#20 of June 2017 Grade 10 CSAT (Korean SAT) mock test

Geometry Level 5

As shown above, there is a half circle whose diameter is segment AB \text{AB} and radius is 5 5 .

Let P \text{P} be on the half circle.

A line segment that has midpoints of chord AP \text{AP} and arc AP \text{AP} as its ends is circle O 1 O_1 's diameter. Similarly, a line segment that has midpoints of chord BP \text{BP} and arc BP \text{BP} as its ends is circle O 2 O_2 's diameter.

Circle O O is an inscribed circle of ABP \triangle\text{ABP} .

The minimum of the sum of the areas of O O , O 1 O_1 , and O 2 O_2 is S S .

Find the value of 144 S π \dfrac{144S}{\pi} .


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


The answer is 675.

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1 solution

Boi (보이)
Jun 9, 2017

Like the picture above, let Q, R \text{Q, R} be midpoints of AP , BP \overline{\text{AP}}, \overline{\text{BP}} respectively.

Also, AP \overline{\text{AP}} , BP \overline{\text{BP}} and AB \overline{\text{AB}} contact with circle O O at S, T, H \text{S, T, H} respectively.

Midpoint of AB \overline{\text{AB}} is C \text{C} .

Radii of O O , O 1 O_1 and O 2 O_2 are r r , r 1 r_1 and r 2 r_2 respectively.

Let PQ = a \overline{\text{PQ}}=a and PR = b \overline{\text{PR}}=b .

.

Since CQ = PR \overline{\text{CQ}}=\overline{\text{PR}} and CR = PQ \overline{\text{CR}}=\overline{\text{PQ}} ,

r 1 = 5 b 2 r_1=\dfrac{5-b}{2} , r 2 = 5 a 2 r_2=\dfrac{5-a}{2} .

.

Meanwhile, since AS = AH \overline{\text{AS}}=\overline{\text{AH}} and BT = BH \overline{\text{BT}}=\overline{\text{BH}} ,

( 2 a r ) + ( 2 b r ) = 5 × 2 ; (2a-r)+(2b-r)=5\times2;

r = a + b 5 r=a+b-5 .

.

The sum of the areas of O O , O 1 O_1 and O 2 O_2 is equal to:

π { ( 5 b 2 ) 2 + ( 5 a 2 ) 2 + ( a + b 5 ) 2 } [ A ] \pi\left\{\left(\frac{5-b}{2}\right)^2+\left(\frac{5-a}{2}\right)^2+(a+b-5)^2\right\}\text{ }\cdots\text{ }[A]

Let a + b = t ( 5 < t 5 2 ) a+b=t\quad (5<t\leq5\sqrt{2}) , and the expression [ A ] [A] is equal to:

π ( a 2 + b 2 10 ( a + b ) + 50 4 + ( t 5 ) 2 ) = π ( 75 10 t 4 + t 2 10 t + 25 ) ( a 2 + b 2 = CP 2 = 25 ) = π ( t 2 25 2 t + 175 4 ) = π ( t 25 4 ) 2 + 75 16 π \begin{aligned} & \pi\left(\frac{a^2+b^2-10(a+b)+50}{4}+(t-5)^2\right) \\ & =\pi\left(\frac{75-10t}{4}+t^2-10t+25\right) \quad (\because a^2+b^2=\overline{\text{CP}}^2=25)\\ & =\pi\left(t^2-\frac{25}{2}t+\frac{175}{4}\right) \\ & =\pi\left(t-\frac{25}{4}\right)^2+\frac{75}{16}\pi \end{aligned}

Therefore, S = 75 16 π S=\dfrac{75}{16}\pi .

144 S π = 144 × 75 16 = 675 \dfrac{144S}{\pi}=144\times\dfrac{75}{16}=\boxed{675} .

A very complex problem, but it kind of falls into place when you do it. Though I got stuck. So a ~= 1.47 and b ~= 4.78.

Alex Li - 3 years, 11 months ago

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Nice, and yea, it's a bit complex for grade 10 students to solve.

Boi (보이) - 3 years, 11 months ago

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I can't believe it, if this is a SAT question does that mean many of them can solve this in like 2 minutes

Alex Li - 3 years, 11 months ago

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@Alex Li Nah this is like... #20, so it should take less than around 7 minutes, but not many of the students are able to solve these kinds of problems.

Boi (보이) - 3 years, 11 months ago

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@Boi (보이) Honestly, I can't see any 10th grader I know solving this problem within a day. There are a couple of 12th graders who could MAYBE do it (and none in 7 minutes), but it's very clear that the average American spends almost no time studying compared to the average Korean.

Alex Li - 3 years, 11 months ago

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@Alex Li Yea... but though, Korean education system is not... advisable. Students get a lot of stress, they're pushed to study... and those stuff. -sigh-

Boi (보이) - 3 years, 11 months ago

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@Boi (보이) Yeah, you guys have it rough. It's not good to push people too much.

Alex Li - 3 years, 11 months ago

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