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# Tutor profile: Katherine H.

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Katherine H.
High school math tutor for three years, current student at Princeton University
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## Questions

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Katherine H.

### Subject:Geometry

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Question:

Semicircles with centers at $$A$$ and $$B$$ and with radii $$2$$ and $$1$$, respectively, are drawn in the interior of, and sharing bases with, a semicircle with diameter $$\overline{JK}$$. The two smaller semicircles are externally tangent to each other and internally tangent to the largest semicircle. A circle centered at $$P$$ is drawn externally tangent to the two smaller semicircles and internally tangent to the largest semicircle. What is the radius of the circle centered at $$P$$?

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Katherine H.

Connect $$AP$$, $$BP$$, and $$OP$$. Using laws of cosines on $$\triangle BOP$$ with $$\angle B$$ as the reference angle, we obtain $$(3-r)^2 = (1+r)^2 + 2^2 - 2 \times 2(1+r) \cos B$$. Solving for $$\cos B$$, we get $$\cos B = \frac{2r-1}{r+1}$$. Using laws of cosines on $$\triangle BPA$$ with $$\angle B$$ as the reference angle, we obtain $$(2+r)^2 = (1+r)^2 + 3^2 - 2 \times 3(1+r) \cos B$$. Solving for $$\cos B$$, we get $$\cos B = \frac{3-r}{3(r+1})$$. Equate the two expressions of $$\cos B$$, we find $$r = \frac{6}{7}$$.

### Subject:Calculus

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Question:

Calculate $$I = \int_{-\infty}^{\infty} e^{-x^2} dx$$

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Katherine H.

In this problem, we use a double integral to calculate the single integral. Because $$e^{-x^2}$$ is an even function, we can take $$\frac{I}{2} = \int_{0}^{\infty} e^{-x^2} dx$$. Since x is just a dummy variable, $$\int_{0}^{\infty} e^{-x^2} dx = \int_{0}^{\infty} e^{-y^2} dy$$. So we can multiply the two integral of same value but different dummy variables: $$\frac{I}{2} \cdot \frac{I}{2} = \int_{0}^{\infty} e^{-x^2} dx \cdot \int_{0}^{\infty} e^{-y^2} dy = \int_{0}^{\infty} \int_{0}^{\infty} e^{-(x^2+y^2)} dxdy$$. The form $$x^2 + y^2$$ inspires us to change variables x and y into polar coordinates: $$x = r\cos(\theta), y = r\sin(\theta)$$. After change of variables, we obtain $$\frac{I^2}{4} = \int_{0}^{\frac{\pi}{2}} \int_{0}^{\infty} re^{-r^2}drd\theta = \int_{0}^{\frac{\pi}{2}} -\frac{1}{2} e^{-r^2} |_{0}^{\infty}d\theta = \int_{0}^{\frac{\pi}{2}} \frac{1}{2} d\theta = \frac{\pi}{4}$$.

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