2IIM Quant Notes - Triangles


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012 and 2014.


    I am going to have a go at simple ideas in Geometry. Since this is a vast topic I am going to focus on 1-2 very simple ideas and go to town with this. We all know the two basic ideas in triangles. One with angles and one with sides. Sum of three angles of a triangle is 180 degrees and sum of any two sides is greater than the third. These two are very important and get tested in almost every exam. I am going to spend most of the time on the second property.

    Let us straightaway start with a question.

    Triangle PQR has integer sides and a perimeter of 15. How many such triangles are possible ?

    Let us say the three sides are p, q and r, and we write p < q < r (let's say, they are in an ascending order).
    Then sum of any two sides must be larger than the third. It is enough to check if the sum of the two smaller sides is greater than the third. So let us try to find the largest possible value for r.
    Now, p+q > r, so p+q must be greater than half of the perimeter. Or, we can say that p+q = 8, and r = 7, is the maximum possible value for r.
    Let r = 7, (p+q = 8 )
    then p = 1, q = 7 OR
    p =2, q = 6 OR
    p = 3, q = 5 OR
    p = 4, q = 4
    Let r = 6, (p + q = 9)

    then p =1, q = 8 (don't count this. q must be less than or equal to r)
    p = 2, q = 7 (nope)
    p = 3, q = 6 (possible)
    p = 4, q = 5 (possible)
    Let r = 5, (p+q = 10)
    only one value works. equilateral triangle, (5, 5, 5)
    So 7 possible values - (1, 7, 7), (2, 6, 7), (3, 5, 7), (4, 4, 7), (3, 6, 6), (4, 5, 6) and (5, 5, 5)
    Systematic thinking can come to the rescue

    Bonus shortcut :

    Number of Triangles with Integer sides for a given perimeter.
    If the perimeter p is even then, total triangles is [p2]/48.
    If the perimeter p is odd then, total triangles is [(p+3)2]/48
    Where [x] represents nearest integer function.

    Here p = 15, so required answer is [182 /48] = 7

    If a triangle is right angled at angle C, then we can say that
    a2 + b2 = c2. This is Pythagorus theorem and we all know that. There is an extension to this.

    If a triangle has an acute-angled at C, then
    a2 + b2 > c2

    And if the triangle is obtuse-angled at C, then
    a2 + b2 < c2

    Now, using this property, one can identify whether a given triangle is acute-angled, obtuse-angled or right-angled.

    Now, let us move to a quick exercise.
    Classify the following triangles as acute, right or obtuse angled
    3, 3, 4
    3, 4, 5
    2, 2, 3
    5, 5, 2
    3, 7, 4

    Answers :
    Acute
    Right
    Obtuse
    Acute
    Triangle not possible

    Next, we know the classification of triangles as Equilateral, Isosceles and scalene. If three sides are equal triangle is equilateral. if two are equal triangle is isosceles. If no two are equal, the triangle is scalene.

    Now, let us use this idea also and have a go at a question.

    How many triangles exist with integer sides, perimeter 14 that are not isosceles ?

    Triangle is not isosceles - triangle is either equilateral or scalene.
    But since the sides have to be integers, an equilateral triangle is only possible if the perimeter is a multiple of 3 (can't divide 14 equally into three integers)
    So we are looking for any possible scalene triangle. I am a one trick pony. Same method (for those who were patient enough to read my solution to the first question) which starts by considering the largest possible value for r (let the sides be p, q, r)
    Max possible value is 6 (r cannot be 7, because then p+q = 7, and we don't have a triangle)
    when r = 6, p + q = 8
    p = 1, q = 7 (not possible, remember we consider them in an ascending order to be sure we are counting all possible cases)
    p = 2, q = 6 (this is an isosceles triangle)
    p = 3, q = 5 (bingo! scalene triangle)
    p = 4, q = 4 (isosceles again)
    when r = 5, p + q = 9
    only possible triangle is (4,5,5) clearly isosceles.
    r cannot drop farther and still be the largest side. So we've exhausted all possible triangles.
    Only one triangle exists - (3,5,6)

    Bonus shortcut :

    Number of scalene Triangles with Integer sides for a given perimeter.
    If the perimeter p is even then, total triangles is [(p - 6)2]/48.
    If the perimeter p is odd then, total triangles is [(p - 3)2]/48
    Where [x] represents nearest integer function.

    Main point to remember is that an equilateral triangle is considered isosceles

    How many triangles with perimeter 18 are isosceles ?

    Pick the equal side first. We cannot haev 9, 9, 0.. So, we start from 8, 8, 2. Then we move to 7, 7, 4..then 6, 6, 6,.. Then 5, 5, 8.. 4, 4, 10 is not possible. 4 possible triangles.

