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The familiar right angle from the orthocenter
buratinogigle   3
N 12 minutes ago by luutrongphuc
Source: Own, HSGSO P6
Let $ABC$ be a triangle inscribed in a circle $\omega$ with orthocenter $H$ and altitude $BE$. Let $M$ be the midpoint of $AH$. Line $BM$ meets $\omega$ again at $P$. Line $PE$ meets $\omega$ again at $Q$. Let $K$ be the orthogonal projection of $E$ on the line $BC$. Line $QK$ meets $\omega$ again at $G$. Prove that $GA\perp GH$.
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buratinogigle
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Calculus
youochange   13
N 13 minutes ago by wh0nix
Find the area enclosed by the curves $e^x,e^{-x},x^2+y^2=1$
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youochange
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Prove the equality
HHT23   3
N Mar 14, 2015 by TelvCohl
Given convex quadrilateral $ABCD$ inscribed circle $(O)$. Let $I$ be the intersection of $AC$ and $BC$. Line $\Delta$ passes through $I,$ intersects the segments $AB$, $CD$ at $M$, $N$ and intersects $(O)$ at $P$, $Q$, respectively ($M, N$ lie on the segments $IQ, IP$, respectively). Prove that $$\frac{1}{IM}+\frac{1}{IP}=\frac{1}{IN}+\frac{1}{IQ}$$
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HHT23
Mar 14, 2015
TelvCohl
Mar 14, 2015
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Prove the equality
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HHT23
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HHT23
#1 h  Mar 14, 2015, 3:57 PM • 2 Y
Y by Adventure10, Mango247
Given convex quadrilateral $ABCD$ inscribed circle $(O)$. Let $I$ be the intersection of $AC$ and $BC$. Line $\Delta$ passes through $I,$ intersects the segments $AB$, $CD$ at $M$, $N$ and intersects $(O)$ at $P$, $Q$, respectively ($M, N$ lie on the segments $IQ, IP$, respectively). Prove that $$\frac{1}{IM}+\frac{1}{IP}=\frac{1}{IN}+\frac{1}{IQ}$$
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Luis González
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Luis González
#3 h  Mar 14, 2015, 4:21 PM • 3 Y
Y by HHT23, Adventure10, Mango247
By Desargues involution theorem, $\Delta$ cuts the opposite sidelines of $ABCD$ and its circumcircle $(O)$ at pairs of points in involution $\Longrightarrow$ $(P,M,I,N)=(Q,N,I,M)$ $\Longrightarrow$

$\frac{PM}{PI} \cdot \frac{NI}{NM}=\frac{QN}{QI} \cdot \frac{MI}{MN} \Longrightarrow \frac{MP}{IM \cdot IP}=\frac{QN}{IN \cdot IQ} \Longrightarrow$

$\frac{IM+IP}{IM \cdot IP}=\frac{IN +IQ}{IN \cdot IQ} \Longrightarrow \frac{1}{IM}+\frac{1}{IP}=\frac{1}{IN}+\frac{1}{IQ}.$

P.S. Note that the relation still holds for any $ABCD$ and a conic through $A,B,C,D.$
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HHT23
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HHT23
#4 h  Mar 14, 2015, 4:32 PM • 2 Y
Y by Adventure10, Mango247
Can you tell me what Desargues involution theorem is and how to prove it? Thank you.
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TelvCohl
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TelvCohl
#5 h  Mar 14, 2015, 4:50 PM • 3 Y
Y by HHT23, Adventure10, Mango247
HHT23 wrote:
Given convex quadrilateral $ABCD$ inscribed circle $(O)$. Let $I$ be the intersection of $AC$ and $B \color{red}{ D }\normalcolor $. Line $\Delta$ passes through $I,$ intersects the segments $AB$, $CD$ at $M$, $N$ and intersects $(O)$ at $P$, $Q$, respectively ($M, N$ lie on the segments $IQ, IP$, respectively). Prove that $$\frac{1}{IM}+\frac{1}{IP}=\frac{1}{IN}+\frac{1}{IQ}$$
Typo corrected :)

My solution:

Let $ X=\Delta \cap \odot (ICD), Y=\Delta \cap \odot (IAB) $ .
Let $ \Psi $ be the inversion with center $ I $ which swaps $ P, Q $ .
Let $ A^*, D^* $ be the reflection of $ A, D $ in the perpendicular bisector $ \ell $ of $ PQ $, respectively .

Since $ \angle AA^*D=180^{\circ}-\angle DCA=180^{\circ}-\angle DXY $ ,
so combine with $ AA^* \parallel XY $ we get $ A^*, D, X $ are collinear .
Similarly we can prove $ Y \in D^*A \Longrightarrow X, Y $ are symmetry WRT $ \ell \Longrightarrow QX=PY $ .

Since $ X, Y, P, Q $ is the image of $ M, N, Q, P $ under $ \Psi $, respectively ,

so from $ IQ+IX=QX=PY=IP+IY \Longrightarrow \frac{1}{IP}+\frac{1}{IM}=\frac{1}{IQ}+\frac{1}{IN} $ .

Q.E.D
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P.S. You can find some information about Desargue involution theorem at here :)
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