Problem 1 of 16
Consider the sequence $1,\ 3,\ 3,\ 3,\ 5,\ 5,\ 5,\ 5,\ 5,\ 7,\ 7,\ 7,\ 7,\ 7,\ 7,\ 7,\ ...$ and evaluate its $2016\xasuper{th}$ term.Solution:
We can group the given series such that each group contains one unique integer. Thus the groups can be:
$(1)$, $(3,\ 3,\ 3)$, $(5,\ 5,\ 5,\ 5,\ 5)$...
The $n\xasuper{th}$ group contains $1 + (n-1)2$ terms, each term equal to $2n - 1$.
Therefore the count of terms up to the $n\xasuper{th}$ group is:
$\dfrac{n}{2}\{2\times 1 + (n-1)2\}$
$= \dfrac{n}{2}(2n)$
$= n^2$
Let $2016$ be a term in the $t\xasuper{th}$ group.
Therefore, $(t-1)^2$ is the largest perfect square $\le 2016$
Therefore, $t-1 = 44$ $(44^2 = 1936$ and $45^2 = 2025)$
$\Rightarrow t = 45$
Therefore, each of the terms in the $45\xasuper{th}$ group is $2\times 45 - 1 = 89$
Problem 2 of 16
The five digit number $2a9b1$ is a perfect square. Find the value of $a^{b−1} + b^{a−1}$Solution:
Let $n^2 = 2a9b1$
Since $a$ and $b$ are digits, their minimum value can be $0$ and the maximum value can be $9$, therefore:
$\sqrt{20901} \le n \le \sqrt{29991}$
$\Rightarrow 144 \le n \le 173$
Since the last digit of $n^2$ is $1$, the last digit of $n$ can only be $1$ or $9$.
Therefore, the possible values of $n$ are:
$149,\ 151,\ 159,\ 161,\ 169,\ 171$
Now we can square each of the six possible numbers, and verify with the middle digit.
$149^2 = 22201$
$151^2 = 22801$
$159^2 = 25281$
$161^2 = 25921$
Since the middle digit of $161^2$ is $9$, this is the required number.
Therefore, $a = 5$ and $b = 2$
Therefore, $a^{b-1} + b^{a-1}$
$= 5^1 + 2^4 = 5 + 16 = 21$
Problem 3 of 16
The date index of a date is defined as $(12 \times month\ number + day\ number)$. Three events each with a frequency of once in $21$ days, $32$ days and $9$ days, respectively, occurred simultaneously for the first time on $July\ 31,\ 1961$ (Ireland joining the European Economic Community). Find the date index of the date when they occur simultaneously for the eleventh time.Solution:
The lcm of $21$, $32$ and $9$ is $63 \times 32 = 2016$. They will repeat for the first time after $2016$ days.
They will occur together for the $11\xasuper{th}$ time after $2016 \times 11 = 22176$ days.
A period of $4$ years (not having a non-leap century year), will have $365 \times 4 + 1 = 1461$ days.
$\dfrac{2016 \times 11}{1461} = 15\dfrac{87}{487}$
After completing $15$ sets, each consisting of four-years, $15 \times 1461 = 21915$ days would have elapsed.
After $21915$ days, it will be $July, 31, 2021$. And we will be left with $22176 - 21915 = 261$ days.
The number months that will be covered with the $216$ days will be:
$Aug\ (31), Sep\ (30),\ Oct\ (31),\ Nov,\ (30),\ Dec,\ (31),\ Jan,\ (31),\ Feb\ (28),\ Mar,\ (31)$ (tolat $243$ days)
Remaining days $= 261 - 243 = 18$ days
Therefore, the date will be $April\ 18,\ 2022$
Therefore, the index will be $4 \times 12 + 18 = 66$
Problem 4 of 16
There are three kinds of fruits in the market. How many ways are there to purchase $25$ fruits from among them if each kind has at least $25$ of its fruit available?Solution:
To better visualise this problem, we will assume that we have $25$ identical coins, and each fruit costs one coin.
So all we need to find is how to distribute these $25$ coins amongst each type of fruit, that is, each unique distribution of coins will give us a unique combination of fruits.
We will use the method described , to find the number of ways $n$ identical objects (coins) amongst $m$ distinct bins (fruits).
