A geometry problem by A Former Brilliant Member

Geometry Level 5

Consider a triangle $ABC$ with $N$ as the center of its nine point circle , and $R$ as its circumradius.

Let $a,b,c$ denote the side opposite to the vertices $A,B,C$ , respectively.

With $AN = \sqrt{12}$ and $R = \sqrt{13}$ .

If the product of length $b\times c$ is 35, what is the value of $a^2$ ?

The answer is 39.

2 solutions

A Former Brilliant Member
Oct 9, 2016

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Indraneel Mukhopadhyaya - 4 years, 8 months ago

its a great pencil , uniball shalaku

A Former Brilliant Member - 4 years, 7 months ago

Why is $AH = 2R \cos A$ true?

Pi Han Goh - 4 years, 8 months ago

First go through my solution.We see that AH=|B+C|.Also |B|=|C|=R.Square both sides to get AH^2=(B+C). (B+C)=|B|^2+|C|^2+2B.C=2R^2+2R^2*cos(2A)=2R^2 (1+cos(2A))=4R^2 (cosA)^2.Hence AH=2RcosA.

Indraneel Mukhopadhyaya - 4 years, 8 months ago

I mean AH^2=(B+C).(B+C)

Indraneel Mukhopadhyaya - 4 years, 8 months ago

Indraneel Mukhopadhyaya
Oct 11, 2016

Let A,B,C denote position vectors of the vertices.Place the origin at O so that H is A+B+C and hence N is (A+B+C)/2.Also, |A|=|B|=|C|=R.Given:|(A+B+C)/2 - A|=sqrt(12). Rewrite it as |B+C-A|=2sqrt(12). Squaring both sides and simplifying a little gives (B.C)-(A.B)-(A.C)=4.5 ; Rewrite this as (B-A).(C-A)=17.5 and use the given information |(B-A)||(C-A)|=35 to get cos(angle BAC)=1/2 or angle BAC=60 degrees.Using a^2=4(R^2)*(sin(angle BAC))^2, we get the answer as 39.

A geometry problem by A Former Brilliant Member

The answer is 39.

2 solutions

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