The triplets - 5, 12, 13

Geometry Level 5

Let $A(5,12)$ , $B(-13\cos { \theta } ,13\sin { \theta } )$ and $C(13\sin { \theta } ,-13\cos { \theta } )$ are angular points of $\triangle ABC$ where $\theta \in \Re$ . The locus of the orthocentre of the $\triangle ABC$ is $ax+by+c=0$ where $a,b,c$ are integers. Find the minimum positive value of $a+b+c$ .

The answer is 7.

1 solution

Niranjan Khanderia
Dec 23, 2014

Let B be (-p,q) so C is (q, -p). So the the slope of line passing through them is $\dfrac{q+p}{-p-q}=-1. \\ \therefore ~the ~slope~of~the~ locus ~of~ the~ orthocenter~ is ~=-\dfrac{ 1}{-1} = 1.\\But~ax+by+c=0 \implies~the~slope~is~\dfrac{-a}{b}\\\therefore~ a=-b~~~~~~~~~....(1)\\ Since~ A~~ is ~on~this~line\implies~12=1(5)+c. \therefore~c=7~~....(2)\\ From ~(1)~ and~(2)....a+b+c= -b+b+c=7 \\ So ~for~ any~+ tive ~integer~k, ~the~ line~ is~~ k(ax+by+c)=0. \\ \implies ~k(a+b+c)=7k. ~So ~minimum~+tive~value~of~the~sum ~is\\~when~k=1. \\So ~the~required~answer~is~\boxed{ 7 }$
Note that B and C are on the circle $X^2 + Y^2 =13^2$ . A is also on this circle.
B and C are reflections of one another about Y = X.

Note: The value of $a+b+c$ is not unique, because it could be $x -y + 7 = 0$ or $2x - 2y + 14 = 0$ .

I have edited it slightly and I think I fixed this issue. Can you check?

Calvin Lin Staff - 6 years, 5 months ago

Thank you. I did not realise this before. I have made corrections.

Niranjan Khanderia - 6 years, 5 months ago

One can also do as : After Concluding That these three Point's Lies on circle , we Can Also Use Standard Relation of Concurrency of Three Point's O , G , C (OG/GC= 2/1) which give orthocenter's Locus easily

Deepanshu Gupta - 6 years, 4 months ago

The triplets - 5, 12, 13

The answer is 7.

1 solution

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