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# Xuming's Geometry Synthetic Group - Shivam's proposal

1. The circumcircle of the quadrilateral $$ABCD$$ has a radius 2. $$AC, BD$$ cut at $$E$$ such that $$AE=EC$$. If $$AB=\sqrt{2} AE, BD=2\sqrt {3}$$ . Find the area of quadrilateral $$ABCD$$.
2. Triangle ABC is an isosceles triangle in which AB=A C. A circle is drawn passing through B and touching AC at its midpoint M. The circle cuts AB at P . Prove that BP=3AP.
3. In an isosceles triangle the altitude drawn to the base is $$\frac{2}{3}$$ times the radius of the circumcircle . Prove the base angle of the triangle is $$\text{arccos}\sqrt{\frac{2}{3}}$$.
4. The sides of a right angled triangle are all integers . Two sides are prime that differ by 50.Find the smallest value of the third side.
5. In triangle ABC , D is on BC such that BD=3DC. E is on AC such that 3AE=2EC. AD and BE cut F. Area of triangle AFE=4 and area of triangle BFD=30 . Find area of triangle ABC.
6. In a circle AB is a diameter . AB is produced to P such that BP=Radius of the circle . PC is tangent to the circle . The tangent at B amd AC produced cut at E . Then describe triangle CDE.

2 years, 3 months ago

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Q5:

We note that $$\triangle CFE$$ has the same altitude as $$\triangle AFE$$, therefore their areas are in the ratio of the base lengths. That is:

$$\dfrac{[\triangle CFE]}{[\triangle AFE]} = \dfrac{EC}{AE} = \dfrac{3}{2} \quad \Rightarrow [\triangle CFE] = \dfrac{3}{2} \times [\triangle AFE] = \dfrac{3}{2} \times 4 = 6$$

Similarly, $$[\triangle CFD] = \dfrac{DC}{BD} \times [\triangle BFD] = \dfrac{1}{3} \times 30 = 10$$

And, $$[\triangle ABD] = \dfrac{BD}{DC} \times [\triangle ACD] = 3 \times ([\triangle AFE]+[\triangle CFE]+[\triangle CFD]) = 3 \times (4+6+10) = 60$$

Therefore, $$[\triangle ABC] = [\triangle ABD] + [\triangle ACD] = 60 + 20 = \boxed{80}$$

- 2 years, 3 months ago

Q3:

Let the isosceles $$\triangle ABC$$ has $$AB=AC$$, its base angle be $$\theta$$ and the center and radius of the circumcircle be $$O$$ and $$r$$ respectively.

Then the extension of altitude $$AP$$ to $$BC$$ passes through $$O$$; $$AP = \frac{2}{3}r$$, $$OP = \frac{1}{3}r$$ and then $$BP = \sqrt{OB^2-OP^2} = \sqrt{r^2 - \left(\frac{1}{3}r\right)^2} = \frac{2 \sqrt{2}}{3}$$.

Therefore, $$\tan{\theta} = \dfrac{AP}{BP} = \dfrac{\frac{2}{3}r}{\frac{2\sqrt{2}}{3}r} = \dfrac{1}{\sqrt{2}} \quad \Rightarrow \cos \theta = \sqrt{\frac {2}{3}} \quad \Rightarrow \theta = \boxed{\cos^{-1} \left( \sqrt{\frac {2}{3}}\right)}$$

- 2 years, 3 months ago

Q:1

Q:2

- 8 months, 1 week ago