The number is:
$\xacomb{25 + 2}{2} = \xacomb{27}{2}$
$= \dfrac{27 \times 26}{2} = 27 \times 13 = 351$ ways.
Problem 5 of 16
In a school there are $500$ students. Two-thirds of the students who do not wear glasses, do not bring lunch. Three-quarters of the students who do not bring lunch do not wear glasses. Altogether, $60$ students who wear glasses bring lunch. How many students do not wear glasses and do not bring lunch?Solution:
We will use the following Venn diagram to solve this problem.
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Let $G$ denote the set of students who wear glasses, and $L$ denote the number of students who bring lunch.
Therefore, the number of students who wear glasses and bring lunch is $|G \cap L| = 60$
Number of students who do not wear glasses neither bring lunch is
$500 - |G \cup L|$
$= 500 - (|G| + |L| - |G \cap L|)$
$= 500 - (|G| + |L| - 60)$
$= 500 - |G| - |L| + 60$
This count is also given as:
$\dfrac{2}{3}|G'|$ or $\dfrac{3}{4}|L'|$
Therefore,
$\dfrac{2}{3}|G'| = 500 - |G| - |L| + 60$
$\Rightarrow \dfrac{2}{3}(500 - |G|) = 500 - |G| - |L| + 60$
$\Rightarrow 2(500 - |G|) = 3(560 - |G| - |L|)$
$\Rightarrow 1000 - 2|G| = 1680 - 3|G| - 3|L|$
$\Rightarrow$ $3|L| + |G| = 680$ $...eqn\ (i)$
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Similarly,
$\dfrac{3}{4}|L'| = 500 - |G| - |L| + 60$
$\Rightarrow \dfrac{3}{4}(500 - |L|) = 500 - |G| - |L| + 60$
$\Rightarrow 1500 - 3|L| = 4(560 - |G| - |L|)$
$\Rightarrow 1500 - 3|L| = 2240 - 4|G| - 4|L|$
$\Rightarrow$ $4|G| + |L| = 740$ $...eqn\ (ii)$
Subtracting $eqn\ (i)$ from $3 \times$ $eqn\ (ii)$, we get:
$11|G| = 1540$
$\Rightarrow |G| = 140$
Substituting this value of $|G|$ in $eqn\ (ii)$, we get:
$|L| = 180$
Therefore, number of students to do not wear glasses not bring lunch is:
$500 - (140 + 180 - 60)$
$= 240$
Problem 6 of 16
Let $AD$ be an altitude in a right triangle $ABC$ with $\angle A = 90^\circ$ and $D$ on $BC$. Suppose that the radii of the incircles of the triangles $ABD$ and $ACD$ are $33$ and $56$ respectively. Let $r$ be the radius of the incircle of triangle $ABC$. Find the value of $3(r + 7)$.Solution:
We can draw the diagram for this problem as follows:
$\triangle ABD \sim \triangle ACD$, therefore, the corresponding sides and all other dimensions, like inradius, circumradius are in proportion.
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The hypotenuses, $AB$ and $AC$, and the two inradii $r_1$ and $r_2$ are related by:
$\dfrac{AB}{AC} = \dfrac{r_1}{r_2} = \dfrac{33}{56}$
Let $AB = 33k$, $AC = 56k$
$\therefore BC = \sqrt{(33k)^2 + (56k)^2}$
$= \sqrt{(33k)^2 + (56k)^2} = \sqrt{4225k^2}$
$= 65k$
We can show that $\triangle ABC \sim \triangle ADC$
Therefore, the inradius, $r$ of $\triangle ABC$ is given by:
$\dfrac{r_2}{r} = \dfrac{AC}{BC} = \dfrac{56k}{65k}$
$\Rightarrow \dfrac{56}{r} = \dfrac{56}{65}$
$\Rightarrow r = 65$
Therefore,
$3(r + 7) = 3 (65 + 7) = 3 \times 72 = 216$
Problem 7 of 16
Find the sum of digits in decimal form of the number $(999 . . . 9)^3$.
