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

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 $ABCD$ .
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.
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 $\text{arccos}\sqrt{\frac{2}{3}}$ .
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.
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.
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.

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Comments

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

Chew-Seong Cheong - 5 years, 8 months ago

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

Q:1

Q:2

Ahmad Saad - 4 years, 1 month ago

Answer 4 = 60

Saksham Rastogi - 5 years, 8 months ago

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Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
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`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$