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 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

John Mark 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.


John Mark Guimba geomTriangle 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.




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.

John Mark Guimba geom sol'nNow, we let AC=3y and BC=3x

Moreover, using the Pythagorean Theorem,



49=x^2+4y^2  (i)





81=4x^2+y^2  (ii)

Adding (i) and (ii), we have


26=x^2+y^2  (iii)

Now, since the area of  is 6,

[\Delta ABC]=\displaystyle\frac{1}{2}(AC)(BC)




\displaystyle\frac{8}{3}=2xy  (iv)

Adding (iii) and (iv),








AC+BC=3\cdot \displaystyle\frac{\sqrt{258}}{3}






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