(There are $12$ nines)
Solution:
The number $999...9$ with $12$ nines, can be written as $10^{12} - 1$
Therefore,
$(10^{12} - 1)^3$
$= 10^{36} - 3\times 10^{24} + 3\times10^{12} - 1$
$= (10^{36} - 1) - 3(10^{24} - 10^{12})$
$= (10^{36} - 1) - 10^{12}\left\{3 \times (10^{12} - 1)\right\}$
Let us observe the pattern,
$99 \times 3 = 297$
$999 \times 3 = 2997$
$9999 \times 3 = 29997$
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Therefore,
$3 \times (10^{12} - 1)$ will have a $2$ followed by $11$ nines followed by a $7$
Therefore,
$10^{12}\left\{3 \times (10^{12} - 1)\right\}$ will have
$29...97000..0$ with $11$ nines and $12$ zeros.
$10^{36} - 1) = 999...99$ with $36$ nines
When we subtract $10^{12}\left\{3 \times (10^{12} - 1)\right\}$ from this number, we will get:
$999...997000..002999...99$
that is:
$11$ nines, followed by a $7$, followed by $11$ zeros, followed by a $2$, followed by $12$ nines
Therefore, the sum of the digits is:
$11 \times 9 + 7 + 11 \times 0 + 2 + 12 \times 9$
$= 24 \times 9 = 216$
Problem 8 of 16
Let $s(n)$ and $p(n)$ denote the sum of all digits of $n$ and the product of all digits of $n$ (when written in decimal form), respectively. Find the sum of all two-digit natural numbers $n$ such that $n = s(n) + p(n)$.Solution:
Let the ten's digit and the units digit be $x$ and $y$ respectively.
Therefore, the number is $10x + y$
The sum of the digits is $x + y$
The product of the digits is $xy$
Based on the given conditions:
$x + y + xy = 10x + y$
$\Rightarrow x + xy = 10x$
$\Rightarrow 1 + y = 10$
$\Rightarrow y = 9$
We can see that the equation is not dependent on $x$, therefore, is valid for all non-zero digit $x$, where $y = 9$
Therefore, the numbers are $19,\ 29,\ 39,\ ...\ 99$
Therefore, there are $9$ such two digit numbers.
Problem 9 of 16
Suppose that $a$ and $b$ are real numbers such that $ab \ne 1$ and the equations $120a^2 − 120a + 1 = 0$ and $b^2 − 120b + 120 = 0$ hold. Find the value of $\dfrac{1+b+ab}{a}$.Solution:
The roots of the given equations are:
$a = \dfrac{120 \pm \sqrt{120^2 - 480 }}{240}$
and
$b = \dfrac{120 \pm \sqrt{120^2 - 480}}{2}$
Considering $a = \dfrac{120 - \sqrt{120^2 - 480 }}{240}$
$\Rightarrow \dfrac{1}{a} = \dfrac{240}{120 - \sqrt{120^2 - 480 }}$
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$\Rightarrow \dfrac{1}{a} = \dfrac{240 \times ({120 + \sqrt{120^2 - 480 }})}{(120 - \sqrt{120^2 - 480 })({120 + \sqrt{120^2 - 480 }})}$
$\Rightarrow \dfrac{1}{a} = \dfrac{240 \times ({120 + \sqrt{120^2 - 480 }})}{120^2 - 120^2 + 480 }$
$\Rightarrow \dfrac{1}{a} = \dfrac{120 + \sqrt{120^2 - 480 }}{2}$
This is one of the roots of the second equation.
Similarly, if we consider the case where $a = \dfrac{120 + \sqrt{120^2 - 480 }}{240}$
we will get:
$\dfrac{1}{a} = \dfrac{120 - \sqrt{120^2 - 480 }}{2}$
We can take the root of the first equation as $a = \dfrac{120 - \sqrt{120^2 - 480 }}{240}$ and that of the second equation as $b = \dfrac{120 - \sqrt{120^2 - 480 }}{2}$
$\dfrac{1+b+ab}{a}$
$= \dfrac{1 + b + ab}{a}$
$= \dfrac{1}{a} + \dfrac{b}{a} + b$
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$= \dfrac{240}{120 - \sqrt{120^2 - 480 }} + \dfrac{240}{2} + \dfrac{120 - \sqrt{120^2 - 480 }}{2}$
$= \dfrac{120 + \sqrt{120^2 - 480 }}{2} + \dfrac{240}{2} + \dfrac{120 - \sqrt{120^2 - 480 }}{2}$
$= 240$
Problem 10 of 16
Between $5$pm and $6$pm, I looked at my watch. Mistaking the hour hand for the minute hand and the minute hand for the hour hand, I mistook the time to be $57$ minutes earlier than the actual time. Find the number of minutes past $5$ when I looked at my watch.Solution:
Based on the given information, the actual position of the hour hand was between $5$ and $6$.
