#30 of Sept 2015 Grade 10 CSAT (Korean SAT) Mock test

Geometry Level 3

As the image above shows, a right triangle A B C ABC has O 1 O_1 as its inscribed circle, and O 2 , O 3 , O 4 O_2,~O_3,~O_4 as its escribed circle. The radius of a circle O i O_i is r i . ( i { 1 , 2 , 3 , 4 } ) r_i.~(i\in \{1,~2,~3,~4\})

Given that r 1 = 1 r_1=1 and that the area of triangle A B C ABC is 15 2 , \dfrac{15}{2}, find the value of r 2 + r 3 + r 4 . r_2+r_3+r_4.


This problem is a part of <Grade 10 CSAT Mock test> series .


The answer is 14.

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1 solution

Boi (보이)
Jul 20, 2017

Let A B , B C , C A \overline{AB},~\overline{BC},~\overline{CA} be c , a , b c,~a,~b respectively.


Then by above picture, we know that ( b r 1 ) + ( a r 1 ) = c . (b-r_1)+(a-r_1)=c.

r 1 = a + b c 2 . r_1=\dfrac{a+b-c}{2}.


Also, by above picture, we know that b + r 2 = c + ( a r 2 ) . b+r_2=c+(a-r_2).

r 2 = a b + c 2 . r_2=\dfrac{a-b+c}{2}.


Using similar method with r 3 r_3 and r 4 , r_4, we get

r 3 = a + b + c 2 ; r_3=\dfrac{-a+b+c}{2};

r 4 = a + b + c 2 . r_4=\dfrac{a+b+c}{2}.


Therefore,

r 1 + r 2 + r 3 + r 4 = a + b + c = 15. r_1+r_2+r_3+r_4=a+b+c=15.

r 2 + r 3 + r 4 = 14 . \therefore~r_2+r_3+r_4=\boxed{14}.

These 3 large circles are also called excircles. There's plenty of useful formula that is relevant here.

Pi Han Goh - 3 years, 10 months ago

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Mhm. Those are very useful. I used them once or twice on some geometry problems, and they're strikingly effective when the given information doesn't seem enough!

Boi (보이) - 3 years, 10 months ago

I guess this is wrong. How can r4 be s? I got answer as 7. Please check

mohan nayak - 3 years, 10 months ago

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If you want to see why, here's an explanation.

C R = r 4 , C P = r 4 . 2 r 4 = C P + C R = A C + A P + B C + B R = a + b + A P + B R = a + b + A Q + B Q = a + b + A B = a + b + c \overline{CR}=r_4,~\overline{CP}=r_4.\\ \begin{aligned} \therefore~2r_4 &=\overline{CP}+\overline{CR} \\ & =\overline{AC}+\overline{AP}+\overline{BC}+\overline{BR} \\ & =a+b+\overline{AP}+\overline{BR} \\ & =a+b+\overline{AQ}+\overline{BQ} \\ & =a+b+\overline{AB} \\ & =a+b+c \end{aligned}

And therefore, r 4 = a + b + c 2 . r_4=\dfrac{a+b+c}{2}.

Boi (보이) - 3 years, 10 months ago

r 4 r_4 is indeed the half of the perimeter of A B C . \triangle ABC.

This is the solution of Korea Institute for Curriculum and Evaluation, formed by several college professors.

Also I got the same answer when I solved it.

Boi (보이) - 3 years, 10 months ago

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Thanks a lot. My bad. Is it semiperimeter only when it's right angle or general.??

mohan nayak - 3 years, 10 months ago

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@Mohan Nayak It's okay!

And yeah. It's the semiperimeter only because the opposite angle is π 2 . \dfrac{\pi}{2}.

Boi (보이) - 3 years, 10 months ago

I think your solution might take some time to carefully observe and answer, Exams like jee and other competitive exams conducted in india requires on readymade formulas to crack such type of questions. I appreciate for your good geometrical skill.

Syed Shahabudeen - 3 years, 10 months ago

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Yah, this is #30 of the exam, the hardest one of all because in here, we don't learn geometry that seriously.

Boi (보이) - 3 years, 10 months ago

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