    Going with the above formula : [182/48] - [(18 - 6)2/48] = 7-3 =4.

    Triangle with sides 8, 10 and x is obtuse-angled. How many integer values can x take ?

    x could be greater than 10, in which case we are solving for 8^2 + 10^2 < x^2. or, x could be less than 8 in which case we are solving for x^2 + 8^2 < 10^2. In both scenarios, we need to remember to count only legit triangles

    Triangle with sides integer sides x, y and z is obtuse angled. Further, we know that xy = 6. How many different values can z take ?

    xy could be 6 * 1 or 2 * 3. . If it were 6*1, only one triangle is possible. list down all possible triangles with sides 2, 3 adn z and then see which ones are obtuse.

    Triangle ABC has integer sides with product of two sides being 12. Further we know that the triangle is isosceles and acute. How many such triangles are possible? ( Slightly tougher one. So, perhaps worth taking some time over )

    Let the three sides be a,b,c, written in an ascending order (a < =b < =c)
    Since the triangle is acute, a^2 + b^2 > c^2
    Also, the triangle is isosceles, so we have to worry only about those triangles which have two sides equal.
    Now, since the product of two sides is given to be 12, let us look at possible integer sides that can multiply to give 12. We can make this list:
    12 x 1
    6 x 2
    4 x 3
    So two of the sides must be among these values. For the first one, clearly we cannot have (1, 1, 12), since that will not be a triangle. So we can have (12, 12, 1) only. we can verify that 12^2 + 1^2 is indeed greater than 12^2.
    For the second one, (2,2,6) is not possible. (6,6,2) is indeed an acute angled triangle.
    For the last one, we have (4,4,3) or (3,3,4). Both are acute angled triangles. So four triangles possible in total.

    Real numbers a, b, c are all greater than 10. There exists a triangle with sides a, b, c. Can we say for sure that there definitely exists a triangle with sides a +1, b + 1, c + 1 ?

    Let a < =b < =c (this ascending order framework seems to work rather well :))
    Then if a,b,c form a triangle, a + b > c
    for a+1 and b+ 1 and c+1,
    a+b+2 will definitely be greater than c + 1. We are adding two to the LHS, but only one to the RHS, so the condition will always be satisfied.

    Real numbers a, b, c are all greater than 10. There exists a triangle with sides a, b, c. Can we say for sure that there definitely exists a triangle with sides a -1, b - 1, c - 1 ?

    In this case, we have a + b > c,
    and a+b - 2, and c - 1 in the second case. We are taking two away from the LHS, but only one away from the RHS. Intuitively i think there must be an example where a-1, b-1, c-1 is not a triangle.
    Let us take 11, 12, 22
    New triangle is 10, 11, 21, We find that a + b = c . This is not a triangle.
    There are lots of triangles for which a-1, b-1 and c-1 will not be a triangle. We cannot say definitely
    !

    Real numbers a, b, c are all greater than 10. There exists a triangle with sides a, b, c. Can we say for sure that there definitely exists a triangle with sides a^2, b^2, c^2 ?

    Consider a triangle 11, 12, 22. Squaring each of these, we get
    121, 144, 484. sum of first two sides is lesser than the third, so it is not a triangle.
    Thinking about this slightly differently, we have a triangle a,b,c
    to check if it is obtuse, we show that a^2 + b^2 < c^2. clearly this is not a triangle.
    So this is actually not true for any obtuse angled triangle with sides greater than 10.

    Real numbers a, b, c are all greater than 10. There exists a triangle with sides a, b, c. Can we say for sure that there definitely exists a triangle with sides sqrt(a), sqrt(b), sqrt(c) ?

    The question restated becomes given that a + b > c, can we say for sure that sqrt(a) + sqrt(b) > sqrt(c). Turns out we can. :)

    Now, a bunch of true or false questions!

    1. Can an isosceles triangle be obtuse-angled ?
    2. Can a side of a triangle be greater than its semi-perimeter ?
    3. If the ratio of angles of a triangle is x: x: y can we say the triangle is isosceles and acute-angled if x < y ?
    4. If the ratio of angles of a triangle is x: x: y can we say the triangle is isosceles and acute-angled if x > y ?
    5. If we had an isosceles acute angled triangle with sides in the ratio x : x : y. Can we say x < y ?
    6. If a triangle were isosceles with sides in the ratio x : x : y and x > y, can we say the triangle is acute-angled ?
    7. If a triangle were isosceles with sides in the ratio x : x : y and x < y, can we say the triangle is acute-angled ?

    Share your answers as comments :)

     

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