# John’s Geometric Algebra

The fire is still burning in search for Math wizards. This time, John Mark Año Guimba, third place in the recently concluded 2014 Metrobank –MTAP – Dep –Ed Math Challenge decided to share a problem in geometry. He is also in level 4 on his brilliant.org account. The largest online problem solving community in the world wide web.

I was born with a huge interest in Math. Since Grade 1, I was chosen as a Math quizzer, but I always settle around the 5th or 6th spot. My dream is to make it to the National Finals of the MMC, and luckily I made it in my Fourth Year in high school. I always wanted to share my knowledge to others. I usually help in tutorial sessions conducted by my coach every summer. I am now an incoming freshman taking up BS in Civil Engineering at UP-Diliman. I was a salutatorian graduate of Dasmariñas National High School.”

–John Mark Año Guimba Here are some of his achievements

MMC National Finals – Third Placer

PMO Luzon Area Stage Qualifier

2011,2012,2013 MMC Team Orals Champion (Reg IV-A Finals),

2011-2014 MMC Team Orals Champion (Dasmarinas City Division FInals)

2013 and 2014 Team and Individual Category Champion Cavitewide Search for Math Wizards.

Problem: Triangle ABC is right angled at C. Points D and E are trisection points of the hypotenuse AB. If CD = 7 and CE = 9 and the area of triangle ACB is 6, find AC + BC.

Solution:

Our first approach is to produce segments from D and E which are parallel to AC and BC respectively. Also, note that the points of intersections of those lines on the legs of  trisect AC and BC. Now, we let $AC=3y$ and $BC=3x$

Moreover, using the Pythagorean Theorem, $CD^2=x^2+(2y)^2$ $7^2=x^2+4y^2$ $49=x^2+4y^2$  (i)

Similarly, $CE^2=(2x)^2+y^2$ $9^2=4x^2+y^2$ $81=4x^2+y^2$  (ii)

Adding (i) and (ii), we have $130=5x^2+5y^2$ $26=x^2+y^2$  (iii)

Now, since the area of  is 6, $[\Delta ABC]=\displaystyle\frac{1}{2}(AC)(BC)$ $6=\displaystyle\frac{1}{2}(3x)(3y)$ $6=\displaystyle\frac{9xy}{2}$ $\displaystyle\frac{4}{3}=xy$ $\displaystyle\frac{8}{3}=2xy$  (iv) $\displaystyle\frac{86}{3}=x^2+2xy+y^2$ $\displaystyle\frac{86}{3}=(x+y)^2$ $\displaystyle\frac{\sqrt{86}}{\sqrt{3}}=x+y$ $\displaystyle\frac{\sqrt{258}}{3}=x+y$

Finally, $AC+BC=3x+3y$ $AC+BC=3(x+y)$ $AC+BC=3\cdot \displaystyle\frac{\sqrt{258}}{3}$ $AC+BC=\sqrt{258}$