Since inverting the hour and the minute hand caused a reading of $57$ mins early, the actual position of the minute hand must have been before the actual position of the hour hand, which means the actual time was less than $5:30$. Therefore, $57$ mins earlier the time must have been some mins past $4$ o'clock.
Let the incorrect time reading be $m$ mins past $4$.
At $m$ mins past $4$ the minute hand will be at an angle of $6m^\circ$ with the $12$ o'clock position, and the hour hand will be at and angle of $4 \times 30 + \dfrac{m}{2} = \left(120 + \dfrac{m}{2}\right)^\circ$
After $57$ mins the minute hand will be at a position $(6m - 18)^\circ$
In $57$ mins the hour hand would have moved by $57 \times \dfrac{30}{60} = \left(\dfrac{57}{2}\right)^\circ$
$120 + \dfrac{m}{2} + \dfrac{57}{2} = 6m$
$\Rightarrow 120 + \dfrac{57}{2} = \dfrac{11m}{2}$
$\Rightarrow 240 + 57 = 11m$
$\Rightarrow m = 27$
Therefore, $57$ mins earlier the count of minutes was $27$, therefore at the current time the minute count will be $27 + 57 (mod\ 60) = 24$
Problem 11 of 16
In triangle $ABC$ right angled at vertex $B$, a point $O$ is chosen on the side $BC$ such that the circle $\gamma$ centered at $O$ of radius $OB$ touches the side $AC$. Let $AB = 63$ and $BC = 16$, and the radius of $\gamma$ be of the form $\dfrac{m}{n}$ where $m$, $n$ are relatively prime positive integers. Find the value of $m + n$.Solution:
We can draw the diagram for the given problem as shown below.
We have drawn a tangent from $C$ to the circle $\gamma$, and extended it to meet $AB$, extended, at $D$.
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Since $\triangle ABC$ is right-angled at $B$,
$AC = \sqrt{AB^2 + BC^2} = \sqrt{63^2 + 16^2} = \sqrt{4225} = 65$
Since, $CA$ and $CD$ are tangents to $\gamma$, and $CB$ passes through the centre $O$,
$\angle BCA = \angle BCD$
Considering $\triangle CBA$ and $\triangle CBD$
$\angle BCA = \angle BCD$,
$\angle CBA = \angle CBD$
$BC$ is common.
$\therefore \triangle CBA \cong \triangle CBD$
Therefore,
$\triangle CAD$ is isosceles with $CB$ as an altitude, and $\gamma$ as the incircle.
We can use relation between inradius and area of a triangle explained .
$AD = 2 \times 63 = 126$
$ar[CAD] = \dfrac{1}{2} \times 126 \times 16 = 63 \times 16$
Semi-perimeter of $\triangle CAD = \dfrac{126 + 65 + 65}{2} = 128$
$\therefore r \times 128 = 63 \times 16$ where $r$ is the inradius of $\gamma$
$\Rightarrow r = \dfrac{63}{8} = \dfrac{m}{n}$
Therefore, $m + n = 63 + 8 = 71$
Problem 12 of 16
Consider the 50 term sums:
$S = \dfrac{1}{1 \times 2} + \dfrac{1}{3 \times 4} + ... + \dfrac{1}{99 \times 100}$
$T = \dfrac{1}{51 \times 100} + \dfrac{1}{52 \times 99} + ... + \dfrac{1}{100 \times 51}$
The ratio $\dfrac{S}{T}$ is written in lowest form $\dfrac{m}{n}$ where $m$, $n$ are relatively prime natural numbers. Find the value of $m + n$.
