# 5 Math Problems That Can Be Solved Mentally

MMC Elimination just finished. Congratulations to all winners and qualifiers for the next round. Winning the first level of competition is just opening another door to a way cooler type of mathematics. In division level, it’s not just about your skill to think critically but also how well you solve problems in the most efficient manner and the fastest way.

As a math enthusiast, I would like to share some problems that looked hard but very easy to solve if you know the core concept. Let us start counting from 1.

Problem: $ABCD$ is a quadrilateral inscribed in a circle, if $\angle A=121$ degrees, what is the measure $\angle C$?

Solution: Opposite angles of cyclic quadrilateral are supplementary. Hence, $\angle A+\angle C=180^\circ$ $121^\circ+\angle C=180^\circ$ $\angle C=180^\circ-121^\circ$ $\angle C=59^\circ$

2. Slope of the Chord Passing a Circle

Problem: The line $x-2y=10$  intersects the circle with equation $x^2+y^2=81$  at points M and N, what is the slope of chord $\overline{MN}$?

Solution: If we draw the figure, we can easily see that $\overline{MN}$ lies on the given line $x-2y=10$. Thus, the slope we are looking for is the same as the slope of the given line. The answer in this problem is 1/2.

3. Tangents of Angles in a Triangle

Problem: Points A, B, and C are vertices of an isosceles triangle. If tanA+tanB+tanC=4, what is the exact value of tanAtanBtanC?

Solution: Terms like isosceles triangle and exact value are used to cause confusions in this problem. To answer this problem though, we provide the proof and derivation first.

Since A, B, and C are vertices of a triangle, we  have (note: angle is in degrees) $A+B+C=180$ $A+B=180-C$

Taking the tangents of both sides of equation we have, $tan(A+B)=tan(180-C)$

Recall that $tan(x\pm y)=\displaystyle\frac{tanx\pm tany}{1\mp tanxtany}$. Thus, $tan(A+B)=tan(180-C)$ $\displaystyle\frac{tanA+tanB}{1-tanAtanB}=\displaystyle\frac{tan180-tanC}{1+tan180tanc}$

But $tan180=0$ $\displaystyle\frac{tanA+tanB}{1-tanAtanB}=\displaystyle\frac{0-tanC}{1+(0)tanc}$ $\displaystyle\frac{tanA+tanB}{1-tanAtanB}=-tanC$ $tanA+tanB=-tanC(1-tanAtanB)$ $tanA+tanB=-tanC+tanAtanBtanC$ $tanA+tanB+tanC=tanAtanBtanC$

Going back to our problem,

Since tanA+tanB+tanC=4 and tanA+tanB+tanC=tanCtanAtanB, then tanCtanAtanB=4.

4. Series

Problem: Find the sum of $\displaystyle\frac{1}{1\cdot 2}+\displaystyle\frac{1}{2\cdot 3}+\cdots+\displaystyle\frac{1}{2014\cdot 2015}$

Solution: This problem is not an ordinary series, this can be solved using the concept of telescoping series. Full tutorial can be seen here.

To make this story short, $\displaystyle\frac{1}{1\cdot 2}+\displaystyle\frac{1}{2\cdot 3}+\cdots+\displaystyle\frac{1}{2014\cdot 2015}=1-\displaystyle\frac{1}{2015}$ $=\displaystyle\frac{2014}{2015}$

5. Sum and Difference Problem

Problem: The sum of two numbers is 17. If the bigger number is 3 more than the smaller number. What are the numbers?

Solution: This problem can be done by adding the sum and difference of two numbers and divide it by 2. That is 17+3=20. 20/2=10. The first number is 10. Since their sum is 17, it is easy to know that the other number must be 7.