midpoints of its sides as $(0,1)$, $(1,1)$ and $(1,0)$ is
(1) $1 + \sqrt{2}$
(2) $1 - \sqrt{2}$
(3) $2 + \sqrt{2}$
(4) $2 - \sqrt{2}$

Given three points $P , Q , R$ with $P ( 5,3 )$ and $R$ lies on the $x$-axis. If the equation of $RQ$ is $x - 2 y = 2$ and $PQ$ is parallel to the $x$-axis, then the centroid of $\triangle PQR$ lies on the line
(1) $x - 2 y + 1 = 0$
(2) $2 x + y - 9 = 0$
(3) $2 x - 5 y = 0$
(4) $5 x - 2 y = 0$

The circumcentre of a triangle lies at the origin and its centroid is the midpoint of the line segment joining the points $\left( a ^ { 2 } + 1 , a ^ { 2 } + 1 \right)$ and $( 2 a , - 2 a ) , a \neq 0$. Then for any $a$, the orthocentre of this triangle lies on the line
(1) $y - \left( a ^ { 2 } + 1 \right) x = 0$
(2) $y - 2 a x = 0$
(3) $y + x = 0$
(4) $( a - 1 ) ^ { 2 } x - ( a + 1 ) ^ { 2 } y = 0$

Let $k$ be an integer such that the triangle with vertices $(k, -3)$, $(5, k)$ and $(-k, 2)$ has area 28 sq. units. Then the orthocenter of this triangle is at the point:
(1) $\left(2, -\dfrac{1}{2}\right)$
(2) $\left(1, \dfrac{3}{4}\right)$
(3) $\left(1, -\dfrac{3}{4}\right)$
(4) $\left(2, \dfrac{1}{2}\right)$

Let the orthocentre and centroid of a triangle be $A ( - 3,5 )$ and $B ( 3,3 )$ respectively. If $C$ is the circumcentre of this triangle, then the radius of the circle having line segment $A C$ as diameter, is:
(1) $\frac { 3 \sqrt { } 5 } { 2 }$
(2) $\sqrt { 10 }$
(3) $2 \sqrt { 10 }$
(4) $3 \sqrt { \frac { 5 } { 2 } }$

Two vertices of a triangle are $( 0,2 )$ and $( 4,3 )$. If its orthocenter is at the origin, then its third vertex lies in which quadrant?

Let the equations of two sides of a triangle be $3x - 2y + 6 = 0$ and $4x + 5y - 20 = 0$. If the orthocenter of this triangle is at $(1,1)$ then the equation of its third side is:
(1) $122y + 26x + 1675 = 0$
(2) $26x - 122y - 1675 = 0$
(3) $26x + 61y + 1675 = 0$
(4) $122y - 26x - 1675 = 0$

If the line $3 x + 4 y - 24 = 0$ intersects the $x$-axis is at the point $A$ and the $y$-axis at the point $B$, then the incentre of the triangle $O A B$, where $O$ is the origin, is:
(1) $( 4,4 )$
(2) $( 3,4 )$
(3) $( 4,3 )$
(4) $( 2,2 )$

Let $A(1,0)$, $B(6,2)$ and $C \left( \frac { 3 } { 2 } , 6 \right)$ be the vertices of a triangle $ABC$. If $P$ is a point inside the triangle $ABC$ such that the triangles $APC$, $APB$ and $BPC$ have equal areas, then the length of the line segment $PQ$, where $Q$ is the point $\left( - \frac { 7 } { 6 } , - \frac { 1 } { 3 } \right)$, is

Let $D$ be the centroid of the triangle with vertices $( 3 , - 1 ) , ( 1,3 )$ and $( 2,4 )$. Let P be the point of intersection of the lines $x + 3 y - 1 = 10$ and $3 x - y + 1 = 0$. Then, the line passing through the points $D$ and P also passes through the point:
(1) $( - 9 , - 6 )$
(2) $( 9,7 )$
(3) $( 7,6 )$
(4) $( - 9 , - 7 )$

If a $\triangle ABC$ has vertices $A ( - 1,7 ) , B ( - 7,1 )$ and $C ( 5 , - 5 )$, then its orthocentre has coordinates:
(1) $( - 3,3 )$
(2) $( 3 , - 3 )$
(3) $\left( - \frac { 3 } { 5 } , \frac { 3 } { 5 } \right)$
(4) $\left( \frac { 3 } { 5 } , - \frac { 3 } { 5 } \right)$