Solution:
The first series $S$ can be broken up using a method similar to telescoping method described ,
$S = \dfrac{1}{1 \times 2} + \dfrac{1}{3 \times 4} + ... + \dfrac{1}{99 \times 100}$
$= \left(\dfrac{1}{1} - \dfrac{1}{2}\right) + \left(\dfrac{1}{3} - \dfrac{1}{4}\right) + ... + \left(\dfrac{1}{99} - \dfrac{1}{100}\right)$
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$= \left(\dfrac{1}{1} + \dfrac{1}{3} + ... + \dfrac{1}{99}\right) - \left(\dfrac{1}{2} + \dfrac{1}{4} + ... + \dfrac{1}{100}\right)$
$= \left(\dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3}+ ... + \dfrac{1}{100}\right) - 2\left(\dfrac{1}{2} + \dfrac{1}{4} + ... + \dfrac{1}{100}\right)$
$= \left(\dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3}+ ... + \dfrac{1}{100}\right) - \left(\dfrac{1}{1} + \dfrac{1}{2} + ... + \dfrac{1}{50}\right)$
$= \dfrac{1}{51} + \dfrac{1}{52} + ... + \dfrac{1}{100}$
For the series $T$, let us observe that the sum of the two factors of the denominator for all the terms is $151$. We can write this series as:
$T = \dfrac{1}{51 \times 100} + \dfrac{1}{52 \times 99} + ... + \dfrac{1}{100 \times 51}$
$= \dfrac{1}{151}\left(\dfrac{100 + 51}{51 \times 100} + \dfrac{99 + 52}{52 \times 99} + ... + \dfrac{51 + 100}{100 \times 51}\right)$
$= \dfrac{1}{151}\left(\dfrac{1}{51} + \dfrac{1}{100} + \dfrac{1}{52} + \dfrac{1}{99} + ... + \dfrac{1}{100} + \dfrac{1}{51}\right)$
$= \dfrac{2}{151}\left(\dfrac{1}{51} + \dfrac{1}{52} + \dfrac{1}{53} + ... + \dfrac{1}{99} + \dfrac{1}{100}\right)$
Therefore,
$\dfrac{S}{T} = \dfrac{151}{2} = \dfrac{m}{n}$
Therefore,
$m + n = 153$
Problem 13 of 16
Find the value of the expression:
$\dfrac{(3^4 + 3^2 + 1).(5^4 + 5^2 + 1).(7^4 + 7^2 + 1).(9^4 + 9^2 + 1)}{(2^4 + 2^2 + 1).(4^4 + 4^2 + 1).(6^4 + 6^2 + 1).(8^4 + 8^2 + 1)}$
$\ \ \ \times \dfrac{(11^4 + 11^2 + 1).(13^4 + 13^2 + 1)}{(10^4 + 10^2 + 1).(12^4 + 12^2 + 1)}$
when written in the lowest form.
Solution:
We write any term of the form:
$n^4 + n^2 + 1$ as:
$n^4 + 2.n^2 + 1 - n^2$
$= (n^2 + 1)^2 - n^2$
$= (n^2 - n + 1)(n^2 + n + 1)$
For even $n$, we can write $n = 2m$,
$(n^2 - n + 1)(n^2 + n + 1)$
$= ((2m)^2 - (2m) + 1)((2m)^2 + (2m) + 1)$
$(4m^2 - 2m + 1)(4m^2 + 2m + 1)$
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For the odd number preceding $2m$, we can write $n = 2m - 1$
$(n^2 - n + 1)(n^2 + n + 1)$
$= \{(2m - 1)^2 - (2m - 1) + 1\}\{(2m - 1)^2 + (2m - 1) + 1\}$
$= \{(2m + 1)^2 - (2m - 1) + 1\}\{(2m - 1)^2 + (2m - 1) + 1\}$
$= (4m^2 - 4m + 1 - 2m + 1 + 1)(4m^2 - 4m + 1 + 2m - 1 + 1)$
$= (4m^2 - 6m + 3)(4m^2 - 2m + 1)$
$= \{(2m-2)^2 + (2m - 2) + 1\}(4m^2 - 2m + 1)$
And for the odd number succeeding $n$, we can write $n = 2m + 1$:
$(n^2 - n + 1)(n^2 + n + 1)$
$= \{(2m + 1)^2 - (2m + 1) + 1\}\{(2m + 1)^2 + (2m + 1) + 1\}$
$= \{(2m + 1)^2 - (2m + 1) + 1\}\{(2m + 1)^2 + (2m + 1) + 1\}$
$= (4m^2 + 4m + 1 - 2m - 1 + 1)(4m^2 + 4m + 1 + 2m + 1 + 1)$
$= (4m^2 + 2m + 1)(4m^2 + 6m + 3)$
$= (4m^2 + 2m + 1)\{(2m + 2)^2 + (2m + 2) + 1\}$
We can see that for each of the numerator/denominator the smaller factor will cancel out with the previous term, and the larger factor will cancel out with the next term, leaving us with:
$\dfrac{13^2 + 13 + 1}{2^2 - 2 + 1}$
$= \dfrac{169 + 13 + 1}{4 - 2 + 1}$
$= \dfrac{183}{3}$
$= 61$
Problem 14 of 16
The hexagon $OLYMPI$, with all sides equal, has a reflex angle at $O$ and convex at every other vertex. Suppose that $LP = 3\sqrt{2}$ units and the condition $\angle O = 10\angle L = 2\angle Y = 5\angle M = 2\angle P = 10\angle I$ holds. Find the area (in sq units) of the hexagon.Solution:
Based on initial observation, $\angle O$ is the largest angle and $\angle L$ and $\angle I$ are the smallest angles.
Let $\angle L = \angle I = \theta$
Therefore,
$\angle O = 10\theta$
$\angle Y = 5\theta$
$\angle M = 2\theta$
$\angle P = 5\theta$
Therefore, $10\theta + \theta + 5\theta + 2\theta + 5\theta + \theta = 720$
$\Rightarrow 24\theta = 720$
$\Rightarrow \theta = 30^\circ$
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$\angle O = 300^\circ$
$\angle L = \angle I = 30^\circ$
$\angle Y = \angle P = 150^\circ$
$\angle M = 60^\circ$
We can draw the diagram as shown below.
It is easy to show that $LY \parallel PI$, by drawing a line parallel to $LY$ passing through $O$ and then looking at the alternated angles.
We will leave this for the student as an exercise.
In $\triangle MPY$, $MY= MP$ and $\angle M = 60^\circ$.
Therefore, $\triangle MPY$ is equilateral.
Therefore, $\angle PYL = 150 - 60 = 90^\circ$
Similarly, we can show that $\triangle OLI$ is equilateral with $\angle OLI = 60^\circ$
Therefore, $\angle YLI = 30 + 60 = 90^\circ$
Therefore, $YP \parallel LP$, and $LY \parallel PI$, $\angle PYL = 90^\circ$,
$PYLI$ is a square.
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Therfore, $YP = LI$, and $\triangle PMY \cong \triangle IOL$
Therefore, $ar[OLYMPI] = ar[PYLI]$
The diagonal of $PYLI$, $LP = 3\sqrt{2}$
$\therefore LI = \dfrac{3\sqrt{2}}{\sqrt{2}} = 3$
Therefore, the $ar[PYLI] = 3^2 = 9$
Note: This problem, as originally published for this exam, did not specify the condition $\unicode{0x201C}$all sides equal$\unicode{0x201D}$. Without this condition this problem does not have a unique solution.
We can come with the contradiction very easily. If we extend $PM$ to $M'$ and draw a line parallel to $MY$ passing through $M'$, meeting $LY$ extended at $Y'$, as shown in the figure below, then the hexagon $OLY'M'PI$ will have the same angles as $OLYMPI$ and the length of $LP$ will remain unaltered.
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But evidently the area of $OLY'M'PI$ is greater than the area of $OLYMPI$. Similarly we can find many more cases with different areas but the same angles, and $LP$.
Thus, the equal side condition is needed for a unique solution.
Problem 15 of 16
A natural number $a$ has four digits and $a^2$ ends with the same four digits as that of $a$. Find the value of $(10,080 − a)$.Solution:
Using this for finding automorphic numbers, we can find out the $4$-digit numbers ending with $5$ and $6$.
The automorphic numbers ending with $5$ are:
$5$, $25$, $625$, $0625$
Similarly the automorphic numbers ending with $6$ are:
$6$, $76$, $376$, $9376$
Since $0625$ has a leading zero, it will not count as a $4$-digit number.
Therefore, the only proper $4$-digit automorphic number is $9376$ which is the value of $a$.
Therefore, $10080 - a = 10080 - 9376 = 704$
Problem 16 of 16
Points $G$ and $O$ denote the centroid and the circumcenter of the triangle $ABC$. Suppose that $\angle AGO = 90^\circ$ and $AB = 17$, $AC = 19$. Find the value of $BC^2$.Solution:
Let the lengths of $AB,\ AC,\ BC$ be $c,\ b,\ a$ respectively. $O$ and $G$ are the circumcenter and the centroid of $\triangle ABC$ and
$OA$ is the circumradius $(R)$. The diagram is shown below:
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Using Stewart's Theorem explained , we know that if a cevian of length $d$, divides a side of length $a$ into to lengths $m$ and $n$ then,
$b^2m + c^2n = a(d^2+mn)$
As the cevian $AD$ is a median, $m=n=\dfrac{1}{2}a.$
Substituting $m=n$ and $a=2m$ in the above equation, we get,
$\Rightarrow b^2m + c^2m = 2m(d^2+m^2)$
$\Rightarrow m(b^2 + c^2) = 2m(d^2+m^2)$
$\Rightarrow b^2 + c^2 = 2(d^2+m^2)$
$\Rightarrow b^2 + c^2 = 2d^2+2(m^2)$
$\Rightarrow b^2 + c^2 = 2d^2+2\left(\dfrac{1}{2}a\right)^2$
$\Rightarrow b^2 + c^2 = 2d^2+\dfrac{1}{2}a^2$
$\Rightarrow 2b^2 +2c^2 = 4d^2+a^2$
$\Rightarrow 2b^2 +2c^2-a^2 = 4d^2$
$\Rightarrow \dfrac{2b^2 +2c^2-a^2}{4} = d^2$
$\Rightarrow d=\dfrac{\sqrt{2b^2 +2c^2-a^2}}{2}$
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$\therefore AD=\dfrac{1}{2} \sqrt{2b^2+2c^2-a^2}$
$AG= \dfrac{2}{3}AD=\dfrac{2}{3} \times \dfrac{1}{2} \sqrt{2b^2+2c^2-a^2}=\dfrac{1}{3} \sqrt{2b^2+2c^2-a^2}$
$O$ and $G$ are part of the Euler Line, therefore, using the $Corollary\ 2$ of Euler Line from , we get:
$OG^2=R^2-\dfrac{1}{9}(a^2+b^2+c^2)$ $... eqn\ (i)$
$R$ is the circumcentre, $\therefore R=AO=BO=CO.$
Replacing $R$ with $AO$ in $eqn\ (i),$ we get-
$OG^2=AO^2-\dfrac{1}{9}(a^2+b^2+c^2)$
As $\angle AGO=90^\circ,$ using Pythagorean Theorem we get-
$AG^2+GO^2=AO^2$
$\Rightarrow \left\{\dfrac{1}{3} \sqrt{2b^2+2c^2-a^2}\right\}^2+AO^2-\dfrac{1}{9}(a^2+b^2+c^2)=AO^2$
$\Rightarrow \dfrac{1}{9}(2b^2+2c^2-a^2)+AO^2-\dfrac{1}{9}(a^2+b^2+c^2)=AO^2$
$\Rightarrow \dfrac{1}{9}(2b^2+2c^2-a^2)-\dfrac{1}{9}(a^2+b^2+c^2)=0$
$\Rightarrow \dfrac{1}{9}(2b^2+2c^2-a^2)=\dfrac{1}{9}(a^2+b^2+c^2)$
$\Rightarrow 2b^2+2c^2-a^2=a^2+b^2+c^2$
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$\Rightarrow 2a^2=b^2+c^2$
$\Rightarrow 2BC^2=AC^2+AB^2$
$\Rightarrow 2BC^2=(19)^2+(17)^2$
$\Rightarrow 2BC^2=361+289=650$
$\Rightarrow BC^2=\dfrac{650}{2}=325$